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.

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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<p 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}\}.
(2) Is 7 irreducible? Prove your answer.
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}.

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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 eachc_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} <x^2 = \phi (x). 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.

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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.

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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 havea-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.

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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<p. Then (np) \cdot p = np^2 = 0. If there was some m <n 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 haveG \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.

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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.

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