# Modules

## 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 non-finitely 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 zero-divisors, prove that every non-zero $$R{\hbox{-}}$$module has non-zero torsion elements.

• One-step 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 $$_6\curvearrowright_6 = _6$$ and $$_6\curvearrowright_6 = _6$$, but $$_6 + _6 = _6$$, where 5 is coprime to 6, and thus $$[n]_6\curvearrowright_6 =  \implies [n]_6 = _6$$. So $$_6$$ 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 non-principal ideal of $$R$$. Prove that $$M$$ is torsion-free 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 torsion-free:

• 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]}{(x-1)^{3}} \oplus \frac{k[x]}{\left(x^{2}+1\right)^{2}} \oplus \frac{k[x]}{(x-1)\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 finite-dimensional $$k{\hbox{-}}$$vector space and $$T:V\to V$$ a non-invertible $$k{\hbox{-}}$$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{ (x-1)^3 }\right\rangle } \oplus { {\mathbf{R}}[x] \over \left\langle{ (x^2+1)^2 }\right\rangle } \oplus { {\mathbf{R}}[x] \over \left\langle{ (x-1)(x^2-1)(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 Atiyah-MacDonald, 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{\mkern-16mu}\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{\mkern-16mu}\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 non-units 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 $$n-P(n) \in \ker g$$. 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 $$0\leq i \leq n-1$$, 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 $$M' \subsetneq M$$ intersection-decomposable 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 intersection-indecomposable 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.

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