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 \([3]_6\curvearrowright[2]_6 = [0]_6\) 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 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):

Link to Diagram

  • 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 \(A\).


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:

Link to Diagram

  • 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 \(N\).


  • \(M\) 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.

#6 #algebra/qual/work #5 #algebra/qual/completed #3 #4 #7 #10 #2 #8 #1