Annihilators
Fall 2021 #6 #algebra/qual/work
Let \(R\) be a commutative ring with unit and let \(M\) be an \(R\)module. Define the annihilator of \(M\) to be \begin{align*} \operatorname{Ann}(M):=\{r \in R \mathrel{\Big}r \cdot m=0 \text { for all } m \in M\} \end{align*}

Prove that \(\operatorname{Ann}(M)\) is an ideal in \(R\).

Conversely, prove that every ideal in \(R\) is the annihilator of some \(R\)module.
 Give an example of a module \(M\) over a ring \(R\) such that each element \(m \in M\) has a nontrivial annihilator \(\operatorname{Ann}(m):=\{r \in R \mathrel{\Big}r \cdot m=0\}\), but \(\operatorname{Ann}(M)=\{0\}\)
Spring 2017 #5 #algebra/qual/work
Let \(R\) be an integral domain and let \(M\) be a nonzero torsion \(R{\hbox{}}\)module.

Prove that if \(M\) is finitely generated then the annihilator in \(R\) of \(M\) is nonzero.

Give an example of a nonfinitely generated torsion \(R{\hbox{}}\)module whose annihilator is \((0)\), and justify your answer.
Torsion and the Structure Theorem
\(\star\) Fall 2019 #5 #algebra/qual/completed
Let \(R\) be a ring and \(M\) an \(R{\hbox{}}\)module.
Recall that the set of torsion elements in M is defined by \begin{align*} \operatorname{Tor}(M) = \{m \in M {~\mathrel{\Big\vert}~}\exists r \in R, ~r \neq 0, ~rm = 0\} .\end{align*}

Prove that if \(R\) is an integral domain, then \(\operatorname{Tor}(M )\) is a submodule of \(M\) .

Give an example where \(\operatorname{Tor}(M )\) is not a submodule of \(M\).
 If \(R\) has zerodivisors, prove that every nonzero \(R{\hbox{}}\)module has nonzero torsion elements.
 Onestep submodule test.
It suffices to show that \begin{align*} r\in R, ~t_1, t_2\in \operatorname{Tor}(M) \implies rt_1 + t_2 \in \operatorname{Tor}(M) .\end{align*}
We have \begin{align*} t_1 \in \operatorname{Tor}(M) &\implies \exists s_1 \neq 0 \text{ such that } s_1 t_1 = 0 \\ t_2 \in \operatorname{Tor}(M) &\implies \exists s_2 \neq 0 \text{ such that } s_2 t_2 = 0 .\end{align*}
Since \(R\) is an integral domain, \(s_1 s_2 \neq 0\). Then \begin{align*} s_1 s_2(rt_1 + t_2) &= s_1 s_2 r t_1 + s_1 s_2t_2 \\ &= s_2 r (s_1 t_1) + s_1 (s_2 t_2) \quad\text{since $R$ is commutative} \\ &= s_2 r(0) + s_1(0) \\ &= 0 .\end{align*}
Let \(R = {\mathbf{Z}}/6{\mathbf{Z}}\) as a \({\mathbf{Z}}/6{\mathbf{Z}}{\hbox{}}\)module, which is not an integral domain as a ring.
Then \([3]_6\curvearrowright[2]_6 = [0]_6\) and \([2]_6\curvearrowright[3]_6 = [0]_6\), but \([2]_6 + [3]_6 = [5]_6\), where 5 is coprime to 6, and thus \([n]_6\curvearrowright[5]_6 = [0] \implies [n]_6 = [0]_6\). So \([5]_6\) is not a torsion element.
So the set of torsion elements are not closed under addition, and thus not a submodule.
Suppose \(R\) has zero divisors \(a,b \neq 0\) where \(ab = 0\). Then for any \(m\in M\), we have \(b\curvearrowright m \coloneqq bm \in M\) as well, but then \begin{align*} a\curvearrowright bm = (ab)\curvearrowright m = 0\curvearrowright m = 0_M ,\end{align*} so \(m\) is a torsion element for any \(m\).
\(\star\) Spring 2019 #5 #algebra/qual/completed
Let \(R\) be an integral domain. Recall that if \(M\) is an \(R{\hbox{}}\)module, the rank of \(M\) is defined to be the maximum number of \(R{\hbox{}}\)linearly independent elements of \(M\) .

Prove that for any \(R{\hbox{}}\)module \(M\), the rank of \(\operatorname{Tor}(M )\) is \(0\).

Prove that the rank of \(M\) is equal to the rank of of \(M/\operatorname{Tor}(M )\).
 Suppose that \(M\) is a nonprincipal ideal of \(R\). Prove that \(M\) is torsionfree of rank 1 but not free.
 Suppose toward a contradiction \(\operatorname{Tor}(M)\) has rank \(n \geq 1\).
 Then \(\operatorname{Tor}(M)\) has a linearly independent generating set \(B = \left\{{\mathbf{r}_1, \cdots, \mathbf{r}_n}\right\}\), so in particular \begin{align*} \sum_{i=1}^n s_i \mathbf{r}_i = 0 \implies s_i = 0_R \,\forall i .\end{align*}
 Let \(\mathbf{r}\) be any of of these generating elements.
 Since \(\mathbf{r}\in \operatorname{Tor}(M)\), there exists an \(s\in R\setminus 0_R\) such that \(s\mathbf{r} = 0_M\).
 Then \(s\mathbf{r} = 0\) with \(s\neq 0\), so \(\left\{{\mathbf{r}}\right\} \subseteq B\) is not a linearly independent set, a contradiction.
 Let \(n = \operatorname{rank}M\), and let \(\mathcal B = \left\{{\mathbf{r}_i}\right\}_{i=1}^n \subseteq R\) be a generating set.
 Let \(\tilde M \coloneqq M/\operatorname{Tor}(M)\) and \(\pi: M \to M'\) be the canonical quotient map.
\begin{align*} \tilde {\mathcal{B}}\coloneqq\pi(\mathcal B) = \left\{{\mathbf{r}_i + \operatorname{Tor}(M)}\right\} \end{align*} is a basis for \(\tilde M\).
Note that the proof follows immediately.

Suppose that \begin{align*} \sum_{i=1}^n s_i (\mathbf{r}_i + \operatorname{Tor}(M)) = \mathbf{0}_{\tilde M} .\end{align*}

Then using the definition of coset addition/multiplication, we can write this as \begin{align*} \sum_{i=1}^n \qty { s_i \mathbf{r}_i + \operatorname{Tor}(M)} = \qty{ \sum_{i=1}^n s_i \mathbf{r}_i} + \operatorname{Tor}(M) = 0_{\tilde M} .\end{align*}

Since \(\tilde{\mathbf{x}} = 0 \in \tilde M \iff \tilde{\mathbf{x}} = \mathbf{x} + \operatorname{Tor}(M)\) where \(\mathbf{x} \in \operatorname{Tor}(M)\), this forces \(\sum s_i \mathbf{r}_i \in \operatorname{Tor}(M)\).

Then there exists a scalar \(\alpha\in R^{\bullet}\) such that \(\alpha \sum s_i \mathbf{r}_i = 0_M\).

Since \(R\) is an integral domain and \(\alpha \neq 0\), we must have \(\sum s_i \mathbf{r}_i = 0_M\).

Since \(\left\{{\mathbf{r}_i}\right\}\) was linearly independent in \(M\), we must have \(s_i = 0_R\) for all \(i\).

Write \(\pi(\mathcal B) = \left\{{\mathbf{r}_i + \operatorname{Tor}(M)}\right\}_{i=1}^n\) as a set of cosets.

Letting \(\mathbf{x} \in M'\) be arbitrary, we can write \(\mathbf{x} = \mathbf{m} + \operatorname{Tor}(M)\) for some \(\mathbf{m} \in M\) where \(\pi(\mathbf{m}) = \mathbf{x}\) by surjectivity of \(\pi\).

Since \(\mathcal B\) is a basis for \(M\), we have \(\mathbf{m} = \sum_{i=1}^n s_i \mathbf{r}_i\), and so \begin{align*} \mathbf{x} &= \pi(\mathbf{m}) \\ &\coloneqq\pi\qty{ \sum_{i=1}^n s_i \mathbf{r}_i} \\ &= \sum_{i=1}^n s_i \pi(\mathbf{r}_i) \quad\text{since $\pi$ is an $R{\hbox{}}$module morphism}\\ &\coloneqq\sum_{i=1}^n s_i \mathbf{(}\mathbf{r}_i + \operatorname{Tor}(M)) ,\end{align*} which expresses \(\mathbf{x}\) as a linear combination of elements in \(\mathcal B'\).
Notation: Let \(0_R\) denote \(0\in R\) regarded as a ring element, and \(\mathbf{0} \in R\) denoted \(0_R\) regarded as a module element (where \(R\) is regarded as an \(R{\hbox{}}\)module over itself)

Claim: If \(I\subseteq R\) is an ideal and a free \(R{\hbox{}}\)module, then \(I\) is principal .

Suppose \(I\) is free and let \(I = \left\langle{B}\right\rangle\) for some basis, we will show \({\left\lvert {B} \right\rvert} = 1\)>

Toward a contradiction, suppose \({\left\lvert {B} \right\rvert} \geq 2\) and let \(m_1, m_2\in B\).

Then since \(R\) is commutative, \(m_2 m_1  m_1 m_2 = 0\) and this yields a linear dependence

So \(B\) has only one element \(m\).

But then \(I = \left\langle{m}\right\rangle = R_m\) is cyclic as an \(R{\hbox{}}\) module and thus principal as an ideal of \(R\).

Now since \(M\) was assumed to not be principal, \(M\) is not free (using the contrapositive of the claim).


For any module, we can take an element \(\mathbf{m}\in M^{\bullet}\) and consider the cyclic submodule \(R\mathbf{m}\).

Since \(M\) is not principle, it is not the zero ideal, and contains at least two elements. So we can consider an element \(\mathbf{m}\in M\).

We have \(\operatorname{rank}_R(M) \geq 1\), since \(R\mathbf{m} \leq M\) and \(\left\{{m}\right\}\) is a subset of some spanning set.

\(R\mathbf{m}\) can not be linearly dependent, since \(R\) is an integral domain and \(M\subseteq R\), so \(\alpha \mathbf{m} = \mathbf{0} \implies \alpha = 0_R\).

Claim: since \(R\) is commutative, \(\operatorname{rank}_R(M) \leq 1\).
 If we take two elements \(\mathbf{m}, \mathbf{n} \in M^{\bullet}\), then since \(m, n\in R\) as well, we have \(nm = mn\) and so \begin{align*} (n)\mathbf{m} + (m)\mathbf{n} = 0_R = \mathbf{0} \end{align*} is a linear dependence.
\(M\) is torsionfree:

Let \(\mathbf{x} \in \operatorname{Tor}M\), then there exists some \(r\neq 0\in R\) such that \(r\mathbf{x} = \mathbf{0}\).

But \(\mathbf{x}\in R\) as well and \(R\) is an integral domain, so \(\mathbf{x}=0_R\), and thus \(\operatorname{Tor}(M) = \left\{{0_R}\right\}\).
\(\star\) Spring 2020 #6 #algebra/qual/completed
Let \(R\) be a ring with unity.

Give a definition for a free module over \(R\).

Define what it means for an \(R{\hbox{}}\)module to be torsion free.

Prove that if \(F\) is a free module, then any short exact sequence of \(R{\hbox{}}\)modules of the following form splits: \begin{align*} 0 \to N \to M \to F \to 0 .\end{align*}

Let \(R\) be a PID. Show that any finitely generated \(R{\hbox{}}\)module \(M\) can be expressed as a direct sum of a torsion module and a free module.
You may assume that a finitely generated torsionfree module over a PID is free.
Let \(R\) be a ring with 1.
An \(R{\hbox{}}\)module \(M\) is free if any of the following conditions hold:
 \(M\) admits an \(R{\hbox{}}\)linearly independent spanning set \(\left\{{\mathbf{b}_\alpha}\right\}\), so \begin{align*}m\in M \implies m = \sum_\alpha r_\alpha \mathbf{b}_\alpha\end{align*} and \begin{align*}\sum_\alpha r_\alpha \mathbf{b}_\alpha = 0_M \implies r_\alpha = 0_R\end{align*} for all \(\alpha\).
 \(M\) admits a decomposition \(M \cong \bigoplus_{\alpha} R\) as a direct sum of \(R{\hbox{}}\)submodules.
 There is a nonempty set \(X\) an monomorphism \(X\hookrightarrow M\) of sets such that for every \(R{\hbox{}}\)module \(N\), every set map \(X\to N\) lifts to a unique \(R{\hbox{}}\)module morphism \(M\to N\), so the following diagram commutes:
Equivalently, \begin{align*} \mathop{\mathrm{Hom}}_{\mathsf{Set}}(X, \mathop{\mathrm{Forget}}(N)) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{ {}_{R}{\mathsf{Mod}}}(M, N) .\end{align*}

Define the annihilator:
\begin{align*}
\operatorname{Ann}(m) \coloneqq\left\{{r\in R {~\mathrel{\Big\vert}~}r\cdot m = 0_M}\right\} {~\trianglelefteq~}R
.\end{align*}
 Note that \(mR \cong R/\operatorname{Ann}(m)\).
 Define the torsion submodule: \begin{align*} M_t \coloneqq\left\{{m\in M {~\mathrel{\Big\vert}~}\operatorname{Ann}(m) \neq 0}\right\} \leq M \end{align*}
 \(M\) is torsionfree iff \(M_t = 0\) is the trivial submodule.

Let the following be an SES where \(F\) is a free \(R{\hbox{}}\)module: \begin{align*} 0 \to N \to M \xrightarrow{\pi} F \to 0 .\end{align*}

Since \(F\) is free, there is a generating set \(X = \left\{{x_\alpha}\right\}\) and a map \(\iota:X\hookrightarrow F\) satisfying the 3rd property from (a).
 If we construct any map \(f: X\to M\), the universal property modules will give a lift \(\tilde f: F\to M\)

Identify \(X\) with \(\iota(X) \subseteq F\).

For every \(x\in X\), the preimage \(\pi^{1}(x)\) is nonempty by surjectivity. So arbitrarily pick any preimage.

\(\left\{{\iota(x_\alpha)}\right\} \subseteq F\) and \(\pi\) is surjective, so choose fibers \(\left\{{y_\alpha}\right\} \subseteq M\) such that \(\pi(y_\alpha) = \iota(x_\alpha)\) and define \begin{align*} f: X&\to M \\ x_\alpha &\mapsto y_\alpha .\end{align*}

The universal property yields \(h: F\to M\):

It remains to check that it’s a section.
 Write \(f= \sum r_i x_i\), then since both maps are \(R{\hbox{}}\)module morphism, by \(R{\hbox{}}\)linearity we can write \begin{align*} (\pi \circ h)(f) &= (\pi \circ h)\qty{ \sum r_i x_i } \\ &= \sum r_i (\pi \circ h)(x_i) ,\end{align*} but since \(h(x_i) \in \pi^{1}(x_i)\), we have \((\pi \circ h)(x_i) = x_i\). So this recovers \(f\).

Free implies projective
 Universal property of projective objects: for every epimorphism \(\pi:M\twoheadrightarrow N\) and every \(f:P\to N\) there exists a unique lift \(\tilde f: P\to M\):
 Construct \(\phi\) in the following diagram using the same method as above (surjectivity to pick elements in preimage):
 Now take the identity map, then commutativity is equivalent to being a section.
There is a SES
\(M/M_t\) is a free \(R{\hbox{}}\)module, so this sequence splits and \(M\cong M_t \oplus {M\over M_t}\), where \(M_t\) is a torsion \(R{\hbox{}}\)module.
Note that by the hint, since \(R\) is a PID, it suffices to show that \(M/M_t\) is torsionfree.
 Let \(m+M_t \in M/M_t\) be arbitrary. Suppose this is a torsion element, the claim is that it must be the trivial coset. This will follow if \(m\in M_t\)
 Since this is torsion, there exists \(r\in R\) such that \begin{align*} M_t = r(m + M_t) \coloneqq(rm) + M_t \implies rm\in M_t .\end{align*}
 Then \(rm\) is torsion in \(M\), so there exists some \(s\in R\) such \(s(rm) = 0_M\).
 Then \((sr)m = 0_M\) which forces \(m\in M_t\)
Spring 2012 #5 #algebra/qual/work
Let \(M\) be a finitely generated module over a PID \(R\).

\(M_t\) be the set of torsion elements of \(M\), and show that \(M_t\) is a submodule of \(M\).

Show that \(M/M_t\) is torsion free.
 Prove that \(M \cong M_t \oplus F\) where \(F\) is a free module.
Fall 2019 Final #3 #algebra/qual/work
Let \(R = k[x]\) for \(k\) a field and let \(M\) be the \(R{\hbox{}}\)module given by \begin{align*} M=\frac{k[x]}{(x1)^{3}} \oplus \frac{k[x]}{\left(x^{2}+1\right)^{2}} \oplus \frac{k[x]}{(x1)\left(x^{2}+1\right)^{4}} \oplus \frac{k[x]}{(x+2)\left(x^{2}+1\right)^{2}} .\end{align*} Describe the elementary divisors and invariant factors of \(M\).
Fall 2019 Final #4 #algebra/qual/work
Let \(I = (2, x)\) be an ideal in \(R = {\mathbf{Z}}[x]\), and show that \(I\) is not a direct sum of nontrivial cyclic \(R{\hbox{}}\)modules.
Fall 2019 Final #5 #algebra/qual/work
Let \(R\) be a PID.

Classify irreducible \(R{\hbox{}}\)modules up to isomorphism.

Classify indecomposable \(R{\hbox{}}\)modules up to isomorphism.
Fall 2019 Final #6 #algebra/qual/work
Let \(V\) be a finitedimensional \(k{\hbox{}}\)vector space and \(T:V\to V\) a noninvertible \(k{\hbox{}}\)linear map. Show that there exists a \(k{\hbox{}}\)linear map \(S:V\to V\) with \(T\circ S = 0\) but \(S\circ T\neq 0\).
Fall 2019 Final #7 #algebra/qual/work
Let \(A\in M_n({\mathbf{C}})\) with \(A^2 = A\). Show that \(A\) is similar to a diagonal matrix, and exhibit an explicit diagonal matrix similar to \(A\).
Fall 2019 Final #10 #algebra/qual/work
Show that the eigenvalues of a Hermitian matrix \(A\) are real and that \(A = PDP^{1}\) where \(P\) is an invertible matrix with orthogonal columns.
Fall 2020 #7 #algebra/qual/work
Let \(A \in \operatorname{Mat}(n\times n, {\mathbf{R}})\) be arbitrary. Make \({\mathbf{R}}^n\) into an \({\mathbf{R}}[x]{\hbox{}}\)module by letting \(f(x).\mathbf{v} \coloneqq f(A)(\mathbf{v})\) for \(f(\mathbf{v})\in {\mathbf{R}}[x]\) and \(\mathbf{v} \in {\mathbf{R}}^n\). Suppose that this induces the following direct sum decomposition: \begin{align*} {\mathbf{R}}^n \cong { {\mathbf{R}}[x] \over \left\langle{ (x1)^3 }\right\rangle } \oplus { {\mathbf{R}}[x] \over \left\langle{ (x^2+1)^2 }\right\rangle } \oplus { {\mathbf{R}}[x] \over \left\langle{ (x1)(x^21)(x^2+1)^4 }\right\rangle } \oplus { {\mathbf{R}}[x] \over \left\langle{ (x+2)(x^2+1)^2 }\right\rangle } .\end{align*}

Determine the elementary divisors and invariant factors of \(A\).

Determine the minimal polynomial of \(A\).
 Determine the characteristic polynomial of \(A\).
Misc/Unsorted
Spring 2017 #3 #algebra/qual/completed
Let \(R\) be a commutative ring with 1. Suppose that \(M\) is a free \(R{\hbox{}}\)module with a finite basis \(X\).

Let \(I {~\trianglelefteq~}R\) be a proper ideal. Prove that \(M/IM\) is a free \(R/I{\hbox{}}\)module with basis \(X'\), where \(X'\) is the image of \(X\) under the canonical map \(M\to M/IM\).

Prove that any two bases of \(M\) have the same number of elements. You may assume that the result is true when \(R\) is a field.
Part a: First, a slightly more advanced argument that gives some intuition as to why this should be true. Let \(X = \left\{{g_1,\cdots, g_n}\right\}\subseteq M\) be a generating set so that \(\operatorname{rank}_R M = n\) and every \(m\in M\) can be written as \(m = \sum_{i=1}^n r_i m_i\) for some \(r_i\in R\). Then \(M = Rg_1 \oplus Rg_2 \oplus \cdots Rg_n\), using the notation of AtiyahMacDonald, where \(Rm\) is the cyclic submodule generated by \(m\). In particular, \(M\cong R{ {}^{ \scriptscriptstyle\oplus^{n} } }\), and this is true iff \(M\) is a free \(R{\hbox{}}\)module of rank \(n\). So it suffices to show that \(M/IM \cong (R/I){ {}^{ \scriptscriptstyle\oplus^{\ell} } }\) for some \(\ell\) and that \(\ell = n\). Since free modules are flat, the functor \(({})\otimes_R M\) is left and right exact, so take the short exact sequence \begin{align*} 0 \to I \hookrightarrow R \twoheadrightarrow R/I \to 0 \end{align*} and tensor with \(M\) to get \begin{align*} 0 \to I \otimes_R M \xhookrightarrow{\iota} R\otimes_R M \xrightarrow[]{\pi} { \mathrel{\mkern16mu}\rightarrow }\, (R/I)\otimes_R M \to 0 .\end{align*} Noting that \(R\otimes_R M \cong M\) by the canonical map \(r\otimes m \mapsto rm\) (extended linearly), we have \begin{align*} 0 \to I \otimes_R M \xhookrightarrow{\iota} M \xrightarrow[]{\pi} { \mathrel{\mkern16mu}\rightarrow }\, (R/I)\otimes_R M \to 0 .\end{align*} Now \(\iota(i\otimes m) = im\), so the image of \(\iota\) is precisely \(IM\), and \(\operatorname{coker}(\iota) = M/IM\) by definition. By exactness, we must have \(\operatorname{coker}(\iota) \cong (R/I)\otimes_R M\), so \begin{align*} M/IM \cong (R/I)\otimes_R M .\end{align*} But now if \(M\cong R{ {}^{ \scriptscriptstyle\oplus^{n} } }\), we can conclude by a direct calculation: \begin{align*} M/IM &\cong (R/I)\otimes_R M \\ &\cong (R/I)\otimes_R (R{ {}^{ \scriptscriptstyle\oplus^{n} } }) \\ &\coloneqq(R/I)\otimes_R \qty{R\oplus R \oplus\cdots\oplus R} \\ &\cong \qty{(R/I)\otimes_R R} \oplus \qty{(R/I)\otimes_R R}\oplus \cdots \oplus \qty{(R/I) \otimes_R R} \\ &\cong (R/I) \oplus (R/I) \oplus \cdots \oplus (R/I) \\ &= (R/I){ {}^{ \scriptscriptstyle\oplus^{n} } } ,\end{align*} where we’ve used that \(A\otimes(B\oplus C)\cong (A\otimes B ) \oplus (A\otimes C)\) and \(A\otimes_R R \cong A\) for any \(R{\hbox{}}\)modules \(A,B,C\).
A more direct proof of a: Let \(X \coloneqq\left\{{g_1,\cdots, g_n}\right\} \subseteq M\) be a free \(R{\hbox{}}\)basis for \(M\), so \(X\) is \(R{\hbox{}}\)linearly independent and spans \(M\). The claim is that the cosets \(\tilde X \coloneqq\left\{{g_1 + IM, \cdots, g_n + IM}\right\} \subseteq IM\) form a basis for \(IM\).

Spanning: we want to show \begin{align*} m + IM \in M/IM \implies \exists r_k + I \in R/I \text{ such that } \\\qquad m + IM = \sum_{k=1}^n (r_K + I)(g_k + IM) .\end{align*}
 Note that the \(R/I{\hbox{}}\)module structure on \(M/IM\) is defined by \((r+I)(m+IM) \coloneqq rm + IM\).
 Fix \(m+IM\), and use the basis \(X\) of \(M\) to write \(M = \sum_{k=1}^n r_k g_k\), since it spans \(M\).
 Then \begin{align*} m + IM = \qty{\sum_{k=1}^n r_k g_k} + IM = \sum_{k=1}^n (r_k g_k + IM ) = \sum_{k=1}^n (r_k + I)(g_k + IM) ,\end{align*} so these \(r_k\) suffice.

Linear independence: we want to show \begin{align*} \sum_{k=1}^n (r_k + I)(g_k + IM) = 0_{IM} \implies r_k + I = 0_{R/I} \coloneqq 0 + I \quad \forall k .\end{align*}
 Expanding the assumption, we have \begin{align*} \sum_{k=1}^n (r_k + I)(g_k + IM) = 0_{IM} &\implies \sum_{k=1}^n (r_k g_k) + IM = 0 + IM \\ &\implies \sum_{k=1}^n r_k g_k \in IM ,\end{align*} and it suffices to show \(r_k \in I\) for all \(k\).
 Since \(IM \coloneqq\left\{{\sum_{k=1}^N i_k m_k {~\mathrel{\Big\vert}~}i_k\in I, m_k\in M, N < \infty}\right\}\) by definition, we have \begin{align*} \sum_{k=1}^n r_k g_k = \sum_{k=1}^N i_k m_k \end{align*} for some \(i_k\in I\) and \(m_k\in M\) and some finite \(N\).
 Since \(X\) is a spanning set for \(M\), we can write expand \(m_k = \sum_{k=1}^n r_k' g_k\) for each \(k\) and write \begin{align*} \sum_{k=1}^n r_k g_k = \sum_{k=1}^N i_k \qty{ \sum_{j=1}^n r_j' g_j} = \sum_{k=1}^n \sum_{j=1}^n i_k r_j' g_j = \sum_{k=1}^n \sum_{j=1}^n i_{kj} g_j ,\end{align*} where \(i_{kj} \coloneqq i_k r_j'\in I\) since \(I\) is an ideal.
 By collecting terms for each \(g_j\), we can thus write \begin{align*} \sum_{k=1}^n r_k g_k = \sum_{k=1}^n i_k' g_k \implies \sum_{k=1}^n (r_k  i_k) g_k = 0_M ,\end{align*} where \(i_k' \in I\). Using that the \(g_k\) are \(R{\hbox{}}\)linearly independent in \(M\), we have \(r_k = i_k\) for each \(k\), so in fact \(r_k\in I\), which is what we wanted to show.
Part b: suppose the result is true for fields. Noting that \(R/I\) is a field precisely when \(I\) is maximal, suppose \(R\) contains a maximal ideal \(I\) and let \(B_1, B_2\) be two \(R{\hbox{}}\)bases for \(M\). By part a, their images \(B_1', B_2'\) are \(R/I{\hbox{}}\)bases for \(M/IM\), but since \(R/I\) is a field and \(M/IM\) is a module over the field \(k \coloneqq R/I\), the sizes of \(B_1'\) and \(B_2'\) must be the same. This forces the sizes of \(B_1\) and \(B_2\) to be the same.
To see that \(R\) does in fact contain a maximal ideal, let \(S \coloneqq R^{\times}\setminus R\) be the set of nonunits in \(R\). Applying Zorn’s lemma shows that \(S\) is contained in a proper maximal ideal \(I\), which can be used in the argument above.
Spring 2020 #5 #algebra/qual/completed
Let \(R\) be a ring and \(f: M\to N\) and \(g: N\to M\) be \(R{\hbox{}}\)module homomorphisms such that \(g\circ f = \operatorname{id}_M\). Show that \(N \cong \operatorname{im}f \oplus \ker g\).
 We have the following situation:

Claim: \(\operatorname{im}f + \ker g \subseteq N\), and this is in fact an equality.
 For \(n\in N\), write \begin{align*} n = n + (f\circ g)(n)  (f\circ g)(n) = \qty{n  (f\circ g)(n) } + (f\circ g)(n) .\end{align*}
 The first term is in \(\ker g\): \begin{align*} g \qty{ n  (f\circ g)(n) } &= g(n)  (g\circ f \circ g)(n)\\ &= g(n)  (\operatorname{id}_N \circ g)(n)\\ &= g(n)  g(n) \\ &= 0 .\end{align*}
 The second term is clearly in \(\operatorname{im}f\).

Claim: the sum is direct.
 Suppose \(n\in \ker(g) \cap\operatorname{im}(f)\), so \(g(n) = 0\) and \(n=f(m)\) for some \(m\in M\). Then \begin{align*} 0 = g(n) = g(f(m)) = (g\circ f)(m) = \operatorname{id}_M(m) = m ,\end{align*} so \(m=0\) and since \(f\) is a morphism in \(R{\hbox{}}\)modules, \(n\coloneqq f(m) = 0\).
Fall 2018 #6 #algebra/qual/completed
Let \(R\) be a commutative ring, and let \(M\) be an \(R{\hbox{}}\)module. An \(R{\hbox{}}\)submodule \(N\) of \(M\) is maximal if there is no \(R{\hbox{}}\)module \(P\) with \(N \subsetneq P \subsetneq M\).

Show that an \(R{\hbox{}}\)submodule \(N\) of \(M\) is maximal \(\iff M /N\) is a simple \(R{\hbox{}}\)module: i.e., \(M /N\) is nonzero and has no proper, nonzero \(R{\hbox{}}\)submodules.

Let \(M\) be a \({\mathbf{Z}}{\hbox{}}\)module. Show that a \({\mathbf{Z}}{\hbox{}}\)submodule \(N\) of \(M\) is maximal \(\iff {\sharp}M /N\) is a prime number.
 Let \(M\) be the \({\mathbf{Z}}{\hbox{}}\)module of all roots of unity in \({\mathbf{C}}\) under multiplication. Show that there is no maximal \({\mathbf{Z}}{\hbox{}}\)submodule of \(M\).
 Todo
By the correspondence theorem, submodules of \(M/N\) biject with submodules \(A\) of \(M\) containing \(N\).
So

\(M\) is maximal:

\(\iff\) no such (proper, nontrivial) submodule \(A\) exists

\(\iff\) there are no (proper, nontrivial) submodules of \(M/N\)

\(\iff M/N\) is simple.
Identify \({\mathbf{Z}}{\hbox{}}\)modules with abelian groups, then by (a), \(N\) is maximal \(\iff\) \(M/N\) is simple \(\iff\) \(M/N\) has no nontrivial proper subgroups.\
By Cauchy’s theorem, if \({\left\lvert {M/N} \right\rvert} = ab\) is a composite number, then \(a\divides ab \implies\) there is an element (and thus a subgroup) of order \(a\). In this case, \(M/N\) contains a nontrivial proper cyclic subgroup, so \(M/N\) is not simple. So \({\left\lvert {M/N} \right\rvert}\) can not be composite, and therefore must be prime.

Let \(G = \left\{{x \in {\mathbf{C}}{~\mathrel{\Big\vert}~}x^n=1 \text{ for some }n\in {\mathbb{N}}}\right\}\), and suppose \(H < G\) is a proper submodule.

Since \(H\neq G\), there is some \(p\) and some \(k\) such that \(\zeta_{p^k}\not\in H\).
 Otherwise, if \(H\) contains every \(\zeta_{p^k}\) it contains every \(\zeta_n\)
Then there must be a prime \(p\) such that the \(\zeta_{p^k} \not \in H\) for all \(k\) greater than some constant \(m\) – otherwise, we can use the fact that if \(\zeta_{p^k} \in H\) then \(\zeta_{p^\ell} \in H\) for all \(\ell \leq k\), and if \(\zeta_{p^k} \in H\) for all \(p\) and all \(k\) then \(H = G\).
But this means there are infinitely many elements in \(G\setminus H\), and so \(\infty = [G: H] = {\left\lvert {G/H} \right\rvert}\) is not a prime. Thus by (b), \(H\) can not be maximal, a contradiction.
Fall 2019 Final #2 #algebra/qual/work
Consider the \({\mathbf{Z}}{\hbox{}}\)submodule \(N\) of \({\mathbf{Z}}^3\) spanned by \begin{align*} f_1 &= [1, 0, 1], \\ f_2 &= [2,3,1], \\ f_3 &= [0, 3, 1], \\ f_4 &= [3,1,5] .\end{align*} Find a basis for \(N\) and describe \({\mathbf{Z}}^3/N\).
Spring 2018 #6 #algebra/qual/work
Let \begin{align*} M &= \{(w, x, y, z) \in {\mathbf{Z}}^4 {~\mathrel{\Big\vert}~}w + x + y + z \in 2{\mathbf{Z}}\} \\ N &= \left\{{ (w, x, y, z) \in {\mathbf{Z}}^4 {~\mathrel{\Big\vert}~}4\divides (w  x),~ 4\divides (x  y),~ 4\divides ( y  z) }\right\} .\end{align*}

Show that \(N\) is a \({\mathbf{Z}}{\hbox{}}\)submodule of \(M\) .

Find vectors \(u_1 , u_2 , u_3 , u_4 \in {\mathbf{Z}}^4\) and integers \(d_1 , d_2 , d_3 , d_4\) such that \begin{align*} \{ u_1 , u_2 , u_3 , u_4 \} && \text{is a free basis for }M \\ \{ d_1 u_1,~ d_2 u_2,~ d_3 u_3,~ d_4 u_4 \} && \text{is a free basis for }N \end{align*}
 Use the previous part to describe \(M/N\) as a direct sum of cyclic \({\mathbf{Z}}{\hbox{}}\)modules.
Spring 2018 #7 #algebra/qual/work
Let \(R\) be a PID and \(M\) be an \(R{\hbox{}}\)module. Let \(p\) be a prime element of \(R\). The module \(M\) is called \(\left\langle{p}\right\rangle{\hbox{}}\)primary if for every \(m \in M\) there exists \(k > 0\) such that \(p^k m = 0\).

Suppose M is \(\left\langle{p}\right\rangle{\hbox{}}\)primary. Show that if \(m \in M\) and \(t \in R, ~t \not\in \left\langle{p}\right\rangle\), then there exists \(a \in R\) such that \(atm = m\).

A submodule \(S\) of \(M\) is said to be pure if \(S \cap r M = rS\) for all \(r \in R\). Show that if \(M\) is \(\left\langle{p}\right\rangle{\hbox{}}\)primary, then \(S\) is pure if and only if \(S \cap p^k M = p^k S\) for all \(k \geq 0\).
Fall 2016 #6 #algebra/qual/completed
Let \(R\) be a ring and \(f: M\to N\) and \(g: N\to M\) be \(R{\hbox{}}\)module homomorphisms such that \(g\circ f = \operatorname{id}_M\). Show that \(N\cong \operatorname{im}f \oplus \ker g\).
The trick: write down a clever choice of an explicit morphism. Let \(P \coloneqq(f\circ g):N\to N\), then \begin{align*} \Phi: N &\to \operatorname{im}f \oplus \ker g \\ n &\mapsto P(n) + (n P(n)) .\end{align*} The claim is that this is an isomorphism. One first needs to show that this makes sense: \(P(n) \coloneqq f(g(n))\) is clearly in \(\operatorname{im}f\) (since \(f\) is the last function applied), but it remains to show that \(nP(n) \in \ker g\). This is a direct computation: \begin{align*} g(n  P(n)) = g(n)  g(f(g(n))) = g(n)  \operatorname{id}_N(g(n)) = g(n)g(n) = 0_M ,\end{align*} where we’ve used that \(g\) is an \(R{\hbox{}}\)module morphism in the first step. Also note that \(\Phi\) is an \(R{\hbox{}}\)module morphism since it’s formed of compositions and sums of such morphisms.
One then has to show that \(\operatorname{im}f \cap\ker g = \left\{{0}\right\}\) – letting \(n\) be in this intersection, there exists an \(m\in M\) with \(f(m) = n\). Then applying \(g\) yields \(gf(m) = g(n)\), and the RHS is zero since \(n\in \ker g\), and the LHS is \(gf(m) = \operatorname{id}_M(m) = m\), so \(m=0\). Since \(f\) is an \(R{\hbox{}}\)module morphism, we must have \(f(0) = 0\), so \(n = f(m) = f(0) = 0\) as desired.
\(\Phi\) is injective: letting \(n\in \ker \Phi\), we have \begin{align*} 0 = \Phi(n) = P(n) + (n  P(n)) = P(n)  P(n) + n = n .\end{align*} We thus get \(N\cong \operatorname{im}\Phi\) and it remains to show \(\Phi\) is surjective. But this follows for the same reason: \begin{align*} n\in N \implies n = n + P(n)  P(n) = P(n) + (n P(n))= \Phi(n) .\end{align*}
Spring 2016 #4 #algebra/qual/work
Let \(R\) be a ring with the following commutative diagram of \(R{\hbox{}}\)modules, where each row represents a short exact sequence of \(R{\hbox{}}\)modules:
Prove that if \(\alpha\) and \(\gamma\) are isomorphisms then \(\beta\) is an isomorphism.
Spring 2015 #8 #algebra/qual/work
Let \(R\) be a PID and \(M\) a finitely generated \(R{\hbox{}}\)module.

Prove that there are \(R{\hbox{}}\)submodules \begin{align*} 0 = M_0 \subset M_1 \subset \cdots \subset M_n = M \end{align*} such that for all \(0\leq i \leq n1\), the module \(M_{i+1}/M_i\) is cyclic.

Is the integer \(n\) in part (a) uniquely determined by \(M\)? Prove your answer.
Fall 2012 #6 #algebra/qual/work
Let \(R\) be a ring and \(M\) an \(R{\hbox{}}\)module. Recall that \(M\) is Noetherian iff any strictly increasing chain of submodule \(M_1 \subsetneq M_2 \subsetneq \cdots\) is finite. Call a proper submodule \(M' \subsetneq M\) intersectiondecomposable if it can not be written as the intersection of two proper submodules \(M' = M_1\cap M_2\) with \(M_i \subsetneq M\).
Prove that for every Noetherian module \(M\), any proper submodule \(N\subsetneq M\) can be written as a finite intersection \(N = N_1 \cap\cdots \cap N_k\) of intersectionindecomposable modules.
Fall 2019 Final #1 #algebra/qual/work
Let \(A\) be an abelian group, and show \(A\) is a \({\mathbf{Z}}{\hbox{}}\)module in a unique way.
Fall 2020 #6 #algebra/qual/work
Let \(R\) be a ring with \(1\) and let \(M\) be a left \(R{\hbox{}}\)module. If \(I\) is a left ideal of \(R\), define \begin{align*} IM \coloneqq\left\{{ \sum_{i=1}^{N < \infty} a_i m_i {~\mathrel{\Big\vert}~}a_i \in I, m_i \in M, n\in {\mathbb{N}}}\right\} ,\end{align*} i.e. the set of finite sums of of elements of the form \(am\) where \(a\in I, m\in M\).

Prove that \(IM \leq M\) is a submodule.

Let \(M, N\) be left \(R{\hbox{}}\)modules, \(I\) a nilpotent left ideal of \(R\), and \(f: M\to N\) an \(R{\hbox{}}\)module morphism. Prove that if the induced morphism \(\overline{f}: M/IM \to N/IN\) is surjective, then \(f\) is surjective.