## Algebra Qual Study Problems: 5

2011 Qual #8 (Hungerford 3.4.4): Suppose $R =\mathbb{Z}_6, S = \{2,4\} \subset R$. Prove that $S^{-1}R$ is a finite field and identify the field.
Proof-
Recall that $S^{-1}R$ is the set of equivalence classes of $R\times S$ under the relation that $(r,s) \sim (r',s') \Leftrightarrow s_1(rs'-r's) = 0$ for some $s_1 \in S$. With the sets R and S given we have that all possible elements of $S^{-1}R$ are $\{(0,2),(1,2),(2,2),(3,2),(4,2),(5,2),(0,4),(1,4),(2,4),(3,4),(4,4),(5,4)\}.$ Now we must see which are equivalent under the relation. We have that:
$(1,2) \sim (4,2) \sim (2,4) \sim (5,4)$ because $8-2=4-4=10-4 = 0$
$(2,2) \sim (5,2) \sim (1,4) \sim (4,4)$ because $4-10=8-2=8-8=0$
$(0,2) \sim (3,2) \sim (3,4) \sim (0,4)$ because $8-8=12=6=0-6=0$
Thus $S^{-1}R = \{(0,2),(1,2),(2,2)\}.$ Having only 3 elements we know that $S^{-1}R \cong \mathbb{Z}_3$ so that it is in fact a finite field.

2013 Qual #1 (Final Exam #8): Let R be a ring and let I, J be ideals. Let $0 \rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0$ be a short exact sequence of R modules.
(a) Prove that $IM_2 = 0 \Rightarrow IM_1 = 0 = IM_3$
(b) Show that $IM_1 = 0 = JM_3 \Rightarrow IJM_2 = 0$. Give an example of a ring R a short exact sequence of R modules as above, and an ideal I of R such that $IM_1=0=IM_3$ but $IM_2 \neq 0$.
Proof-
a) Suppose that $IM_2 = 0$. Let $i\in I, m_1 \in M_1 , m_3\in M_3$. Then we have $f_1 (im_1) = if_1(m_1) \in IM_2 = 0$. Therefore $im_1 \in Ker(f_1)$. But by being a short exact sequence we have that $Ker(f_1) = 0$. Therefore $im_1 = 0$. Since this was arbitrary we have that $IM_1 = 0$. Next, $f_2$ is surjective due to exactness so there exists $m_2 \in M_2$ such that $f_2(m_2) = m_3$. This gives that $if_2(m_2) = im_3$, but because of the R-mod homomorphism we have $im_3 = if_2(m_2) = f_2(im_2)$, but again since $IM_2 = 0$ we have that $im_2 = 0$ so that $im_3 = f_2(0) = 0$. Therefore $IM_3 = 0$.
b) Now suppose that $IM_1 = 0 = JM_3$. Let elements be the same as in part a and let $j\in J$. Since $f_2$ is onto we have there exists some $m_2 \in M_2$ such that $f_2(m_2) = m_3 \Rightarrow f_2(jm_2) = jm_3 = 0.$ So then $jm_2 \in Ker(f_2) = Im(f_1)$ by exactness. This gives that there exists $m_1 \in M_1$ such that $f_1(m_1)= jm_2$. Then $if_1(m_1) = f_1(im_1) = f_1(0) = 0 = ijm_2$. Therefore $IJM_2 = 0$.
For the counterexample take the following short exact sequence:
$0 \rightarrow \mathbb{Z}_2 \xrightarrow{f} \mathbb{Z}_4 \xrightarrow{g} \mathbb{Z}_2 \rightarrow 0$, where the maps are defined as $f(0) = 0, f(1) = 2, g(0) = 0 = g(2), g(1) = 1 = g(3)$. Then clearly f and g are $\mathbb{Z} \text{-module}$ homomorphisms, with the property that $Imf = Kerg$. Then take $I = 2\mathbb{Z}$ to be the ideal of $R = \mathbb{Z}$. We have then that $(2\mathbb{Z} ) \mathbb{Z}_2 = 0$ but $(2\mathbb{Z} ) \mathbb{Z}_4 \neq 0$. Note that this counterexample works because the sequence does not split and that the statement would be true for any short exact sequence that splits. This is because if $0 \rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0$ splits we can write $M_2 \cong M_1 \bigoplus M_3$ And so if an ideal I has the property that $IM_1 = IM_3 = 0$, then $0 = IM_1 \bigoplus IM_3 \cong IM_2$

2006 Qual #1 (Hungerford 4.1.9, Takehome midterm #4c): If $f: A\rightarrow A$ is an R-module homomorphism such that $f^2 = f$ then $A = Kerf \bigoplus Imf$.
Proof-
First recall that: $Kerf \bigoplus Imf = \{a_1 + \ldots +a_n | a_i\in Kerf \cup Imf\}$ Since the kernel and image are submodules of A we have that $\{a_1 + \ldots + a_n | a_i\in Kerf \cup Imf\} \subset A$. To finish we must show the opposite containment. Let $a\in A$. We can write $a = f(a) + (a-f(a))$. Then note that $f(a) \in Imf$ by definition and that $f(a-f(a)) = f(a)-ff(a) = f(a)-f(a) = 0$ because of the assumption. Therefore $a-f(a) \in Kerf$. Thus $a\in \{a_1 + \ldots + a_n | a_i\in Kerf \cup Imf\}$. Next we must show that $Kerf \cap Imf = \{0\}$. Let $a \in Kerf \cap Imf$. Then $f(a) = 0$ and also there exists a $b\in A$ such that $f(b) = a$. But then applying f to both sides of this equality we have $ff(b) = f(a) \Rightarrow a = f(b) = f(a) = 0$. Therefore $Kerf \cap Imf = \{0\}$. This gives that A is the internal direct sum, $Kerf \bigoplus Imf$.

Qual 2011 #6: Let R be a ring and $M_i,N_i$ for $i = 1,2,3$ be R-modules. Consider the diagram with exact rows of R-mod homomorphisms:
$0\rightarrow M_1 \xrightarrow{f_1} M_2 \xrightarrow{f_2} M_3 \rightarrow 0$

$0\rightarrow N_1 \xrightarrow{g_1} N_2 \xrightarrow{g_2} N_3 \rightarrow 0$
Suppose that there exist $\phi_1 \in Hom_R(M_1,N_1), \phi_2 \in Hom_R(M_2,N_2)$ such that $g_1\phi_1 = \phi_2f_1$. Prove that there exists $\phi_3 \in Hom_R(M_3,N_3)$ that makes the diagram commute. Which conditions should $\phi_1$ and/or $\phi_2$ satisfy to ensure that $\phi_3$ is surjective.
Proof-
First we define such a function. Let $\phi_3 : M_3 \rightarrow N_3$ by $\phi (m_3) = g_2\phi_2 (m_2)$ where $m_2 \in \{m | f_2(m) = m_3\}$. This can be done because $f_2$ is assumed to be surjective. We must show that this function is well defined. If $m_3 = m_3'$ then there exist $m_2, m_2' \in \{m \in M_2 | f_2(m) = m_3\}$ which define $\phi_3 (m_3)$ and $\phi_3 (m_3')$. So to show well defined we must show that $g_2\phi_2 (m_2) = g_2\phi_2 (m_2')$. We have $m_3 = m_3'f_2 (m_2) = f_2(m_2') \Rightarrow m_2-m_2'\in Kerf_2 =Imf_1$. So then there exists $m_1 \in M_1$ such that $m_2 -m_2' =f_1(m_1)$. Then $\phi_2(m_2-m_2') = \phi_2f_1(m_1) = g_1 \phi_1 (m_1)$ by the commutativity of the first square. Then we have $g_2\phi_2(m_2-m_2') = g_2g_1\phi_1(m_1) = 0$ because the bottom row is exact and so $g_2g_1 = 0$ always. Therefore $g_2\phi_2(m_2-m_2') = 0 \Rightarrow g_2\phi_2(m_2) = g_2\phi_2(m_2')$. So the map is well defined. It is a composition of R-mod homomorphisms so it is also one. The diagram commutes by construction.
Now, claim that $\phi_2$ surjective gives that $\phi_3$ surjective. Let $n_3 \in N_3$. Since $g_2$ is surjective we have there exists a $n_2 \in N_2$ such that $g_2(n_2) = n_3$. Since $\phi_2$ is surjective there exists a $m_2 \in M_2$ such that $\phi (m_2) = n_2$. Then $n_3 = g_2\phi_2 (m_2)$ and hence $\phi_3f_2 (m_2) = n_3$ which gives the map to be surjective.

Qual 2009 #1 (Final #10): Let R be a ring, M an R-mod, $M_1,M_2 \subset M$. Define a natural homomorphism $M_1 \bigoplus M_2 \rightarrow M$ where this is the external direct sum. Describe it’s image and kernel and use it to prove the formula $dim_D(V_1+V_2) = dim_D(V_1\cap V_2) = dim_DV_1 + dim_DV_2$ where D is a division ring and $V_1,V_2$ are vector subspaces.
Proof-
Let $\phi : M_1\bigoplus M_2 \rightarrow M$ be the natural homomorphism defined by $(m_1,m_2) \mapsto m_1 + m_2$. Then clearly $Im(\phi ) = M_1 + M_2$, and $\phi ((m_1,m_2)) = m_1 + m_2 = 0 \Leftrightarrow m_1 = -m_2 \Leftrightarrow m_1 \in M_1 \cap M_2 = \{0\}$. Therefore the Kernel is trivial. Now let D be a division ring, then any vector space is free and we have $dim_D(V\bigoplus W) = dim_DV + dim_DW$ and $dim_D(V/W) = dim_DV - dim_DW$. Apply $\phi$ to the vector subspaces $V_1/(V_1 \cap V_2) \bigoplus V_2/(V_1 \cap V_2) \rightarrow V/(V_1\cap V_2)$ then by the rank nullity theorem we have $dim_D ( V_1/(V_1\cap V_2) \bigoplus V_2/(V_1 \cap V_2)) = dim_D (Im(\phi )) +dim_D (Ker(\phi ))$ Then using the above facts this gives that:
$dim_DV_1 + dim_DV_2 - 2 dim_D (V_1 \cap V_2 ) = dim_D (V_1+V_2) - dim_D ( V_1 \cap V_2)$ which implies $dim_DV_1 + dim_DV_2 = dim_D(V_1+V_2) + dim_D( V_1 \cap V_2)$.

2011 Final #2(i): Is $\mathbb{Q}$ free as a $\mathbb{Z} \text{-module}$?
Proof-
Suppose that the statement were true. If it were a free $\mathbb{Z} \text{-module}$ then it would have a basis. Let $X = \{\frac{p}{q}\}$ be a basis for $\mathbb{Q}$ as a $\mathbb{Z} \text{-module}$. Let $\frac{p}{q+1} \in \mathbb{Q}$, and suppose there exists an element $m\in \mathbb{Z}$, such that $\frac{p}{q+1} = m \frac{p}{q} \Rightarrow m = \frac{q}{q+1}$ but this is not an integer so contradiction. Thus if the rationals have a basis it cannot have cardinality 1. Suppose then that $X = \{\frac{p}{q}, \frac{r}{s}\}$ is a basis. We can write $(rq)\frac{p}{q} - (ps)\frac{r}{s} = rp-rp = 0$ so that the set is not linearly independent. Thus if $|X| \geq 2$ X will not be linearly independent. Therefore the rationals do not have a basis and so cannot be free as a $\mathbb{Z} \text{-module}$.

Qual 2013 #4 (Hungerford 4.2.14, Final #12): Let R be commutative and let E, F be free R-modules of the same finite rank. Prove that if $f\in Hom_R(E,F)$ is surjective then it is an isomorphism. Why do we need to assume that R is commutative?
Proof-
Let $rankE = rankF =n$. Let $\{x_1,\ldots ,x_k\}$ be a basis for $Ker(f)$. Then because the kernel is a submodule, we can extend this to a basis for E, $X = \{x_1,\ldots , x_k, x_{k+1}, \ldots , x_n\}$. Next, claim that $\{f(x_{k+1}), \ldots , f(x_n)\}$ is a basis for $Im(f) = F$. Let $r_{k+1}f(x_{k+1}) + \ldots + r_nf(x_n) = 0$. Then because f is an R-module homomorphism we have $f( r_{k+1}x_{k+1} + \ldots + r_nx_n) = 0$. This tells us that $r_{k+1}x_{k+1} + \ldots + r_nx_n \in Ker(f)$. Then we can use the basis of the kernel to write: $r_{k+1}x_{k+1} + \ldots + r_nx_n = r_1x_1 + \ldots r_kx_k$. Then subtracting we get $r_1x_1 + \ldots r_kx_k - r_{k+1}x_{k+1} - \ldots - r_nx_n = 0$. Because X is a basis for E we get that $r_i = 0 \forall i$. Thus $\{f(x_{k+1}), \ldots , f(x_n)\}$ is linearly independent. Next, suppose $b\in F,$ then there exists an element $a\in E$ such that $f(a) = b$. Since X is a basis of E we can write $f(r_1x_1 + \ldots + r_nx_n) = b$. Then by the homomorphism we have $r_1f(x_1) + \ldots + r_kf(x_k) + r_{k+1}f(x_{k+1}) + \ldots + r_nf(x_n)= b$. Then because $\{x_1,\ldots ,x_k\} \subset Ker(f)$, we have that the first k elements in the sum are 0. Thus $r_{k+1}f(x_{k+1}) + \ldots + r_nf(x_n)= b$ so that b is written as a linear combination of elements in the set $\{f(x_{k+1}), \ldots , f(x_n)\}$. Thus we have that this is a basis of F. However we know $rank(F) = n \Rightarrow k = 0$. Thus $Ker(f) = \{0\} \Rightarrow$ f is one to one. Thus f is an isomorphism.
We need commutativity in order to have the invariant rank property for R, which allows us to have basis’ for E and F with n elements.

## Algebra Qual Study Problems: 4

Final #4: (a) Find all Sylow subgroups of G, such that $|G| = 16$.
Proof-
Write $|G| = 16 = 2^4$. Then we can see there is only one Sylow 2 subgroup of order 16, which is the entire group

(b) $|G| = pq$ with p, q prime. Say that $q wlog. What is the cardinality of Sylow subgroups of G. Prove that any 2 distinct sylow subgroups intersect in the identity. Suppose in addition that all Sylow subgroups are normal, prove that G is isomorphic to the direct product of its Sylow Subgroups.
Proof- If G is a sylow-p subgroup then it must have order p, and a sylow-q subgroup must have order q. If P is a sylow p-subgroup and Q is a sylow q-subgroup, then $P \cap Q = \emptyset$ because an element cannot have order a power of p and order a power of q at the same time since p and q are distinct primes. Let $P_1, P_2$ be two sylow p-subgroups. Let $a\in P_1 \cap P_2$ then $\langle a\rangle \subset P_1 \cap P_2$. Since $|P_1| = |P_2| = p$ and they are p groups, the element a must also have order p. Thus $|\langle a\rangle | = p \Rightarrow P_1 = P_2$. Now assume that all Sylow subgroups are normal. Let P be a sylow p-subgroup and Q be a sylow q-subgroup. Then we have shown that $P \cap Q = \emptyset$ and $|\langle P\cup Q \rangle | = p\cdot q \Rightarrow \langle P\cup Q \rangle = G$, and since P and Q are normal in G we have by previous theorem that $G \cong P \times Q$.

Qual 95 #4, 08 #2 07 #4 (Hungerford 2.5 #9): If $|G| = p^nq$ with $p>q$ primes, then G contains a unique normal subgroup of index q.
Proof-
From Sylow’s First Theorem, G has a Sylow p-subgroup of order $p^n$. Let m be the number of Sylow p subgroups of G. By Sylows Third Theorem we have that $m | |G|$ and $m\equiv 1 \pmod{p}$. Then since $|G| = p^nq$ we have to have that either $m=1,q$ or $p^i$ for some $1\leq i \leq n$. If $m = q$ then $p|(q-1) \Rightarrow q-1 = pk$ for some $k\in \mathbb{Z}$. Then $pk+1 = q < p \Rightarrow k = 0 \Rightarrow q =1$ which is a contradiction. If instead we have that $m = p^i$ then $p | (p^i-1) \Rightarrow p = 1$, which is another contradiction. Therefore $m =1$ so that there is the only 1 Sylow p subgroup of G, call it H, which must be normal in G. Then by Lagrange, $|G| = [G:H]|H| \Rightarrow p^nq = [G:H] p^k \Rightarrow q = [G:H]$. Thus H has index q.

Qual 2005 #2, Final #5 (part 2 is Hungerford 2.5.11): 1) Exhibit 2 distinct Sylow 2 subgroups of $S_5$ and an element of $S_5$ conjugating one into the other.
Example –
Note that $|S_5|= 5! = 2^3 \cdot 3 \cdot 5$. Thus a sylow 2 subgroup has order 8. $\langle (24), (1234) \rangle \cong D_4$ and so is a Sylow 2 subgroup. Since any 2 Sylow subgroups are conjugate we can finish by working backwards. Conjugate the given subgroup by the element (12). So we have that $(12)(24)(12) = (14)$ and $(12)(1234)(12) = (1242)$ so that $\langle (14), (1342) \rangle$ is another distinct Sylow 2 subgroup with the desired property.

2) How many elements of order 7 are there in a simple group of order 168.
Proof-
Let $|G| = 168 = 2^3\cdot 3 \cdot 7$. By the Third Sylow Theorem, the number of Sylow 7-subgroups must divide $2^3\cdot 3$ and also must be congruent to 1 modulo 7. Thus the only possibilities are that the number of Sylow 7-subgroups is 1 or 8, but since G is simple we cannot have only one Sylow 7-subgroup otherwise it would be normal. Therefore there are 8 of them, each with 7 elements, with each non identity element having order 7 since a group with prime order must be cyclic. Thus there are $8(7-1)=48$ elements of order 7.

Final 2010 9 (Hungerford 2.5.10): Prove that a group of order 12 has a normal sylow subgroup.
Proof-
$|G| = 12 = 2^2\cdot 3$. By 3rd Sylow Theorem we know that there is either 1 Sylow 3 subgroup, or 4 Sylow 3 subgroups. If there is only one, then know it must be normal. Thus, suppose there are 4, $H_1,H_2,H_3,H_4$ each with order 3. Each of these subgroups has 3 elements, 1 being the identity. Thus the other 2 elements are of order 3. Together then we have $4(3-1) = 8$ elements of order 3 in G. Then G can only have one Sylow 2-subgroup which must be of order 4 because this gives us 12 total elements. Again, being the unique Sylow 2-subgroup gives that it is also normal.

Qual 2006 #4, 1999 #6: (1)Determine the units in $\mathbb{Z}[\sqrt{-5} ] = \{ a +b\sqrt{-5} : a,b \in \mathbb{Z}\}$.
Proof-
Define the map $N: \mathbb{Z}[\sqrt{-5} ] \rightarrow \mathbb{Z}$ by $a +b\sqrt{-5} \mapsto ( a +b\sqrt{-5} )( a -b\sqrt{-5}) = a^2 + 5b^2$. Note that $N(uv) = N(u)N(v)$ and claim u is a unit iff $N(u) = \pm 1$. Let u be a unit, then there exists $v$ such that $uv = 1 \Rightarrow 1 = N(1) = N(uv) = N(u)N(v) \in \mathbb{Z} \Rightarrow N(u) = \pm 1$. Then let $N(u) = 1$ where $u = a +b\sqrt{-5}$. Let $v = a -b\sqrt{-5} \Rightarrow N(u) = uv = a^2 + 5b^2 = 1 \Rightarrow$ u is a unit. The proof is similar if $N(u) = -1$ instead. Thus the claim is proved. Now we can prove (1). $a +b\sqrt{-5}$ is a unit iff $N(u) = (a +b\sqrt{-5} ) (a -b\sqrt{-5} ) = \pm 1 \Rightarrow a^2+5b^2 = 1 \Rightarrow b^2 = 0, a^2 = 1 \Rightarrow a = \pm 1$. Thus the only units are $\pm 1$ because the implications can be reversed. For (2), write $7 = (a +b\sqrt{-5}) (c +d\sqrt{-5} )$ Then $N(7) = N(a +b\sqrt{-5}) N(c +b\sqrt{-5}) \Rightarrow 49 = (a^2 + 5b^2) (c^2 + 5d^2)$. Wlog we have 2 possible outcomes: a) $a^2+ 5b^2 = 7$ and $c^2+5d^2 = 7$ or b)$a^2+5b^2 = 1$ and $c^2 + 5d^2 = 49$. The case of (a) gives a contradiction for any choice of b, while the (b) case gives a contradiction unless b = 0, in which case $a = \pm 1$. This gives that the only factorization of 7 is $7 = (7)(1)$ or $(7) = (-7)(-1)$ and so 7 is irreducible since 1 and -1 are units.

Qual 1999 #7: (a) Give an example of an integral domain R and ideals I and J of R such that $IJ \neq I \cap J$.
(b) Show that if $I + J = R$ then $IJ = I \cap J$.
Proof-
(a) Let $R = \mathbb{Z}, I = 2\mathbb{Z}, J = 4\mathbb{Z}$. Then $I \cap J = 4\mathbb{Z}$, but $IJ= 8\mathbb{Z}$. This is because in $\mathbb{Z}$, we have $\langle n \rangle + \langle m \rangle = \langle n,m \rangle = \langle gcd(n,m) \rangle$ but $\langle n \rangle \cap \langle m \rangle = \langle lcm(n,m) \rangle$.

(b) $IJ \subset I \cap J$ is always true because I and J are ideals.
Now let $x \in I \cap J$. Note that since we are given $I + J = R$ we can write $1_R = i + j$ for some $i\in I, j\in J$. Then $x = x\cdot 1_R = x(i + j) = xi + xj$. Then since x is in both I and J, $xi \in JI = IJ, xj \in IJ \Rightarrow x = xi + xj \in IJ$. This completes the proof.

Hungerford 3.1.15: (a) Give an example of a nonzero homomorphism $f: R \rightarrow S$ of rings with identity such that $f(1_R) \neq 1_S$.
(b) If $f:R \rightarrow S$ is an epimorphism then $f(1_r) = 1_S$.
(c) If $f:R \rightarrow S$ is a homomorphism of rings with identity and u is a unit in R such that $f(u)$ is a unit in S, then $f(1_R) = 1_S$ and $f(u^{-1}) = f(u)^{-1}$. Give an example to show that $f(u)$ need not be a unit in S even though u is a unit in R.
Proof-
a) Consider $f: \mathbb{Z}_3 \rightarrow \mathbb{Z}_6$ by $f(k) = 4k$. Then for $a,b\in \mathbb{Z}$ we have $f(a+b) = 4(a+b) = 4a +4b = f(a) + f(b)$ and $f(a)f(b) = 4a4b = 16ab = 4ab \pmod{6} = f(ab)$ but note that $f(1) = 4 \neq 1$ in $\mathbb{Z}_6$.

b) Let $s\in S$. Since f is surjective we have that $s = f(r)$ for some $r\in R$. Then $f(1_R) s = f(1_r) f (r) = f(r) = s$ and similarly $sf(1_R) = s$. Since the identity element is unique and $f(1_R)$ is acting as the identity we must have that $f(1_R) = 1_S$.

c) First note that $f(u) = f(1_R u ) = f(1_R) f(u)$. Then since $f(u)$ is given to be a unit we can multiply both sides by the inverse so that $1_s = f(1_R)$. Then using this fact we have $1_S = f(1_R) = f(uu^{-1}) = f(u) f(u^{-1})$ and similarly $1_S = f(u^{-1}) f(u)$. Then since inverses are unique and $f(u^{-1})$ is acting as the inverse of $f(u)$ we must have that $f(u^{-1}) = f(u)^{-1}$. Lastly, use the same map as in part a) for the counterexample and note that 2 is a unit $\mathbb{Z}_3$, but $f(2)=2$ is not a unit in $\mathbb{Z}_6$.

Qual 2013 #4: Find a maximal ideal in $\mathbb{C}[x]$ containing $f(x) = x^3-3x^2+2$. How many such ideals are there? In $\mathbb{R}[x]$ how many are there? Find $g(x) \in \mathbb{C}[x]$ such that $\langle f(x),g(x) \rangle \mathbb{C}[x]$.
Proof-
Recall some facts: If R is an integral domain and $0 \neq p(x),q(x) \in R[x]$ then 1) $deg(p(x)(q(x)) = deg(p(x)) + deg(q(x))$, 2) units of $R[x]$ are the units in R, 3) $R[x]$ is an ID also. Lastly, if F is a field then $F[x]$ is an ED which means it is a PID, UFD, and ID as well. Now, since $\mathbb{C}$ is a field we have that $\mathbb{C}[x]$ is a PID. Since any ideal is generated by an element, we know that $M =$ is maximal iff m is irreducible in $\mathbb{C}[x]$. Now we have $x^3-3x^2 + 2 = (x-1) (x-(1+\sqrt{3}))(x-(1-\sqrt{3}))$. Then we have $x^3-3x^2+2 \in \langle x-1 \rangle, \langle x-1-\sqrt{3} \rangle, \langle x-1+\sqrt{3} \rangle$. Since each factor is irreducible, each of these ideals is maximal. If $g(x) = 1$, then $\langle f(x),g(x) \rangle = \mathbb{C}[x]$.

Hungerford 3.4.12: A commutative ring with identity is local iff for all $r,x\in R$ we have $r+s =1_R \Rightarrow r \text{ or } s$ is a unit.
Proof-
$(\Leftarrow )$ Suppose that R is local and that $r+s= 1$. Suppose that r is not a unit. Since R is a local ring it has a unique max ideal, call it M. Then $r = 1-s \in M$ since all nonunits must be in M, and r is not a unit. If s was also a nonunit then we would have $(1-s)+s = 1 \in M$ which is a contradiction to the definition of maximal ideal. Therefore s must be a unit.
$(\Leftarrow )$ Let $1 = r + s \Rightarrow r \text{ or } s$ is a unit be true for all $r,s \in R$. Now, let r and s be nonunits. If $r + s$ is a unit then there exists a $x \in R$ such that $x(r+s) = 1 \Rightarrow x r + xs = 1$. Then by the assumption we have either $xr$ or $xs$ is a unit. This tells us that either r or s is a unit, which is a contradiction to assuming neither was a unit. Therefore, $r+s$ is a nonunit. If r is a nonunit then we know $xr$ is a nonunit for all x. Thus, we have that the sum of nonunits is a nonunit, and that nonunits absorb any element through multiplication. This gives that the set of all nonunits forms an ideal, which implies R is a local ring.

Qual 2013 #3: Prove that if R is a local ring and $\phi : R \rightarrow S$ is a surjective ring hom onto nonzero ring S, then S is local.
Proof-
Let R be a local ring and let M be the unique maximal ideal of R. Since the map is an epimorphism, ideals map to ideals and the inverse image of an ideal is an ideal which contains the kernel of the map. Now suppose that $\phi (M ) \subset N \subset S$ then $\phi^{-1}(\phi (M) \subset \phi^{-1}(N) \subset \phi^{-1}(S) =R$. This tells us that $M \subset \phi^{-1}(N) \subset R$. However we know M is maximal so that $\phi^{-1}(N) = M$ or R. Thus $N = \phi (M)$ or $S = \phi (R ) = N$ Therefore $\phi (M)$ is maximal. Next we show it is unique. If N is another maximal ideal in S, then we have $\phi^{-1}(N)$ is also max in R which by uniqueness of M gives that $\phi^{-1} (N) = M \Rightarrow N = \phi (M)$. Therefore S is a local ring.

Let $R = \mathbb{Z}$. Then $n\mathbb{Z}$ is maximal iff n is prime.
Proof-
$(\Rightarrow )$Let $p\mathbb{Z}$ be a maximal ideal and suppose $p\mathbb{Z} \subset J$. Then there exists an element which is in J but not in $p\mathbb{Z}$. Therefore $gcd(r,p) = 1$ so that $rx +py = 1$ for some $x,y \in \mathbb{Z}$. Then $py\in J$ and $rx\in J$ by definition of ideal. But this gives that $1\in J$ which means that $R = J$. Therefore $p\mathbb{Z}$ is maximal.
$(\Leftarrow )$ We know that the integers form a PID so if $m\mathbb{Z} \subset n\mathbb{Z}$ then $n |m$. So if $p\mathbb{Z} \subset n\mathbb{Z} \Rightarrow n|p \Rightarrow n =1 \text{ or } p$. Therefore $n\mathbb{Z} = \mathbb{Z}$ or $p\mathbb{Z}$ so that it is maximal.

Final 2013 #2: (a) Consider the ring $\mathbb{Z}$ of integers. Find a maximal ideal that contains the integer 45.
(b) Find a prime ideal in the integers that contains both 45 and 120
(c) Find an integer r such that $r\mathbb{Z} = 45\mathbb{Z} \cap 120\mathbb{Z}$.
Proof-
a) A maximal ideal in the integers is the same as a prime ideal. Since $45 = 3^2 \cdot 5$ we have that $45\in 3\mathbb{Z}$ which is a max ideal.

b) We can write $45 = 3^2\cdot 5$ and $120 = 2^3 \cdot 3\cdot 5$. Therefore both integers are contained in $3\mathbb{Z}$.

c) $45\mathbb{Z} \cap 120\mathbb{Z} = \{ x\in \mathbb{Z} | x = 45m = 120, \text{ where } n,m\in \mathbb{Z} \}$. But this is the same as the set $\{x\in \mathbb{Z} | 45|x \text{ and } 120 |x\}$. This set is generated by the lcm of 45 and 120. Therefore $45\mathbb{Z} \cap 120\mathbb{Z} = \langle lcm(45,120) \rangle = \langle 2^3\cdot 3^2 \cdot 5\rangle = 360\mathbb{Z}$.

## Algebra Solutions: Hungerford III.3

1) A nonzero ideal in a principle ideal domain is maximal if and only if it is prime.
Proof-
Let R be a principle ideal domain and I be a nonzero ideal in R.
$( \Rightarrow )$ Assume that I is maximal. Then $R/I$ is a field, which means that $R/I$ is an integral domain. Hence, I is prime.
$(\Leftarrow )$ Assume that I is a prime ideal. Since R is a PID, we can write $I = (a)$ for some nonzero $a\in R$. Then we know that the element a is prime and hence irreducible. Then since a is irreducible, $(a)$ is maximal in the set of all proper principle ideals of R. But since every ideal in R is principle, $(a)$ is maximal in R.

2) An integral domain R is a unique factorization domain if and only if every nonzero prime ideal in R contains a nonzero principle ideal that is prime.
Proof-
In order to prove this, we need a lemma: In a UFD, irreducible elements are prime.
Subproof- Let R be a UFD and $c\in R$ be irreducible. Then c is a nonzero nonunit. Now assume that $c | ab$ for some $a,b\in R$. Then we can write $cx = ab$ for some $x\in R$. If $a = 0$ or $b = 0$ then we are done trivially so assume that a, b are nonzero. If either a or b is a unit, then either $b = a^{-1}cx$ or $a = cxb^{-1}$ which gives that $c|b$ or $c|a$ so assume that a, b are nonunits as well. If $x=0$ then since R is an integral domain we would have either a or b is zero, a contradiction. If x is a unit then since c is irreducible, $c = abx^{-1} \Rightarrow$ a is a unit or $bx^{-1}$ is a unit. We know a is not a unit so that $bx^{-1}$ would have to be a unit. Then $a = c(bx^{-1})^{-1} \Rightarrow c|a$ so that c is prime. Now assume that x is not a unit. Being in a UFD means that we can now factor a, b, and x into a product of irreducibles. Let $a = a_1\cdots a_r, b = b_1 \cdots b_s, x = x_1 \cdots x_t$. We have that $a_1\cdots a_r b_1 \cdots b_s = c x_1 \cdots x_t$. Both sides are products of irreducibles and so by the definition of UFD we must have that $r + s = 1 + t$ and that any element on one side must be an associate to some element on the other side. Thus c is an associate of some $a_i$ or $b_j$. In the first case we get that $a_i = cu$ for some unit $u\in R$. Then $c|a_i \Rightarrow c|a$ and similarly if c is an associate of some $b_j$ then $c|b$. Thus c is prime. Now using the lemma we can do the proof:
$(\Rightarrow )$ Assume that R is a UFD. Let I be a nonzero prime ideal in R. Then there exists a nonzero element $c\in I$. This element cannot be a unit or else I would be all of R and this is a contradiction since prime ideals must be proper ideals. So then c is a nonzero nonunit. Since we are in a UFD we can write $c = c_1\cdots c_n$ where each $c_i$ is irreducible. Then by the lemma we have that each$c_i$ is in fact prime. Therefore $(c_i)$ is a prime ideal contained in I.
$(\Leftarrow )$ Assume that every nonzero ideal in R contains a nonzero principle ideal that is prime. Let S be the set of all units in R and all products of prime elements of R. Note that S is a multiplicative set since a product of units is a unit and a product of a product of primes is a product of primes. Also, if $p$ is prime and $u$ is a unit, then we need to know that $pu$ is prime or a unit as well. Suppose it is not a unit, then suppose $pu | ab \Rightarrow ab = pux\Rightarrow abu^{-1} = px \Rightarrow p |abu^{-1}$. But p is prime so then $p|a$ or $p|bu^{-1}$. In the first case we are done and if $p| bu^{-1} \Rightarrow bu^{-1} = py \Rightarrow b= pyu \Rightarrow p|b$. Therefore S is closed. We want to show that $S = R \backslash \{0\}$. This will give that every nonzero nonunit is a product of primes and hence a product of irreducibles since we are in an integral domain. Suppose to the contrary that there exists some nonzero nonunit $x\in R \backslash S$. Then we must have that $S \cap (x) = \emptyset$ since S is also closed under divisors, i.e. whenever a product $ab\in S \Rightarrow a,b \in S$. Next we need a few facts: 1) Since $(x)$ is an ideal, it is contained in a maximal ideal. 2) The set of all ideals of R which are disjoint from S contains a maximal element, and that element is prime. Therefore since $(x)$ is an ideal disjoint from S, it is contained in a maximal ideal, P, which is also a prime ideal and also disjoint from S. However, by assumption this prime ideal P contains a principle ideal which is prime, and therefore contains a prime element, say p. But S contains all prime elements so this contradicts the disjointness. Therefore S consists of units and irreducible elements and hence R is a UFD.

6) (a) If a and n are integers, $n> 0$, then there exist integers q and r such that $a = qn + r$ where $|r| \leq n/2$.
(b) the Gaussian integers $\mathbb{Z}[i]$ form a Euclidean domain with $\phi (a +bi) = a^2 + b^2$.
Proof- a)
Let $a,n\in \mathbb{Z}$ with $n>0$. By the division algorithm there exists $q,r \in \mathbb{Z}$ such that $a = qn + r$ where $0 \leq r \frac{n}{2}$. Then $n - r < \frac{n}{2}$. Pick $q' =q + 1$ and $r' = n-r$. Then $a = qn + r = qn + n-n+r = (q+1)n +r-n$, where $|r-n| = n -r <\frac{n}{2}$. This ends the problem

b) Consider $(\mathbb{Z}[i], \phi )$ where $\phi (a+bi) = a^2+b^2$. We must show that this is a Euclidean Ring. First assume that $x= a+bi, y = c+di \in \mathbb{Z}[i]$ and $xy \neq 0$. Then with some work we can show that $\phi (xy) = \phi (x)\phi (y) \geq \phi (x)$ since $\phi (y) \in \mathbb{N}$. Now let $y = a+bi$ and $x\in \mathbb{Z}$, $x >0$. Then we can find integers $q_1,q_2,r_1,r_2$ such that $a = q_1x + r_1, b = q_2x + r_2$, with $|r_1| < \frac{x}{2}, |r_2| <\frac{x}{2}$. Let $q = q_1 + q_2i, r = r_1+r_2i \Rightarrow y = qx + r$. If r is zero we are done so assume r is nonzero. Then $\phi (r) = r_1^2 + r_2^2 \leq \frac{x^2}{2} . So when $x\in \mathbb{Z}$ for $y \neq 0$ there exists $q,r \in \mathbb{Z}[i]$ such that either $r= 0$ or $r\neq 0$ and $\phi (r) 0$. Then just replace x with $x\bar{x}$ in the previous proof. Find $q,r_0 \in \mathbb{Z}[i]$ such that $y\bar{x} = q(x\bar{x}) + r_0$ with either $r_0 = 0$ or $r_0 \neq 0$ and $\phi (r_0) < \phi (x\bar{x})$. Set $r = y -qx$. Then $r\bar{x} = q(x\bar{x}) + r_0 - q(x\bar{x}) = r_0$. Then $\phi (r) \leq \phi (r_0) < \phi (x)$ so we have a Euclidean domain.

7) What are the units of Gaussian integers $\mathbb{Z}[i]$?
Proof –
Let $U(\mathbb{Z}[i]) = \{ x \in \mathbb{Z}[i] | x \text{ is a unit} \}$. We claim that $U(\mathbb{Z}[i]) = \{\pm 1, \pm i\}$ and show it by subset inclusion. Let $x = a+bi \in U(\mathbb{Z}[i])$. Then there is $y = c+di \in \mathbb{Z}[i]$ such that $xy= 1$. By exercise 6(b) we have that $\phi (x) \phi (y) = \phi (xy) = \phi (y) = 1$. Since $\phi (x), \phi (y) \in \mathbb{N}$ we can only have the case that $\phi (x) = 1$. But then $a^2 + b^2 = \phi (x) = 1$ says that either $a = \pm 1$ and $b = 0$ or $a = 0$ and $b = \pm 1$. Thus $x = \pm 1$ or $x = \pm i$. Therefore we have inclusion. The other inclusion is trivial to check.

9) Let R be a UFD and d a nonzero element of R. There are only a finite number of distinct principal ideals that contain the ideal $(d)$.
Proof-
If d is a unit then we are done trivially, so assume d is a nonunit. Then since we are in a UFD we can write $d=d_1 \cdots d_l$ where $d_i$ is irreducible. Let $k\in R$ be any element satisfying $(d) \subset (k)$. Then k cannot be 0 and if k is a unit then it generates all of R so that $(k) = R$ is the only principle ideal which contains (d). So assume that k is a nonunit. Then we can write $k = k_1 \cdots k_m$ with each $k_j$ irreducible. Now $(d) \subset (k) \Rightarrow k|d$ so that $d = kx$ for some $x\in R$. If $x = 0$ then d would be zero also, which is a contradiction. If x is a unit then $k = dx^{-1} \Rightarrow kr = dx^{-1}r \forall r \in R$ which tells us that $(k) = (d)$ is the only principle ideal containing $(d)$. Therefore we can assume x is a nonunit. Then we can write $x = x_1\cdots x_n$ for some irreducibles $x_k$. Then we have $d_1 \cdots d_l = k_l\cdots k_m x_1 \cdots x_n \Rightarrow l = m+n$ since we are in a UFD. Note that for each $1\leq i \leq m$ we have $(k) \subset (k_i)$ and each $k_i$ is irreducible so that each $(k_i)$ is maximal. Thus $(k)$ is contained in m maximal ideals. If there were infinitely many such principal ideals $(k)$ we would have to have infinitely many m’s which satisfy $m +n = l$. But $l,m,n \in \mathbb{N}$  and l is fixed so this is impossible. Hence there are only finitely many k satisfying the condition.

10) If R is a UFD and $a,b \in$ are relatively prime and $a|bc$ then $a|c$.
Proof-
First note that if $gcd(a,b) = 1$ then $a,b$ have no irreducible factors in common. This means that if $a = a_1 \cdots a_m, b = b_1\cdots b_n$ with $a_i,b_j$ irreducible, then no $a_i$ is an associate of $b_j$. For if they were associates, then we could say $a_i = b_ju$ where u is a unit, then $a = a_1\cdots a_{i-1}b_jua_{i+1}\cdots a_m \Rightarrow b_j |a$. But since $b_j |b$ and gcd(a,b) = 1 this would give that $b_j |1$ which means that $b_j$ is a unit, contradicting that $b_j$ is irreducible. Finally, assume that $a|bc$. Then $bc = ax$for some $x\in R$. Factoring into products of irreducibles we have that $b_1 \cdots b_n c_1 \cdots c_l = a_1 \cdots a_m x$. Now $a_i$ cannot be associate of any $b_j$ so each $a_i$ must be an associate of some $c_k$. Therefore $l\geq m$ so without loss of generality say that $a_i$ is an associate of $c_i$ for $1 \leq i \leq m$. This tells us that $a_i | c_i$ for $1 \leq i \leq m$. Thus we can say $c_1 = a_1 y_1, c_2 = a_2y_2, \ldots c_m = a_m y_m \Rightarrow c_1 \cdots c_m = a_1\cdots a_m y_1 \cdots y_m =a y_1 \cdots y_m$. But then we can multiply by what we need to get all of c on one side, so that $c = c_1 \cdots c_m c_{m+1} \cdots c_l = ay_1\cdots y_m c_{m+1}\cdots c_l \Rightarrow a |c$.

11) Let R be a Euclidean ring and $a\in R$. Then a is a unit iff $\phi (a ) = \phi (1_R)$.
Proof-
$( \Rightarrow )$ Assume that a is a unit. Then $\phi(a) \leq \phi(aa^{-1} )= \phi (1) \leq \phi (1a) = \phi (a)$. Therefore $\phi (a) = \phi (1)$.
$(\Leftarrow )$ Assume now that $\phi (a) = \phi (1)$. There must exist $q,r \in R$ such that $1 = aq +r$ where either r = 0 or $\phi (r) < \phi (a)$. Note that $\phi (1) \leq \phi (1r) = \phi (r)$, so we cannot have that $\phi (r) < \phi (a) = \phi (1)$. Therefore r = 0 and hence $q = a^{-1}$ so that a is a unit.

## Algebra Solutions: Hungerford III.2

1) The set of all nilpotent elements in a commutative ring forms an ideal.
Proof-
Let R be a commutative ring and let $N = \{a \in R | a^n = 0 \text{ for some natural number n} \}$. Since 0 is in N we know that N is nonempty. From exercise 3.1.12 we know that the sum of two nilpotent elements is nilpotent. Let $a,b \in N$, then say that $a^m = 0, b^n=0 \Rightarrow (-b)^n = \pm b^n = \pm 0 = 0$. Therefore $-b$ is nilpotent. Thus for $a,b \in N$ we have that $a-b \in N$ being a sum of nilpotent elements. Lastly, let $r\in R$. Then we must show that $ra \in N$. Note that $(ra)^m = r^ma^m = r^m0 = 0$. Thus $ra= ar \in N$ so that N is an ideal by the ideal test theorem.

2) Let I be an ideal in a commutative ring R and let Rad(I) $= \{ r \in R | r^n \in I \text{ for some n} \}$. Show that Rad(I) is an ideal.
Proof-
0 is in Rad(I) trivially so that Rad(I) is nonempty. Let $a,b \in Rad(I)$ and $r\in R$ with $a^n,b^m\in I$. We want to show that $a-b \in Rad(I)$. Consider then:
$(a-b)^{n+m} = \sum_{k=0}^{n+m} \dbinom{n+m}{k} a^{n+m-k}(-b)^k$. If $k \leq m \Rightarrow n+m-k > n \Rightarrow a^{n+m-k} \in I$. On the other hand if $k\geq m$ then $(-b)^k \in I$. Thus $\forall 0 \leq k \leq n+m$ we have that $\dbinom{n+m}{k} a^{n+m-k}(-b)^k\in I$. Thus $(a-b)^{n+m} \in I \Rightarrow a-b \in Rad(I)$. Lastly $(ra)^n =r^na^n = \in I$ since I is an ideal. Thus $ra=ar \in Rad(I)$. So once again by the ideal test, Rad(I) is an ideal.

3) If R is a ring and $a\in R$, then $J = \{r\in R | ra = 0\}$ is a left ideal and $K = \{r\in R | ar = 0\}$ is a right ideal in R.
Proof-
0 is obviously in J so J is nonempty. Let $b,c\in J$ and $r\in R$. Then $ba = ca = 0$. Then $(b-c)a = ba-ca = 0-0 = 0$ so that $b-c \in J$. Also, $(rb)a = r(ba) = r0 = 0 \Rightarrow rb \in J$. Thus J is a left ideal. The proof for K is exactly the same but on the right side.

4) If I is a left ideal of R, then $A(I) = \{r\in R | rx = 0 \forall x\in I\}$ is an ideal in R.
Proof-
Since $0x=0 \forall x\in I$ we have that $0\in A(I)$ so that A(I) is nonempty. Now let $a,b\in A(I)$ and $r\in R$. Then $ax=bx=0 \forall x \in I.$ Thus $(a-b)x = ax-bx=0-0=0 \forall x \in I$. Therefore $a-b\in A(I)$. Lastly, $(ra)x = r(ax) = r0=0 \Rightarrow ra \in A(I)$ and $(ar)x = a(rx)= a0 = 0$ since $rx\in I$ because $x\in I$ and I is an ideal. Thus $ra,ar \in A(I)$so that A(I) is an ideal in R.

5) If I is an ideal in a ring R, let $[R:I] = \{ r\in R | xr \in I \forall x\in R \}$. Prove that $[R:I]$ is an ideal of R which contains I.
Proof-
First note that $x0=0 \in I \forall x\in R$ so that $0 \in [R:I]$. Thus $[R:I]$ is nonempty. Now let $a,b\in [R:I]$ and $r\in R$. Then $xa, xb \in I \forall x\in R$. Then $x(a-b) = xa-xb \in I \forall x \in R$. Therefore $a-b \in [R:I]$. Also $x(ra) = (xr)a \in I$ because $xr\in R$ and $a\in [R:I]$. Thus $ra \in [R:I]$. Similarly $ar \in [R:I]$ so that $[R:I]$ is an ideal. Lastly, let $c \in I$. Then $c\in R \Rightarrow xc \in I \forall x\in R$ because I is an ideal. Thus $c \in [R:I]$ so that $I \subset [R:I]$.

10) (a) Show that $\mathbb{Z}$ is a principle ideal ring.
(b) Every homomorphic image of a principle ideal ring is also a principle ideal ring.
(c) $\mathbb{Z}_m$ is a principle ideal ring for every $m>0$.
Proof- a)
Let I be an ideal in $\mathbb{Z}$. Then being a subring of the integers it must have a lowest positive integer, call it a. Claim that $I = (a)$. Obviously $(a) \subset I$ so we must show that $I \subset (a)$. Let $b \in I$. Then we can write by the division algorithm, $b = aq + r$, where $0\leq r < a$. Then $r = b-aq$. We have that $b \in I$, and since I is an ideal and $a\in I \Rightarrow aq \in I$. This gives that $b-aq = r\in I$. But a was the least positive integer in I, so that r must be 0. Thus $b = aq \Rightarrow b \in (a)$. Thus $I = (a)$ and since I was arbitrary, any ideal of $\mathbb{Z}$ must be principle.

b) Let R be a principle ideal ring and $f:R \rightarrow S$ be a homomorphism. Want to show that $Im(f)$ is a principle ideal ring, i.e. that every ideal in $Im(f)$ is principle. Let J be an ideal in $Im(f)$. Then consider $I = f^{-1}(J) = \{ r\in R | f(r) \in J\}$. Since R is principle, and the inverse image of an ideal is an ideal, we have that I is a principle ideal, so we can write, $I = (a)$. Since f is onto the image we know that f((a)) is an ideal in S. We have that $(a) = \{ ra | r\in R\}$ so that:
$f((a)) = \{ f(ra) | r \in R\} = \{ f(r)f(a) | r \in R\} = \{ sf(a) | s \in S\} = (f(a))$ since f is surjective onto the image.
Want to show that $f((a)) = (f(a)) = J$. Since $f(a) \in J$ we have that $(f(a)) \subset J$. Now let $y\in J$. Then since f is surjective onto the image, there exist $x\in R$ such that $f(x) = y$. In fact, $x\in f^{-1}(J) = (a)$. Thus we can write $x = r'a$. Then $y = f(x) = f(r'a) = f(r') f(a) \in (f(a))$. Therefore $(f(a)) = J$ so that Im(f) is a principle ideal ring.

c) Consider the canonical homomorphism $\pi : \mathbb{Z} \rightarrow \mathbb{Z}_m$ by $a\mapsto \bar{a}$. Since this is a surjective homomorphism and since the integers are a principle ideal ring, from part b we have that $\mathbb{Z}_m$ is a principle ideal ring.

13) Let $f:R \rightarrow S$ be a homomorphism of rings, I an ideal in R, and J an ideal in S.
(a)$f^{-1}(J)$ is an ideal in R that contains Kerf.
(b) If f is an epimorphism then f(I) is an ideal in S. If f is not surjective then f(I) need not be an ideal in S.
Proof- a
) Let $a,b\in f^{-1}(J)$ and $r\in R$. Then we have that $f(a),f(b) \in J \Rightarrow f(a)-f(b) = f(a-b) \in J \Rightarrow a-b \in f^{-1}(J)$. Also, $f(ra) = f(r)f(a) \in J$ because J is an ideal. Therefore $ra \in f^{-1}(J),$ and so $f^{-1}(J)$ is an ideal. Now let $x\in Kerf$. This gives us that $f(x) = 0 \in J \Rightarrow x\in f^{-1}(J)$. So that the Kernel is contained in $f^{-1}(J)$.

b) Now we assume that f is onto. Let $c,d \in f(I)$, and $s\in S$. Since f is onto there exists, $a,b\in I$ such that $f(a) = c, f(b) = d$. Since I is an ideal, $a-b \in I$ and then $f(a-b) = f(a)-f(b) = c-d \Rightarrow c-d \in f(I)$. Also, since f is onto, there exists an $r\in R$ such that $f(r) = s$. Since I is an ideal, $ra \in I$. Then $f(ra) = f(r)f(a) = sc \Rightarrow sc \in f(I)$. Similarly, $cs \in f(I)$, so that $f(I)$ is an ideal in S.

17) Let $f: R\rightarrow S$ be an epimorphism of rings with kernel K.
(a) If P is a prime ideal in R that contains K, then $f(P)$ is a prime ideal in S.
(b) If Q is a prime ideal in S, then $f^{-1}(Q)$ is a prime ideal in R that contains K.
(c) There is a 1-1 correspondence between the set of all prime ideals in R that contain K and the set of all prime ideals in S, given by $P\mapsto f(P)$.
(d) If I is an ideal in a ring R, then every prime ideal in $R/I$ is of the form $P/I$ where P is a prime ideal in R that contains I.
Proof- a)
Let P be a prime ideal in R that contains K. Then since f is an epimorphism we know that $f(P)$ is an ideal in S. Let $c,d\in S$ and assume that $cd\in f(P)$. Since f is onto there exist $a,b \in R$ such that $f(a) = c, f(b) =d$. Then $f(ab) = f(a)f(b) = cd \in f(P)$. Since $f(P)$ is an ideal this also tells us that $-f(ab) \in f(P)$ so that there exists $p\in P$ such that $f(p) = -f(ab)$. Then $0 = f(ab) + f(p) = f(ab+p) \Rightarrow ab+p \in K \subset P$, by assumption. Therefore there is some $p'\in P$ such that $ab+p = p' \Rightarrow ab = p'-p\in P$. Since P is a prime ideal this tells us that $a\in P$ or $b\in P$. Therefore $f(a) = c \in f(P)$ or $f(b) = d \in f(P)$ so that f(P) is prime.

b) Let Q be a prime ideal in S. Then we have that $f^{-1}(Q)$ is an ideal in R that contains K. Let $a,b\in R$ and assume that $ab\in f^{-1}(Q)$. Then $f(ab) = f(a) f(b) \in Q \Rightarrow f(a) \in Q$ or $f(b) \in Q$ since Q is a prime ideal. Therefore $a\in f^{-1}(Q)$ or $b\in f^{-1}(Q)$ which tells us that $f^{-1}(Q)$ is prime.

c) Let X be the set of all prime ideals in R that contain K and Y be the set of all prime ideals in S. Defined $\phi :X \rightarrow Y$ by $\phi (P) = f(P)$. Claim that $\phi$ is bijective. Let $P,Q \in X$. and assume that $\phi (P) = \phi (Q)$. Then, $f(P) = f(Q)$. Let $p\in P$. Then $f(p) \in f(P) = f(Q)$ gives us that there exists a $q\in Q$ such that $f(q) = f(p)$. Then $0 = f(p-q) \Rightarrow p-q \in K \subset Q$. Which means that $p\in Q$. So then $P \subset Q$, and we can similarly show that $Q \subset P$ so that $P = Q$. Therefore $\phi$ is one to one. Next, let $J \in Y$. From part b we know that $f^{-1}(J)$ is a prime ideal so it is in the set X. Then $\phi (f^{-1}(J)) = f(f^{-1}(J)) = J$ so that $\phi$ is surjective. Thus we have a one to one correspondence.

d) Let $\bar{P}$ be a prime ideal in $R/I$. Consider $\pi : R \rightarrow R/I$ given by $\pi (r) = r+ I$. Since $\pi$ is an epimorphism with Kernel I we have that $\pi^{-1} (\bar{P})$ is a prime ideal in R which contains I by part (b). Then $\bar{P} = \pi ( \pi^{-1} (\bar{P} )) = \{ q+I | q\in \pi^{-1} (\bar{P})\} = \pi^{-1} (\bar{P})/I$. Then call $\pi^{-1} (\bar{P}) = P$ so that $\bar{P} = P/I$, where P has the wanted properties.

## Algebra Solutions: Hungerford III.1

3) A ring R such that $a^2 = a \forall a \in R$ is called a Boolean ring. Prove that every Boolean ring R is commutative and $a+a = 0 \forall a\in R$.
Proof-
Let $a,b \in$. Since R has an underlying additive group structure we know that $a\ +b \in R \Rightarrow (a+b)^2 = a + b$. But also, $(a+b)(a+b) = a^2 + ab + ba + b^2 = a + ab + ba + b$. Thus we have $a + b = a + b + ab + ba \Rightarrow ab + ba = 0 \Rightarrow ab = -ba$. Then we have$a-b = (a-b)^2 \Rightarrow a-b = a - ab - ba + b = a - ab + ab + b = a + b \Rightarrow b = -b$ Thus $ab = -ba = ba$ so that R is commutative. Lastly, $a + a = (a+a)^2 = a^2 + a^2 + a^2 + a^2 = a + a + a + a \Rightarrow a + a = 0$.

5) If A is the abelian group $\mathbb{Z} \bigoplus \mathbb{Z}$, then End A is a noncommutative ring.
Proof-
Recall that End A is defined to be the set of all group homomorphisms from A to A with addition being pointwise and multiplication being composition. Let $f\,g \in End A$ and $a,b\in A$. Then we have $(f+g)(a+b) = f(a+b) + g(a+b) = f(a) + f(b) + g(a) + g(b) = f(a) + g(a) + f(b) + g(b) = (f+g)(a) + (f+g)(b)$ which shows that addition is well defined. Composition is similarly well defined. Addition is clearly associatve and commutative, with additive identity the 0 map and additive inverse of f being $-f$. Composition is also clearly associative and distributive over addition so that End A is a ring. Now we show that End A is noncommutative. Consider the function $f:A \rightarrow A$ defined by $f(a,b) = (b,a)$ and the function $g:A\rightarrow A$ defined by $g(a,b) = (2a,b).$ It is easily checked that they are both homomorphisms. We have that $fg(a,b) = f(2a,b) = (b,2a)$ while $gf(a,b) = g(b,a) = (2b,a)$ so that End A is not commutative.

6) A finite ring with more than one element and no zero divisors is a division ring.
Proof –
Let $a\in R$ be nonzero. Since we have a finite ring, there exists natural numbers $i>j$ such that $a^i = a^j$. First we show that there is a multiplicative identity element. Since we are in a ring we have additive inverses so we can say $a^i + (-a^j) = a^j + (-a^j) = 0$. Then we can rewrite $a^i - a^j = a\cdot a^{j-1}\cdot a^{i-j} - a\cdot a^{j-1} = a^{j-1} ( a\cdot a^{i-j} - a) = 0.$ Lastly, because we have no zero divisors one of the two factors must be 0, but a was assumed to be nonzero so we have $a\cdot a^{i-j} - a = 0 \Rightarrow a\cdot a^{i-j} = a$ which gives us the right identity. We can similarly get the left identity so that R has a multiplicative identity, call it 1. Now we must show that is a a unit, i.e. that it has an inverse. Again we can write $a^i = a^j \Rightarrow a^ja^{i-j} - a^j = 0 \Rightarrow a^j( a^{i-j} -1) = 0 \Rightarrow a^{i-j} = 1$. We can write this as $a \cdot a^{i-j-1} = 1$ so that the right inverse of a is $a^{i-j-1}$. Similarly it is the left inverse of a so that we have a is invertible and thus R is a division ring.

7) Let R be a ring with more than one element such that for each nonzero $a\in R$ there is a unique $b\in R$ such that $aba = a$. Prove:
a)R has no zero divisors
b) $bab= b$
c) R has an identity
d) R is a division ring
Proof-
a) Assume that $a\in R$ is nonzero and that $c\in R$ and assume $ac=0$. Then we can say $aca = 0$. Let $b\in R$ such that $aba = a$. Then we have $aca + aba = 0 + a = a \Rightarrow a(c+b)a = a$. However we know b is the unique element with this property so that $c + b = b \Rightarrow c=0$. Thus R has no zero divisors.
b)$aba = a \Rightarrow abab = ab \Rightarrow abab - ab = 0$. This gives $0 = a(bab-b) \Rightarrow bab - b = 0 \Rightarrow bab = b$ since a was nonzero.
c) $aba = a \Rightarrow abac = ac \Rightarrow bac = c$. This gives the element ba as the left identity. Then $bab = b \Rightarrow cbab = cb \Rightarrow cba = c$. This gives that the element ba is the right identity, and thus is the identity, call it 1.
d) $ba = 1$ gives that b is a left inverse of a and $aba = 1 \cdot a \Rightarrow ab = 1$ gives that b is also a right inverse. Thus a is a unit.

11) Let R be a commutative ring with identity of prime characteristic p. If $a,b \in R$ then $(a\pm b)^{p^n} = a^{p^n} \pm b^{p^n}$ for all integers $n\geq 0$.
Proof-
Since R is commutative with identity we have that $(a+b)^{p^n} = \sum_{k=0}^{p^n} \dbinom{p^n}{k} a^{p^n-k}b^k$. Then since $p | \dbinom{p^n}{k}$ for all $a\leq k \leq p^n-1$ and R has characteristic p, we know that $\dbinom{p^n}{k}a^{p^n-k}b^k = 0 \forall 1 \leq k \leq p^n-1$. Therefore $(a+b)^{p^n} = \dbinom{p^n}{0}a^{p^n-0}b^0 + \dbinom{p^n}{p^n}a^{p^n-p^n}b^{p^n}$. Thus $(a+b)^{p^n} = a^{p^n} + b^{p^n}$. Also $(a-b)^{p^n} = a^{p^n} + (-b)^{p^n}$. When p = 2 we have that $(-b)^{p^n} = b^{p^n}$, otherwise $(-b)^{p^n} =-b^{p^n}$. This completes the proof.

12) An element of a ring is nilpotent if $a^n=0$ for some n. Prove that in a commutative ring a+b is nilpotent if a and b are.
Proof-
Let a and b be two nilpotent elements such that $a^r = b^s = 0$ for some r,s. Take $n = r + s$ and consider $(a+b)^n= \sum_{k=0}^n \dbinom{n^k}{k} a^{n-k}b^k$. We have two cases. If $k \geq s$ then $b^k = 0$. If $k\leq s$ then $s-k \geq 0 \Rightarrow n - k = r + s-k \geq r \Rightarrow a^{n-k} = 0$. Thus $(a+b)^n = 0$ so that a + b is nilpotent.

13) In a ring R, the following are equivalent.
(a) R has no nonzero nilpotent elements
(b) If $a\in R$ and $a^2 = 0$ then $a = 0$.
Proof-
$( \Rightarrow )$ Assume that R has no nonzero nilpotent elements. If $a\in R$ and $a^2 = 0$ then a is nilpotent, but by assumption there are no nonzero nilpotent elements so that a must be 0.
$(\Leftarrow )$ Assume that if $a\in R$ and $a^2=0$ then $a= 0$. Want to show that there are no nonzero nilpotent elements. We do this by showing that $a^n = 0 \Rightarrow a=0 \forall n \in \mathbb{N}$ using induction. If n is one then it is trivially true. Assume that $a^k = 0 \Rightarrow a = 0$, for some $k \geq 1$. Then if $a^{k+1} = 0 \Rightarrow a^{k+1}a^{k-1} = 0 a^{k-1} \Rightarrow a^{2k} = 0\Rightarrow (a^k)^2 = 0 \Rightarrow a^k = 0 \Rightarrow a = 0$ by assumption and by inductive hypothesis. Thus any element which is nilpotent is zero, so the claim is true.

14) Let R be a commutative ring with identity and prime characteristic p. The map $R \rightarrow R$ given by $r \mapsto r^p$ is a homomorphism of rings called the Frobenius homomorphism.
Proof-
Call the map f. Let $a,b \in R$. Then to check that f is a ring homomorphism we note that $f(a+b) = (a+b)^p = a^p + b^p = f(a) + f(b)$ by exercise 11. Also, $f(ab) = (ab)^p = a^pb^p = f(a)f(b)$.

15)(a) Give an example of a nonzero homomorphism $f: R \rightarrow S$ of rings with identity such that $f(1_R) \neq 1_S$.
(b) If $f:R \rightarrow S$ is an epimorphism then $f(1_r) = 1_S$.
(c) If $f:R \rightarrow S$ is a homomorphism of rings with identity and u is a unit in R such that $f(u)$ is a unit in S, then $f(1_R) = 1_S$ and $f(u^{-1}) = f(u)^{-1}$. Give an example to show that $f(u)$ need not be a unit in S even though u is a unit in R.
Proof-
a) Consider $f: \mathbb{Z}_3 \rightarrow \mathbb{Z}_6$ by $f(k) = 4k$. Then for $a,b\in \mathbb{Z}$ we have $f(a+b) = 4(a+b) = 4a +4b = f(a) + f(b)$ and $f(a)f(b) = 4a4b = 16ab = 4ab \pmod{6} = f(ab)$ but note that $f(1) = 4 \neq 1$ in $\mathbb{Z}_6$.

b) Let $s\in S$. Since f is surjective we have that $s = f(r)$ for some $r\in R$. Then $f(1_R) s = f(1_r) f (r) = f(r) = s$ and similarly $sf(1_R) = s$. Since the identity element is unique and $f(1_R)$ is acting as the identity we must have that $f(1_R) = 1_S$.

c) First note that $f(u) = f(1_R u ) = f(1_R) f(u)$. Then since $f(u)$ is given to be a unit we can multiply both sides by the inverse so that $1_s = f(1_R)$. Then using this fact we have $1_S = f(1_R) = f(uu^{-1}) = f(u) f(u^{-1})$ and similarly $1_S = f(u^{-1}) f(u)$. Then since inverses are unique and $f(u^{-1})$ is acting as the inverse of $f(u)$ we must have that $f(u^{-1}) = f(u)^{-1}$. Lastly, use the same map as in part a) for the counterexample and note that 2 is a unit $\mathbb{Z}_3$, but $f(2)=2$ is not a unite in $\mathbb{Z}_6$.

16) Let $f: R \rightarrow S$ be a homomorphism of rings such that $f(r) \neq 0$ for some nonzero $r\in R$. If R has identity and S has no zero divisors, then S is a ring with identity $f(1_R)$.
Proof-
Let $r\in R$ be nonzero such that $f(r) \neq 0$. Then we have $0 = f(r)s-f(r)s = f(r1_R)s - f(r)s = f(r)f(1_R)s-f(r)s = f(r)(f(1_R)s-s)$. Then since S has no zero divisors and it was assumed that $f(r) \neq 0$ we must have that $f(1_R)s -s = 0 \Rightarrow f(1_R)s = s$. Similarly we can get that $s f(1_R) = s$ which means that $f(1_R)$ must be the multiplicative identity element in S, $1_S$.

18) Let $\mathbb{Q}$be the field of rational numbers and R any ring. If $f,g : \mathbb{Q} \rightarrow R$ are ring homomorphisms such that $f|_{\mathbb{Z}} = g|_{\mathbb{Z}}$, then f = g.
Proof –
Let x be an arbitrary rational number so that we can write $x = \frac{m}{n}$ where m and n are integers and n is nonzero. Then note that $g(1) = f(1) = f(\frac{1}{n} n) = f(\frac{1}{n} ) f( n) = f(\frac{1}{n} ) g( n)$. Then also $g(\frac{1}{n} ) = g(1 \frac{1}{n} ) = g(1) g( \frac{1}{n} ) =f(\frac{1}{n} ) g( n) g( \frac{1}{n} ) = f(\frac{1}{n} ).$ Using this finally we have that $f(x) = f(\frac{m}{n}) = mf(\frac{1}{n}) = mg(\frac{1}{n}) = g(\frac{m}{n}) = g(x)$. Thus f and g agree on the rationals.

## Algebra Qual Study Problems: 3

Example 1: How many non isomorphic abelian groups are there of order 1500?
Proof-
We have that $|G| = 1500 = 2^2 \cdot 3 \cdot 5^3$. Using this we can list all families of elementary divisors, which are : $\{2^2,3,5^3\},\{2,2,3,5^3\},\{2^2,3,5,5^2\},\{2^2,3,5,5,5\},\{2,2,3,5,5^2\},\{2,2,3,5,5,5\}$. This gives that there are exactly 6 non isomorphic abelian groups of order 1500 since each set corresponds to a unique (up to isomorphism) abelian group of order 1500.

Example 2: Find the invariant factors of $G = \mathbb{Z}_5 \bigoplus \mathbb{Z}_{15} \bigoplus \mathbb{Z}_{25} \bigoplus \mathbb{Z}_{36} \bigoplus \mathbb{Z}_{54}$.
Proof-
First we rewrite in terms of elementary divisors:
$G \cong \mathbb{Z}_5 \bigoplus ( \mathbb{Z}_{5} \bigoplus \mathbb{Z}_{3} ) \bigoplus \mathbb{Z}_{5^2} \bigoplus ( \mathbb{Z}_{2^2} \bigoplus \mathbb{Z}_{3^2} )\bigoplus ( \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3^3} )$. Thus we have the elementary divisors being $\{2,2^2,3,3^2,3^3,5,5,5^2\}$. Ordering them in the correct matrix formation we get:
$2^0,3,5$
$2^1,3^2,5$
$2^2,3^3,5^2$
Multiplying across gives us the invariant factors, which are $15,90,2700$.

Hungerford 2.2.12:(a) What are the elementary divisors of the group $\mathbb{Z}_2 \bigoplus \mathbb{Z}_9 \bigoplus \mathbb{Z}_{35}$. Do the same for $\mathbb{Z}_{26} \bigoplus \mathbb{Z}_{42} \bigoplus \mathbb{Z}_{49} \bigoplus \mathbb{Z}_{200} \bigoplus \mathbb{Z}_{1000}$.
Proof-
For the group $\mathbb{Z}_2 \bigoplus \mathbb{Z}_9 \bigoplus \mathbb{Z}_{35}$ we have that $2 = 2, 9 = 3^2, 35 = 5 \cdot 7$. Thus the elementary divisors are $2, 3^2,5,7$. To get the invariant factor we need only multiply the elementary divisors together because they are all relatively prime. Therefore the invariant factor is $2 \cdot 3^2 \cdot 5 \cdot 7 = 630$. Then we can say finally that the group $\mathbb{Z}_2 \bigoplus \mathbb{Z}_9 \bigoplus \mathbb{Z}_{35} \cong \mathbb{Z}_2 \bigoplus \mathbb{Z}_{3^2} \bigoplus \mathbb{Z}_{5} \bigoplus \mathbb{Z}_{7} \cong \mathbb{Z}_{630}$.
In a similar manner we have the elementary divisors for $\mathbb{Z}_{26} \bigoplus \mathbb{Z}_{42} \bigoplus \mathbb{Z}_{49} \bigoplus \mathbb{Z}_{200} \bigoplus \mathbb{Z}_{1000}$ are $2,2,2^3,2^3,3,5^2,5^3,7,7^2,13$. Because we have 5 unique primes and 2 appearing the most amount of times, 4, we arrange the elementary factors into 4 rows and 5 columns ordering from least to greatest filling in with 0 powers where necessary:
$2^1 \cdot 3^0 \cdot 5^0 \cdot 7^0 \cdot 13^0$
$2^1 \cdot 3^0 \cdot 5^0 \cdot 7^0 \cdot 13^0$
$2^3 \cdot 3^0 \cdot 5^2 \cdot 7^1 \cdot 13^0$
$2^3 \cdot 3^1 \cdot 5^3 \cdot 7^2 \cdot 13^1$
Lastly, multiply across to get the invariant factors of $2,2,1400,1911000$. Thus we finally have:
$\mathbb{Z}_{26} \bigoplus \mathbb{Z}_{42} \bigoplus \mathbb{Z}_{49} \bigoplus \mathbb{Z}_{200} \bigoplus \mathbb{Z}_{1000} \cong$ $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{13} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{7} \bigoplus \mathbb{Z}_{7^2} \bigoplus \mathbb{Z}_{2^3} \bigoplus \mathbb{Z}_{5^2} \bigoplus \mathbb{Z}_{5^3} \cong$ $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{1400} \bigoplus \mathbb{Z}_{1911000}$.

(b) Determine up to isomorphism all abelian groups of order 64 and 96.
Proof- $64 = 2^6$ so we have the following abelian groups. $\mathbb{Z}_{64}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{32}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{16}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{16}, \mathbb{Z}_{8} \bigoplus \mathbb{Z}_{8}$, $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4}$, $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4}, \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{4}$, $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2}$.
Next we have that $96 = 2^5 \cdot 3$ so that any abelian group of that order is isomorphic to one of the following:
$\mathbb{Z}_{32} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{2} \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{16}, \mathbb{Z}_{3} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{8}, \mathbb{Z}_{8} \bigoplus \mathbb{Z}_{2}$, $\mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{4} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}$, $\mathbb{Z}_{4} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}, \mathbb{Z}_{2} \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{2} \bigoplus \mathbb{Z}_{3}$

Hungerford 2.2.9: How many subgroups of order $p^2$ does the abelian group $\mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}$ have?
Proof-
By the classification theorem we know that if $|G| = p^2$ then $G \cong \mathbb{Z}_p \bigoplus \mathbb{Z}_p$ or $\mathbb{Z}_{p^2}$. So we first need to know how many unique subgroups of $\mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}$ there are which are isomorphic to $\mathbb{Z}_p \bigoplus \mathbb{Z}_p$. Consider the elements in $\mathbb{Z}_{p^2}$
$1,2,3 \ldots ,p ,p+1, ,p+2,p+3, \ldots ,2p, 2p+1, \ldots ,(p-1)p , (p-1)p+1, (p-1)p+2, \ldots p \cdot p.$ Now we need to know which of these has order p. All elements that are multiples of p other than the identity, have order p. Consider an element $np$, where $1\leq n. Then $(np) \cdot p = np^2 = 0$. If there was some $m such that $npm = 0$ we would have that $npm = kp^2 \Rightarrow nm = kp \Rightarrow p | nm$ which is a contradiction because $gcd(p,n)=gcd(p,m) = 1$. There are $p-1$ multiples of p and so there are $p-1$ elements of order p in $\mathbb{Z}_{p^2}$.
Next we find the number of elements of order p in $\mathbb{Z}_{p^3}$. In a similar manner we get that all multiples of $p^2$ have order p and that there are $p-1$ such elements in $\mathbb{Z}_{p^3}$.
Lastly we need how many elements there are of the form $(x,y) \in \mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}$ which have order p. Note that $|(x,y)| = p \text{ iff } lcm(|x|,|y|) = p$. This can only happen in a few cases. If either order is 1, then the other must be p, or they can both be p. If $|x|= p$, then x can be $p-1$ elements. In order for the least common multiple to be p we need that $|y| = 1$ or $|y| = p-1$. So each of the p-1 elements that x can be can be paired with the identity of y, giving p-1 choices. Then each of the p-1 elements that x can be can be paired with the p-1 choices of y, giving $(p-1)(p-1)$ choices. Lastly if x is the identity it can be paired with p-1 more choices from y. This gives a total of $p-1+(p-1)(p-1) + p-1 = (p-1)( 1 + p-1 + 1) = (p-1)(p+1) = p^2-1$ elements of order p in $\mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}$. Note that $\mathbb{Z}_p \bigoplus \mathbb{Z}_p$ has order $p^2$ and has $p^2-1$ elements of order p so that there can only be one subgroup of $\mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}$. There are not enough elements of order p to make up another subgroup.
Next we need to know how many unique subgroups there are of $\mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}$ which are isomorphic to $\mathbb{Z}_{p^2}$. To do this we need a lemma: If there are n elements of order d in a group G, then there are $\frac{n}{\phi (d)}$ cyclic subgroups of order d, where $\phi$ is the Euler Phi function, which counts the number of positive integers relatively prime and less than the input number. Thus by the lemma it is enough to count the number of elements of order $p^2$ in $\mathbb{Z}_{p^3} \bigoplus \mathbb{Z}_{p^2}$. Again an element $(x,y)$ has order $p^2$ iff $lcm(|x|,|y|) = p^2$. This can only happen if either $|x| = p^2$ or $|y| = p^2$. After an annoying counting argument we get that there are $(p^2-p)(p^2+p)$ such elements so that by the lemma we have $\frac{(p^2-p)(p^2+p)}{\phi (p^2)} = p^2 +p$ subgroups of type $\mathbb{Z}_{p^2}$. Thus in total there are $p^2 + p + 1$ subgroups of order $p^2$.

Midterm #5: G an abelian group of order 252. Find all possible primes such that G can contain a subgroup isomorphic to \mathbb{Z}_p \bigoplus \mathbb{Z}_p$for some prime p. Give an example where it doesn’t. Proof- By the classification theorem we can find all such groups G by listing out possible elementary divisors. $|G| = 252 = 2^2 \cdot 3^2 \cdot 7$ which gives the possible elementary divisors to be: $\{2^2,3^2,7\},\{2,2,3^2,7\},\{2^2,3,3,7\},\{2,2,3,3,7\}$. Writing them out in the correct way we can get invariant factors. $\{2^2,3^2,7\} \rightarrow \{252\}$ $\{2,2,3^2,7\} \rightarrow \{2,126\}$ $\{2^2,3,3,7\} \rightarrow \{3,84\}$ $\{2,2,3,3,7\} \rightarrow \{6,42\}$ From this we can see the only possible primes are 2 and 3. As a counterexample, $\mathbb{Z}_{252}$ is cyclic and thus every subgroup is cyclic. Therefore it cannot contain a subgroup isomorphic to $\mathbb{Z}_p \bigoplus \mathbb{Z}_p$ because that group is not cyclic. Hungerford 2.2.1: Show that a finite abelian group that is not cyclic contains a subgroup which is isomorphic to $\mathbb{Z} \bigoplus \mathbb{Z}$ for some prime p. Proof- Let G be a finite abelian group. Then we have$G \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \mathbb{Z}_{p_2^{k_2}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{k_r}}$ where $p_i$ is prime and $k_i$ is a positive integer. If every prime is distinct, then all the powers of the primes are relatively prime so that $G \cong \mathbb{Z}_{p_1^{k_2}p_1^{k_2}\cdots p_r^{k_r}}$, which is cyclic. So instead we must have some case of $p_i = p_j$. WLOG suppose that $p_1 = p_2$ and call them both p. Then consider that $\mathbb{Z}_{p^{k_i}} = \langle 1 \rangle$ where $i = 1,2$. Note that $p^{k_i-1} \in \mathbb{Z}_{p^{k_i}}$ so that $\langle p^{k_i-1} \rangle$ is a cyclic subgroup of $\mathbb{Z}_{p^{k_i}}$. Since $pp^{k_i-1} = 0$ in $\mathbb{Z}_{p^{k_i}}$ the order of $p^{k_i-1}$ must divide p. This is only possible if $| p^{k_i-1}| = p$ which means $\langle p^{k_i-1} \rangle$ is a cyclic subgroup of $\mathbb{Z}_{p^{k_i}}$ of order p, and thus isomorphic to $\mathbb{Z}_p$. Then we have that $\langle p^{k_1-1} \rangle \bigoplus \langle p^{k_2-1} \rangle \bigoplus \{0\} \bigoplus \cdots \bigoplus \{0\}$ is a cyclic subgroup of $G \cong \mathbb{Z}_{p_1^{k_1}} \bigoplus \mathbb{Z}_{p_2^{k_2}} \bigoplus \cdots \bigoplus \mathbb{Z}_{p_r^{k_r}}$. Theorem 2.4.3: If a group G acts on a set S then $| \bar{x} | = [G:G_x]$, where $\bar{x}$ is the orbit of x, and $G_x$ is the stabilizer subgroup of x. Proof- We can exhibit a bijection between the two sets which gives the cardinality to be the same. We have that, $gG_x = hG_x \Leftrightarrow h^{-1}gG_x = G_x \Leftrightarrow h^{-1}g \in G_x \Leftrightarrow h^{-1}gx = x \Leftrightarrow gx = hx$. This tells us that the association: $gG_x \rightarrow gx$ is injective, and it is obviously surjective, so that the cardinalities are equal. Hungerford 2.4.3: If a group G contains an element a having exactly 2 conjugates, then G has a proper normal nontrivial subgroup N. Proof- Let G be a group and let $a\in G$ have exactly 2 conjugates. That is, there are 2 elements in the conjugacy class (orbit) $\bar{a} = \{gag^{-1} | g\in G\}$ of a. By theorem 4.3, $|\bar{a} | = 2 = | \{gag^{-1} | g\in G\}| = [G:G_a]$, but by previous theorem this gives that $G_a$ is normal in G. If $G_a$ were to be trivial then that would mean that G has 2 elements so is the same as $\mathbb{Z}_2$ and so is abelian. But in an abelian group, every conjugacy class has 1 element so this is a contradiction. Thus $G_a = N$ is nontrivial and normal in G. October 2004 Qual #5 (Corollary 2.4.9): Let G be a finite simple group having a subgroup of index n. Show that G is isomorphic to a subgroup of $S_n$. Proof- Let H be a subgroup of G with index n and define the group action of G on the set of left cosets: $G \times G/H \rightarrow G/H$ by $(g_1,g_2H)\mapsto (g_1g_2)H$. This is clearly a group action. Then we have the induced homomorphism: $\phi G \rightarrow A(G/H)$ given by $g\mapsto \tau_g$, where $\tau_g :G/H \rightarrow G/H$ by $\tau_g (xH) = gxH$. Now suppose that $g\in Ker(\phi ).$ Then we have $\tau_g = id_{G/H} \Rightarrow \tau_g (xH) = xH = gxH \forall x\in G$. In particular we have that it is true for $x= e$, so that $geH = eH = H \Rightarrow g \in H$. This tells us that $Ker(\phi ) \leq H$. Note t hat G is simple, so that the Kernel must be trivial since it is a subset of H. Now by the first isomorphism theorem we have $G/Ker(\phi ) \cong Im( \phi ) \leq A(G/H)$. Note that since $|G/H| = n$ we have that $A(G/H) \cong S_n$. Thus $G/Ker(\phi ) \cong G/\{e\} \cong G \cong Im(\phi ) \leq A(G/H) \cong S_n$. April 2004 #2: Let G be a group. If Aut(G) is cyclic then G is abelian. Proof- First recall that the inner automorphisms of G, denoted $Inn(G) = \{\tau_g :G \rightarrow G | \tau_g (h ) = ghg^{-1}\}$ and that $Inn(G) \leq Aut(G)$ so that $Inn(G)$ is also cyclic. If G acts on G by conjugation, then the induced homomorphism $\phi : G \rightarrow Aut(G)$ has $Ker(\phi ) = C(G)$ and $Im(\phi ) = Inn(G)$ so that $G/C(G) \cong Inn(G)$. This gives that $G/C(G)$ is cyclic which tells us that G is abelian. 2011 Qual #5: Suppose that H acts on a set S. Given $s\in S$ define the H-orbit of s. Prove that if $s'\in S$ is another element then either $s'$ is in the orbit of s or that the orbits of s and s’ are disjoint. Proof- Suppose that $s'$ is not in the orbit of s, and suppose there is some $x\in \bar{s'} \cap \bar{s}$. Then we can write $x= h_1s'=h_2 \Rightarrow s' = h_1^{-1}h_2s$. This tells us that $s'$ is in the orbit of s which is a contradiction. 2013 Qual #2(a) (Hungerford 2.4.6): Consider $S_n$ acting on the set $S = \{1,2,\ldots , n\}$ by permutations. What is the orbit of i? What is the stabilizer of i? Prove that the stabilizer subgroup of any two elements of S are conjugate. Proof- Orbit of i: $\bar{i} = \{ \psi i | \psi \in S_n\} = S$ Stabilizer of i: Claim that $G_i \cong S_{n-1}$. First note that $[S_n : G_i] = |\bar{i} | = n \Rightarrow |G_i| = (n-1)!$, which comes from the fact that $|S_n| = [S_n : G_i] |G_i|$. Consider $S_{n-1}$ as the set of permutations on $S\backslash \{i\}$. Then clearly these permutations fix i and are a subgroup isomorphic to $S_{n-1}$. Next, we show that $G_i$ is conjugate to $G_j$. We know that for any i,j that exists $(ij) \in S_n$ such that $(ij) i = j$ \Rightarrow (ji) j = i$, since the inverse of (ij) is (ji). Claim that $(ji)G_j(ij) = G_i$. Let $\tau \in G_i$. Then $\tau i = i$. Note that $\tau = (ji) ( (ij) \tau (ji) ) (ij)$, while $(ij) \tau (ji) \in G_j$ because $(ij) \tau (ji) j = (ij) \tau i = (ij) i = j$. This gives that $\tau \in (ji)G_j(ij)$. Thus $G_i \subset (ji)G_j(ij)$. Now let $\tau \in (ji)G_j(ij)$. Then we can write $\tau (ji) \psi (ij)$ for some $\psi \in G_j$. Then $\tau i = (ji) \psi (ij) i = (ji ) \psi j = (ji) j = i$. This tells us that $\tau \in G_i$. Thus we have subset inclusion so that the two sets are conjugate.

## Algebra Solutions: Hungerford II.5

1) If $N \triangleleft G$ and N, $G/N$ are both p-groups, then G is a p-group.
Proof-
Since N and $G/N$ are p-groups let $|N| = p^i, |G/N| = p^j$ for some i and j. Then by Lagrange we have $|G| = [G:N]|N| = |G/N| |N| = p^ip^j = p^{i+j}$. Thus G is a p-group.

2) If G is a finite p-group, $H\triangleleft G$ and H is nontrivial, then $H\cap C(G) \neq \{e\}.$
Proof-
Note that the center of G is $C(G) = \{ g\in G| gx = xg \forall x \in G\}$. Let G be a finite p-group so that $|G| = p^n$. Then since H is a subgroup we have $|H| = p^m$ where $m \leq n$. We can let G act on H by conjugation and let $H_0 = \{h\in H | ghg^{-1} = h \forall g \in G\}.$ Then we have by theorem that $|H|\equiv |H_0| \pmod{p}$. We can also write $H_0 = \{ h\in H | gh = hg \forall g\in G\}$. Now if $a\in H_0 \Rightarrow a\in H$, and $ga = ag \forall g\in G \Rightarrow a\in H \cap C(G)$. Therefore $H_0 \subset H \cap C(G)$. We also have that p divides the order of $H_0$ so that $1 < |H_0| \leq |H \cap C(G)|$. Thus $H\cap C(G)$ must have more than one element and is therefore nontrivial.

3) Let $|G| = p^n$. For each $k, 0 \leq k\leq n,$ G has a normal subgroup of order $p^k$.
Proof-
First note that any subgroup of the center of a group is automatically normal in the group. Now we prove that G has a normal subgroup of order $p$. Since G is a p-group it has a nontrivial center so that the center is in fact a p-group as well. Then being of order a prime power, we can say by Cauchy’s Theorem that there exists $g \in C(G)$ such that $|g| = p$. Then $\langle g \rangle$ is a subgroup of the center of G, so is automatically normal in G, and has order p.
Now we show for each $k, 0 \leq k\leq n,$ G has a normal subgroup of order $p^k$ by induction on k. If $k = 0$ we have the trivial subgroup which is automatically normal in G. Now assume that G has normal subgroups for each $p^i$ where $0 \leq i \leq k$ and call them $N_i$. Consider the quotient group $G/N_k$. We have from Lagrange that $|G/N_k| = p^{n-k}$ so that $G/N_k$ is a p-group. In the previous step we’ve shown that a p-group must contain a normal subgroup of order p, call it $\langle aN_k \rangle$. Let $\pi : G \rightarrow G/N_k$ be the canonical projection map. Then let $H = \pi^{-1} (\langle aN_k \rangle ) = \{ g\in G| gN_k \in \langle aN_k \rangle \}$. The preimage of a normal subgroup under a homomorphism is also normal, so that H is normal in G. Note that $N_k \leq H$ because $N_k$ is the identity in $G/ N_k$ so that $N_k \in \langle aN_k \rangle$ so that $\pi^{-1} (\{N_k \}) = N_k \leq \pi^{-1}(\langle aN_k \rangle )= H$. Then using Lagrange we have $|H| = [H:N_k] |N_k|,$ but $H/N_k = \langle aN_k \rangle \Rightarrow |H/N_k| = |\langle aN_k \rangle | = p$. Therefore $|H| = p \cdot p^{k} = p^{k+1}$

4) If G is an infinite p-group, then either G has a subgroup of order $p^n$ for each $n\geq 1$ or there exists $m\in \mathbb{N}$ such that every finite subgroup of G has order $\leq p^m$.
Proof-
Suppose that no such m exists. Then for each positive integer n, there exists another positive integer $N>n$ such that G has a subgroup H of order $p^N$. But then by Sylow we know that H has a subgroup of order $p^n$.

5) If P is a normal Sylow p-subgroup of a finite group G and $f: P \rightarrow P$ is an endomorphism, then $f(P) \leq P$.
Proof-
Note that $f(P) \leq G$. Let $y\in f(P)$. Then there exists a $x\in P$ such that $f(x)= y$. Since P is a p-group, $|x| = p^i$ for some i.Then we have $y^{p^i} = (f(x))^{p^i}) = f(x^{p^i}) = f(e) = e$. This tells us that $|y|$ divides $p^i$ so that $|y| = p^j$ for some $j\leq i$. Since y was arbitrary, $f(P)$ is a p-subgroup of G. Therefore by Sylow’s Second Theorem, there exists $x\in G$ such that $f(P) \leq xPx^{-1} = P$.

6) If H is a normal subgroup of order $p^k$ of a finite group G, then H is contained in every Sylow p-subgroup of G.
Proof-
H is a p-subgroup of a finite group G. Let P be any Sylow p-subgroup of G. From Sylow’s Second Theorem we know there exists $x\in G$ such that $H \leq xPx^{-1}$. We also have that $x^{-1}Hx$ is nonempty. Let $y\in x^{-1}Hx$. Then we can write $y = x^{-1}hx$ for some $h\in H$. Since $H\leq xPx^{-1}$ for all $h\in H$, we can write $h = xpx^{-1}$ for some $p\in P$. Then rewrite $y = x^{-1}xpx^{-1}x = p\in P$. Therefore $x^{-1}Hx \subset P$. Lastly, let $a,b \in x^{-1}Hx$. Then we can write $a=x^{-1}h_1x, b = x^{-1}h_2x$ for some $h_1,h_2 \in H$. Then we have $ab^{-1} = x^{-1}h_1xx^{-1}h_2^{-1}x = x^{-1}h_1h_2^{-1}x \in x^{-1}Hx$ so that in fact $x^{-1}Hx$ is a subgroup of P. Now, since H is normal in G we can say that $H = xHx^{-1} = x^{-1}Hx \leq P$, so that H is contained in any arbitrary Sylow p-subgroup P.

7) Find the Sylow 2-subgroups and Sylow 3-subgroups of $S_3,S_4,S_5.$
Proof-
$|S_3| = 6 = 2 \cdot 3$. Thus $S_3$ contains at least one Sylow 2 subgroup of order 2 and one Sylow 3 subgroup of order 3 by the first Sylow Theorem. In fact, from Sylow’s Third Theorem we know that there is either 1 Sylow 2 subgroup or 3 Sylow 2 subgroups. We also know that there can only be one Sylow 3 subgroup.
In the same manner we can say that $S_4$ has either 1 or 3 Sylow 2 subgroups of order 8 and either 1 or 4 Sylow 3-subgroups of order 3. This is because $|S_4| = 12 = 2^3 \cdot 3$.
Lastly, $S_5$ has either 1,3,4, or 15 Sylow 2 subgroups of order 8 and either 1,4, or 10 Sylow 3-subgroups of order 3. This is because $|S_5| = 120 = 2^3 \cdot 3 \cdot 5$.

8) If every Sylow p-subgroup of a finite group G is normal for every prime p, then G is the direct product of its Sylow subgroups.
Proof-
First we write $|G| = p_1^{r_1}p_2^{r_2}\cdots p_1^{r_n}$. where we have distinct primes. Then by Sylows first theorem, for each prime $p_i$, G contains a Sylow $p_i$ subgroup of order $p_i^{r_i}$. From the second Sylow theorem we have that any two Sylow $p_i$ subgroups are conjugate, but we are given that they are all normal so in fact there is a unique Sylow $p_i$ subgroup for each $p_i$. Denote each Sylow $p_i$ subgroup by $H_i$. Then we have that $\{H_i\}_{i=1}^n$ is a family of normal subgroups of G so that $\langle \cup_{i=1}^m H_i \rangle = H_1 \cdots H_m$ for any $1\leq m\leq n$.
Next we need to show that for each $j\in \{1,\ldots ,n\}$ that $H_j \cap \langle \cup_{i\neq j} \rangle = \{e\}$, for all $n\in \mathbb{N}$. This is done by induction on n. Lastly we have to show that $G = \langle \cup_{i=1}^n H_i \rangle$, but we already have that $H_1 \cdots H_n = \langle \cup_{i=1}^n H_i \rangle \leq G$ so that we need only show that $| H_1 \cdots H_n| = |G|$. Then we will have that $G \cong H_1 \times \cdots \times H_n$.

9) If $|G| = p^nq$ with $p>q$ primes, then G contains a unique normal subgroup of index q.
Proof-
From Sylow’s First Theorem, G has a Sylow p-subgroup of order $p^n$. Let m be the number of Sylow p subgroups of G. By Sylows Third Theorem we have that $m | |G|$ and $m\equiv 1 \pmod{p}$. Then since $|G| = p^nq$ we have to have that either $m=1,q$ or $p^i$ for some $1\leq i \leq n$. If $m = q$ then $p|(q-1) \Rightarrow q-1 = pk$ for some $k\in \mathbb{Z}$. Then $pk+1 = q < p \Rightarrow k = 0 \Rightarrow q =1$ which is a contradiction. If instead we have that $m = p^i$ then $p | (p^i-1) \Rightarrow p = 1$, which is another contradiction. Therefore $m =1$ so that there is the only 1 Sylow p subgroup of G, call it H, which must be normal in G. Then by Lagrange, $|G| = [G:H]|H| \Rightarrow p^nq = [G:H] p^k \Rightarrow q = [G:H]$. Thus H has index q.

10) Every group of order 12, 28, 56, and 200 must contain a normal Sylow subgroup and hence is not simple.
Proof- i)
$|G| = 12 = 2^2\cdot 3$. By 3rd Sylow Theorem we know that there is either 1 Sylow 3 subgroup, or 4 Sylow 3 subgroups. If there is only one, then know it must be normal. Thus, suppose there are 4, $H_1,H_2,H_3,H_4$ each with order 3. Each of these subgroups has 3 elements, 1 being the identity. Thus the other 2 elements are of order 3. Together then we have $4(3-1) = 8$ elements of order 3 in G. Then G can only have one Sylow 2-subgroup which must be of order 4 because this gives us 12 total elements. Again, being the unique Sylow 2-subgroup gives that it is also normal.

ii) $|G| = 28 = 2^2 \cdot 7$. The Third Sylow Theorem tells us that the number of Sylow 7 subgroups is either 1 or 4, and that it must also be 1 mod 7, therefore it cannot be 4. So there is only one Sylow 7-subgroup and thus it is normal.

iii) $|G| = 56 = 2^3 \cdot 7$. By the Third Sylow Theorem we know that the number of Sylow 7-subgroups is either 1 or 8. If it is 1 then we are done, so suppose it is 8. Call them $H_1, H_2,\ldots , H_8$. Then similar to part (i) we have that there are $8(7-1) = 48$ elements of order 7 in G. Then G can only have one Sylow 2-subgroup as it must be of order $2^3 = 8$. Therefore it is normal, being unique.

iv) $|G| = 200 = 2^35^2$. By the Third Sylow Theorem we have that the number of Sylow 5-subgroups is a factor of $2^3$ or equal to 1, and must also be congruent to 1 modulo 5. This can only happen if there is exactly one Sylow 5-subgroup. Thus it is normal.

11) How many elements of order 7 are there in a simple group of order 168?
Proof-
Let $|G| = 168 = 2^3\cdot 3 \cdot 7$. By the Third Sylow Theorem, the number of Sylow 7-subgroups must divide $2^3\cdot 3$ and also must be congruent to 1 modulo 7. Thus the only possibilities are that the number of Sylow 7-subgroups is 1 or 8, but since G is simple we cannot have only one Sylow 7-subgroup otherwise it would be normal. Therefore there are 8 of them, each with 7 elements, with each non identity element having order 7 since a group with prime order must be cyclic. Thus there are $8(7-1)=48$ elements of order 7.

13) Every group G of order $p^2$ is abelian.
Proof-
Since G is a p-group, the center $C(G) \neq \{e\}$. By Lagrange we then have either $|C(G)| = p^2$ which gives the center as the entire group, thus making G abelian, or we have $|C(G)| = p$. If $|C(G)| = p$ then we have that $G/C(G)$ is cyclic being of prime order, which tells us that G is abelian by 2.4.9.