## 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.