Problem 1

Part a

Define a map

\begin{align*} \phi_\mathrm{ev}: \hom_{\mathbf{Z}}({\mathbf{Z}}_m, A) \to A \\ (f: {\mathbf{Z}}_m \to A) \mapsto f(1) \end{align*}

Then \(\phi_\mathrm{ev}\) is a \({\mathbf{Z}}{\hbox{-}}\)module homomorphism, since

\begin{align*} \begin{align*} \phi_\mathrm{ev}(nf + g) &= (nf + g)(1) \\ &= nf(1) + g(2) \\ &= n\phi_\mathrm{ev}(f) + \phi_\mathrm{ev}(g) \end{align*} \end{align*}

But this forces \(f(\overline 0) = 0_A\) (where \(\overline 0: {\mathbf{Z}}_m \to A\) is the zero map), we have \begin{align*} 0 = f(0) = f(m) = m f(1), \end{align*}

we must have \(mf(1) = 0\) in \(A\). So \begin{align*} \operatorname{im}\phi_\mathrm{ev} = \left\{{a\in A \mathrel{\Big|}ma = 0}\right\} \coloneqq A[m]. \end{align*}

It is also the case that \begin{align*} \begin{align*} \ker \phi_{\mathrm{ev}} &= \left\{{f \in \hom_{\mathbf{Z}}({\mathbf{Z}}_m, A) \mathrel{\Big|}f(1) = 0}\right\} = \left\{{\overline 0}\right\}, \end{align*} \end{align*}

which follows from the fact that \({\mathbf{Z}}_m = \left\langle{1 \operatorname{mod}m}\right\rangle\) and \(A = \left\langle{1_A}\right\rangle\) as \({\mathbf{Z}}{\hbox{-}}\)modules, so if \(f(1 \operatorname{mod}m) = 0_A\) then \begin{align*} f(n \operatorname{mod}m) = nf( 1 \operatorname{mod}m) = 0 \end{align*}

and so \(f\) is necessarily the zero map. So \(\ker \phi = \overline 0\).

We can then apply the first isomorphism theorem, \begin{align*} \frac{\hom_{\mathbf{Z}}({\mathbf{Z}}_m, A)}{\ker \phi_{\mathrm{ev}}} \cong \operatorname{im}\phi_{\mathrm{ev}} \implies \hom_{\mathbf{Z}}({\mathbf{Z}}_m, A) \cong A[m]. \end{align*}

Part 2

Lemma: If \(x \divides n\) and \(x\divides m\) then \(x\divides \gcd(m, n)\)

Proof: We have \(x\divides km + \ell n\) for any integers \(k, \ell\). So let \(d = \gcd(m, n)\), then there exist integers \(a, b\) such that \(am + bn = d\). But we can now just take \(k=a\) and \(\ell = b\). \(\hfill\blacksquare\)

We claim that \({\mathbf{Z}}_n[m] \cong {\mathbf{Z}}_{(m, n)}\), from which the result immediately follows by part 1.

Define a map \begin{align*} \begin{align*} \phi: {\mathbf{Z}}&\to {\mathbf{Z}}_n[m] \\ 1 &\mapsto [1]_{\operatorname{mod}n} ,\end{align*} \end{align*}

which we claim is an isomorphism. \(\phi\) is clearly surjective since \({\mathbf{Z}}\to {\mathbf{Z}}_n\) is a quotient map and \({\mathbf{Z}}_n[m]\) is a subgroup of \({\mathbf{Z}}_n\), and if we let \(d \coloneqq\gcd(m, n)\), we have

\begin{align*} \begin{align*} \ker \phi &= \left\{{x\in {\mathbf{Z}}_n {~\mathrel{\Big\vert}~}mx = 0}\right\} \\ &= \left\{{x \in {\mathbf{Z}}_n {~\mathrel{\Big\vert}~}x \divides m}\right\} \\ &= \left\{{x \in {\mathbf{Z}}{~\mathrel{\Big\vert}~}x \divides n ~\text{ and }~ x \divides m}\right\} \\ &= \left\{{x \in {\mathbf{Z}}{~\mathrel{\Big\vert}~}x \divides d}\right\} \quad \text{by the lemma} \\ &= d{\mathbf{Z}}~ .\end{align*} \end{align*}

Then by the first isomorphism theorem, we have \begin{align*} \frac{{\mathbf{Z}}}{\ker \phi} \cong \operatorname{im}\phi \implies \frac{{\mathbf{Z}}}{d{\mathbf{Z}}} \cong {\mathbf{Z}}_n[m]. \end{align*}

Part 3

Note: let \([x]_m\) denote the equivalence class of \(x \operatorname{mod}m\).

Let \(f\in {\mathbf{Z}}^* = \hom_{\mathbf{Z}}({\mathbf{Z}}_m, {\mathbf{Z}})\), so \(f: {\mathbf{Z}}_m \to {\mathbf{Z}}\). These are both \({\mathbf{Z}}{\hbox{-}}\)modules generated by their identity elements, so such a map is determined by where it send \([1]_m\).

So let \(f([1]_m) = n \in {\mathbf{Z}}\). Since \(f\) is a module homomorphism, we have \(f([0]_m) = 0\), and in particular we have \begin{align*} \begin{align*} 0 &= f([0]_m) \\ &= f([m]_m) \\ &= f([1 m]_m) \\ &= m f([1]_m), \end{align*} \end{align*}

which forces \(f([1]) \in {\mathbf{Z}}[m] = \left\{{0}\right\}\), so \(f\) must be the zero map and \({\mathbf{Z}}^* = 0\).

Note: \({\mathbf{Z}}[m] = 0\) because \({\mathbf{Z}}\) is an integral domain, so \(mx = 0\) forces \(m=0\) or \(x=0\).

Part 4

To see that \({\mathbf{Z}}_m\) is a \({\mathbf{Z}}_{mk}\) module, we define an action

\begin{align*} \begin{align*} {\mathbf{Z}}_{mk} &\curvearrowright{\mathbf{Z}}_{m} \\ [x]_{mk} \curvearrowright[y]_m &\coloneqq[xy]_m \end{align*} \end{align*}

This is a well-defined action:

If \([x_1]_{mk} = [x_2]_{mk}\) are two representatives of the same equivalence class, then \begin{align*} [x_1]_{mk} - [x_2]_{mk} = [x_1-x_2]_{mk} = [0]_{mk} \implies m \divides x_1 - x_2. \end{align*}

But then \begin{align*} \begin{align*} ([x_1]_{mk}\curvearrowright[y]_m) - ([x_2]_{mk} \curvearrowright[y]_m) &= [x_1 y]_m - [x_2 y]_m \\ &= [(x_1 - x_2)y]_m \\ &= [0]_m, \end{align*} \end{align*}

which shows that their resulting actions on \({\mathbf{Z}}_m\) are equal.

This action yields a module structure:

• \(r.(x+y) = r.x + r.y\): \begin{align*} [r]_{mk} \curvearrowright([x]_m + [y]_m) = [r]_{mk} \curvearrowright[x+y]_m = [r(x+y)]_m = [rx]_m + [ry]_m. \end{align*}

• \((r+s).x = r.x + s.x\): \begin{align*} [r]_{mk} + [s]_{mk} \curvearrowright[x]_m = [r + s]_{mk} \curvearrowright[x]_m = [(r+s)x]_m = [rx]_m + [sx]_m. \end{align*}

• \((rs).x = r.s.x\): \begin{align*} \begin{align*} [r]_{mk} \cdot [s]_{mk} \curvearrowright[x]_m &= [rs]_{mk} \curvearrowright[x]_m \\ &= [(rs)x]_m \\ &= [r]_{mk} \curvearrowright[sx]_m \\ &= [r]_{mk} \curvearrowright( [s]_{mk} \curvearrowright[x]_m). \end{align*} \end{align*}

• \(1.x = x\): \begin{align*} [1]_{mk} \curvearrowright[x]_m = [1x]_m = [x]_m. \end{align*}

\({\mathbf{Z}}_m^* \coloneqq\hom_{{\mathbf{Z}}_{mk}}({\mathbf{Z}}_m, {\mathbf{Z}}_{mk}) \cong {\mathbf{Z}}_m\):

Define a map \begin{align*} \begin{align*} \phi: \hom_{{\mathbf{Z}}_{mk}}({\mathbf{Z}}_m, {\mathbf{Z}}_{mk}) &\to {\mathbf{Z}}_m \\ f \mapsto [f([1]_m)]_m \end{align*} \end{align*}

\(\phi\) is a homomorphism, as

\begin{align*} \begin{align*} \phi(f + g) &= [(f+g)([1]_m)]_m \\ &= [f([1]_m) + g([1]_m)]_m \\ &= [f([1]_m)]_m + [g([1]_m)]_m \\ \\ \phi([r]_{mk} \curvearrowright f) &= [[r]_{mk} f([1]_m)]_m \\ &= [r]_m \cdot [f([1]_m)]_m \\ &= [r]_{mk}\curvearrowright\phi(f). \end{align*} \end{align*}

\(\phi\) is injective, as \([f([1]_m)]_m = [0]_m\), then for any \(1 \leq \ell \leq m\), we have \begin{align*} \begin{align*} [f([\ell]_m)]_m &= [\ell f([1]_m)]_m \\ &= \ell [f([1]_m)]_m \\ &= \ell[0]_m \\ &= [0]_m ,\end{align*} \end{align*}

so \(f\) must be the zero map.

\(\phi\) is surjective, since if \([\ell]_m \in {\mathbf{Z}}_m\), we can define \begin{align*} \begin{align*} f_\ell: {\mathbf{Z}}_m &\to {\mathbf{Z}}_{mk} \\ [1]_m &\mapsto [\ell]_{mk} \end{align*} \end{align*}

which makes sense and is well-defined because \({\mathbf{Z}}_m \hookrightarrow{\mathbf{Z}}_{mk}\), and the map is defined on the generator.

So we have the desired bijection. \(\hfill\blacksquare\)

Problem 2

We have the map

\begin{align*} \begin{align*} \pi: {\mathbf{Z}}&\to {\mathbf{Z}}_2 \\ x &\mapsto [x]_2 \end{align*} \end{align*}

which is a surjection and thus an epimorphism in the category \({{\mathbf{Z}}}{\hbox{-}}\mathrm{Mod}\), and if we apply the functor \(\hom_{\mathbf{Z}}({\mathbf{Z}}_2, {-})\) to \(\pi\) we obtain an induced map

\begin{align*} \begin{align*} \overline{\pi}: \hom_{{\mathbf{Z}}}({\mathbf{Z}}_2, {\mathbf{Z}}) &\to \hom_{{\mathbf{Z}}}({\mathbf{Z}}_2, {\mathbf{Z}}_2) \\ f &\mapsto \pi \circ f .\end{align*} \end{align*}

The claim is that \(\overline{\pi}\) is not a surjection, and thus not an epimorphism (in the same category).

To see that this is the case, we can simply note that \(\hom_{\mathbf{Z}}({\mathbf{Z}}_2, {\mathbf{Z}}) = 0\) by part 3 of Problem 1, whereas \(\hom_{\mathbf{Z}}({\mathbf{Z}}_2, {\mathbf{Z}}_2) \neq 0\).

For example, one can define \begin{align*} \begin{align*} \operatorname{id}_{{\mathbf{Z}}_2}: {\mathbf{Z}}_2 &\to {\mathbf{Z}}_2\\ [x]_2 &\mapsto [x]_2 ,\end{align*} \end{align*} which is a nontrivial module homomorphisms.

So any such \(f\) appearing must be the zero map, and thus \(\overline{\pi}\) is also the zero map. \(\hfill\blacksquare\)

Problem 3

Let \(f: R \to R\) be an endomorphism of \(R\) in the category of rings. We can then check that for any \(r\in R\), we have \(f(r) = f(r 1_R) = rf(1_R)\), which says that \(f\) is given by right-multiplication by some fixed element \(x_f \coloneqq f(1_R)\), i.e. \begin{align*} \begin{align*} f: R &\to R \\ r &\mapsto r \cdot x_f \end{align*} \end{align*}

and so we can attempt to define \begin{align*} \begin{align*} \phi_1: \hom_R(R, R) &\to R \\ f &\mapsto x_f \coloneqq f(1_R) \end{align*} \end{align*}

We can check that \begin{align*} (g\circ f(r)) = g(f(r)) = g(r\cdot x_f) = r \cdot x_f \cdot x_g, \end{align*}

which shows that in fact \begin{align*} \phi(g \circ f) = x_f \cdot x_g, \end{align*} which reverses the multiplication. So the correct codomain is \(R^{op}\), and we amend the definition: \begin{align*} \begin{align*} \phi_2: \hom_R(R, R) &\to R^{op} \\ f &\mapsto x_f \coloneqq f(1_R) \end{align*} \end{align*}

By construction, \(\phi_s\) is a ring homomorphism. If \(R\) is commutative, then \(x_f \cdot x_g = x_g \cdot x_f\), which makes \(\phi_1\) a ring homomorphism as well. It remains to check that it is an isomorphism/

\(\phi_1\) is in injective: We can check that \(\ker \phi_1 = 0\) as a ring. To that end, suppose \(\phi_1(f) = x_f = 0\). Then \(f(r) = r \cdot 0 = 0\), so \(f\) can only be the zero map.

\(\phi_1\) is surjective: Let \(x\in R\) be arbitrary, then we can define \(f: R \to R\) by \(f(1_R) = x\), so \(f(r) = r\cdot x\). This is an endomorphism of \(R\), and thus an element of \(\hom_R(R, R)\).

By the first isomorphism theorem for rings, we thus have \(\hom_R(R, R) \cong R\). \(\hfill\blacksquare\)

Problem 4

Note: Let \(X {}^{ \vee }\coloneqq\hom_R(X, R)\) denote the dual.

We have maps

\begin{align*} \begin{align*} \theta_A: A &\to (A {}^{ \vee }) {}^{ \vee }\\ a &\mapsto (\mathrm{ev}_a: f \mapsto f(a) ) \end{align*} \end{align*}

\begin{align*} \begin{align*} \theta_B: B &\to (B {}^{ \vee }) {}^{ \vee }\\ b &\mapsto (\mathrm{ev}_b: g \mapsto g(b) ) \end{align*} \end{align*}

\begin{align*} \begin{align*} f: A &\to B \\ a &\mapsto f(a) \end{align*} \end{align*}

\begin{align*} \begin{align*} f {}^{ \vee }: B {}^{ \vee }&\to A {}^{ \vee }\\ g &\mapsto g \circ f \end{align*} \end{align*}

\begin{align*} \begin{align*} f^{\vee\vee}: A^{\vee\vee} &\to B^{\vee\vee} \\ h &\mapsto h \circ f {}^{ \vee } \end{align*} \end{align*}

We can now check that \(f^{\vee\vee} \circ \theta_A = \theta_B \circ f\) as maps from \(A\) to \(B^{\vee\vee}\). Letting \(a\in A\), and \(h\in B^{\vee\vee}\) (so \(h: B {}^{ \vee }\to R\)), we will show that both maps act on \(h\) in the same way.

For notational convenience, write \(\phi \curvearrowright h \coloneqq h\circ \phi\). We then have

\begin{align*} \begin{align*} (f^{\vee\vee} \circ \theta_A)(a) \curvearrowright h &\coloneqq f^{\vee\vee}(\theta_A(a)) \curvearrowright h \\ &\coloneqq f^{\vee\vee}(\mathrm{ev}_a)\curvearrowright h \\ &=(\mathrm{ev}_a \circ f {}^{ \vee })\curvearrowright h \\ &\coloneqq h \circ (\mathrm{ev}_a \circ f) \\ &\coloneqq h(f(a)) \\ &= \mathrm{ev}_{f(a)}\curvearrowright h \\ &\coloneqq\theta_B(f(a))\curvearrowright h \\ &\coloneqq(\theta_B \circ f)(a)\curvearrowright h, \end{align*} \end{align*}

which shows that these actions agree, and thus the diagram commutes.

Problem 5

Let \(E\) be a free module over \(R\) an integral domain. Then \(E\) has a basis \(\left\{{\mathbf{e}_i}\right\} \subseteq F\), so if \(x \neq 0 \in E\), we have \begin{align*} x = \sum_i r_i \mathbf{e}_i \end{align*}

where each \(r_i \in R\). Moreover, since \(x\neq 0\), at least one \(r_i \neq 0\), so let \(r_j\) denote one of the nonzero coefficients.

Now suppose \(x\) is a torsion element, so \(mx = 0\) for some \(m\neq 0 \in E\). We can then write \begin{align*} mx = m\sum_i r_i \mathbf{e}_i = \sum_i mr_i \mathbf{e}_i = 0 \end{align*}

But by linear independence, this forces \(mr_i = 0\) for all \(i\). In particular, \(mr_j = 0\) where \(r_j \neq 0\). But this exhibits either \(m\) or \(r_j\) as a zero divisor, and since the only zero divisor in an integral domain is zero, we must have \(m=0\) or \(r_j = 0\), a contradiction.

So \(x\) can not be a torsion element. But since \(x \in E\) was arbitrary, \(E\) must be torsion-free.

For an example of a torsion-free module over an integral domain that is not free, consider \({\mathbf{Q}}\) as a \({\mathbf{Z}}{\hbox{-}}\)module. Then \({\mathbf{Q}}\) is clearly torsion-free, since it is an integral domain and the same argument as above applies.

But \({\mathbf{Q}}\) is not free as \({\mathbf{Z}}{\hbox{-}}\)module. Supposing that \(\mathcal B =\left\{{\mathbf{b}_1, \mathbf{b}_2, \cdots}\right\} \subset {\mathbf{Q}}\) was a \({\mathbf{Z}}{\hbox{-}}\)basis, consider \(\mathbf{b}_1 = \frac {p_1} {q_1}\) and \(\mathbf{b}_2 = \frac {p_2} {q_2}\). Then \(\mathbf{b}_1, \mathbf{b}_2\) can not be linearly independent over \({\mathbf{Z}}\), which follows from the fact that \begin{align*} q_1 p_2 \mathbf{b}_1 + q_2 p_1 \mathbf{b}_2 = p_2 p_1 - p_1 p_2 = 0, \end{align*}

while \(q_1 p_2, ~q_2 p_1 \neq 0 \in {\mathbf{Z}}\). \(\hfill\blacksquare\)

Problem 6

If \(A\) is a cyclic module over a commutative ring \(R\), so we have \(A = Ra\) for some \(a\in A\). By Hungerford’s definition, the submodule \(A\) has order \(r\) \(\iff\) the element \(a\) has order \(r\) \(\iff\) the order ideal \(\mathcal{O}_a \coloneqq\left\{{x\in R \mathrel{\Big|}xa = 0}\right\} = (r)\).

In particular, \(ra = 0\).

Part 1

Since \((r, s) = (1)\), we can find \(t_1, t_2 \in R\) such that \begin{align*} \begin{align*} t_1r + t_2 s = 1 &\implies t_1ra + t_2 sa = 1a &\\ &\implies t_1(ra) + t_2 sa = a &\\ &\implies t_2 sa = a &\text{since $ra=0$}\\ &\implies s(t_2 a) = a &\text{since $R$ is commutative} ,\end{align*} \end{align*}

which implies that \(a \in sA\) and thus \(A \subseteq sA\). However, we always have \(sA \subseteq A\) for modules, so this shows that \(A = sA\).

To see that \(A[s] = \left\{{x\in A \mathrel{\Big|}sx = 0}\right\} = 0\), let \(x\in A[s]\); we will show \(x=0\). Since \(x\in A = Ra\), we have \(x = r_1 a\), and in particular \begin{align*} ra = 0 \implies rx = r r_1 a = r_1 (ra) = 0. \end{align*}

So we now have \(rx = 0\) and \(sx=0\), and we can write \begin{align*} \begin{align*} x &= (t_1 r + t_2 s)x \\ &= t_1 (rx) + t_2 (sx) \\ &= t_1 0 + t_2 0 \\ &= 0 .\end{align*} \end{align*}

So \(x = 0\) and thus \(A[s] = 0\). \(\hfill\blacksquare\)

Part 2

Suppose \(r = sk\). Toward an application of the first isomorphism theorem, define a map \begin{align*} \begin{align*} \phi: R &\to sA = sRa \\ x &\mapsto sxa .\end{align*} \end{align*}

\(\phi\) is well-defined:

This follows from that fact that \(a\in A \implies xA \in A\) for any \(x\in R\), so the codomain is in fact \(sA\).

\(\phi\) is an \(R{\hbox{-}}\)module homomorphism:

We have \begin{align*} \begin{align*} t\in R \implies \phi(tx) &= s(tx)a = t(sxa) = t\phi(x) \\ x,y \in R \implies \phi(x + y) &= s(x+y)a = sxa + sya + \phi(x) + \phi(y) \end{align*} \end{align*}

\(\ker \phi = (k)\):

Suppose \(x\in \ker\phi\) so \(sxa = 0_A\); we’d like to show \(x \in (k)\).

By definition \(sx \in \mathcal O_a\), and by assumption \(\mathcal O_a = (r)\), so \(sx = t_1 r\) for some \(t_1 \in R\). \begin{align*} \begin{align*} & sxa = 0_A \\ \implies sx &= t_1 r &\text{since $sx \in \mathcal O_a$} \\ \implies sx &= t_1 (sk) &\text{since $r=sk$ by assumption}\\ \implies sx &= s (t_1 k) &\text{since elements in $R$ and $A$ commute}\\ \implies x &= t_1 k &\text{since $R$ is a domain, so $sm = sn, s\neq 0 \implies m=n$} ,\end{align*} \end{align*}

which exhibits \(x = t_1 k \implies x\in (k)\) as desired.

\(\phi\) is surjective:

Since \(A=Ra\), we have \(sA =sRA\) and thus \(x\in sA \implies x = sra\) for some \(r\in R\); but then \(\phi(r) = sra = x\).

We thus have \begin{align*} \begin{align*} R/\ker \phi \cong \operatorname{im}\phi \implies R/(k) \cong sA. \end{align*} \end{align*}

Similarly, define a map \begin{align*} \begin{align*} \psi: R &\to A[s] \\ x &\mapsto kxa \end{align*} \end{align*}

\(\psi\) is well-defined:

It suffices to check that \(\operatorname{im}\psi \subseteq A[s]\) (since we will show surjectivity shortly), i.e. that \(s\) annihilates anything in the image. This follows from \begin{align*} s (kxa) = (sk)xa = rxa = x(ra) = 0, \end{align*} since \(ra = 0\) by assumption.

\(\psi\) is an \(R{\hbox{-}}\)module homomorphism:

We can check \begin{align*} \psi(tr_1 + r_2) = k(tr_1 + r_2)s = tkr_1s + kr_2 s = t\psi(r_1) + \psi(r_2) \end{align*}

which follows because elements of \(R\) commute with those from \(A\) under multiplication.

\(\ker \psi = (s)\):

Suppose \(x\in \ker\psi\), so \(kxa = 0\). Then \(kx \in \mathcal O_a = (r)\), so \(kx = r t_1\). Then

\begin{align*} \begin{align*} & kxa = 0_A \\ &\implies kx = r t_1 & \text{since } kx \in \mathcal O_a\\ &\implies kx = (sk)t_1 &\text{since } r = sk\\ &\implies kx = k(st_1) &\text{since $R$ is commutative} \\ &\implies x = s t_1 &\text{since $R$ is a domain}, \end{align*} \end{align*} and so \(x\in (s)\) as desired.

\(\psi\) is surjective:

Letting \(y \in A[s]\) be arbitrary. We have \begin{align*} \begin{align*} y \in A[s] &\implies x = t_1 a,\quad sx = 0 \\ &\implies s(t_1 a) = 0 \\ &\implies st_1 \in \mathcal O_a \implies \exists x\in R {~\mathrel{\Big\vert}~}st_1 = xr = x (sk) \\ &\implies st_1 = sxk \\ &\implies t_1 = xk \qquad\qquad\text{since $R$ is a domain} \\ &\implies y = t_1 a = (x k)a = kxa, \end{align*} \end{align*}

so \(\psi(x) = y\).

We can then apply the first isomorphism theorem \begin{align*} R/\ker \psi \cong \operatorname{im}\psi \implies R/(s) \cong A[s]. \end{align*}

\(\hfill\blacksquare\)

Problem 7

Lemma: If \(M\) is a cyclic module over a PID, then \(M\) has exactly 1 invariant factor.

Lemma: Let \(A\) be a cyclic module, so \(A = Ra\). If the order of \(A\) is \(r\), so \(\mathcal O_a = (r)\), then \(A \cong R/(r)\).

This means that we can write \(A = R/(a)\) and \(B = R/(b)\), and \(a, b\) are the invariant factors of \(A, B\) respectively, and \(M\coloneqq A \oplus B \cong R/(ab)\).

Since \(R\) is a PID, there is unique factorization, so we can write \begin{align*} \begin{align*} r &= \prod_{i=1}^n p_i^{k_i} \\ s &= \prod_{i=1}^n p_i^{\ell_i} \\ \implies rs &= \prod_{i=1}^n p_i^{k_i + \ell_i}, \end{align*} \end{align*}

where we allow some \(k_i, \ell_i = 0\) so that we can take the product over the same set of primes.

However, means that the elementary divisors of \(M\) are given by the multiset \(L \coloneqq\left\{{p_i^{k_i}}\right\} \cup\left\{{p_i^{\ell_i}}\right\}\).

The largest invariant factor \(d_1\) of \(M\) is obtained from the elementary divisors by

• Forming the multiset \(L\) of elementary divisors,
• Selecting the highest power of each prime occurring, say \(s_i \coloneqq p_i^{\max(k_i, \ell_i)}\),

• Removing \(s_i\) from \(L\),
• Then letting \(d_1 = \prod s_i\).

However, this process yields \(d_1 = \mathrm{lcm}(r, s)\) by construction, since \begin{align*} d_1 = \prod_{i=1}^n s_i = \prod_{i=1}^n p_i^{\max(k_i, \ell_i)} \coloneqq\mathrm{lcm}(r_s). \end{align*}

The next largest invariant factor is obtained by performing the same process on the remaining prime powers in \(L\). However, we can note that after obtaining \(d_1\), we have \(L = \left\{{p_i^{\min(k_i, \ell_i)}}\right\}\), since there were only two choices for each \(p_i\) occurring and we chose the copy with the maximal exponent.

But this means when we perform step (b) to obtain \(d_2\), there is now only one choice, and thus each \(s_i = p_i^{\min(k_i, \ell_i)}\) and we have \begin{align*} d_2 = \prod_{i=1}^n s_i = \prod_i p_i^{\min(k_i, \ell_i)} \coloneqq\gcd(r, s). \end{align*}

Note: by construction, \(d_2 \divides d_1\), since we are choosing from the same prime powers but with smaller exponents.

Since there were only at most two copies of each prime occurring in \(L\), where one of them was chosen for \(d_1\) and the other was chosen for \(d_2\), this exhausts all of the elements in \(L\). But this means \(M\) has only two invariant divisors, \begin{align*} \begin{align*} d_1 &= \operatorname{lcm}(r, s) \\ d_2 &= \gcd(r, s) ,\end{align*} \end{align*}

which is what we wanted to show. \(\hfill\blacksquare\)

Note: the indexing convention for \(d_i\) is opposite the usual one here, since we are choosing the largest invariant factor first, and so we have \(d_n \divides d_{n-1} \divides \cdots \divides d_1\).

Qual Problems

Problem 8

Part 1

The claim is that every element in \(M \coloneqq R^n/\operatorname{im}A\) is torsion \(\iff\) the matrix rank of \(A\) is exactly \(n \iff\) the Smith normal form of \(A\) has exactly \(n\) nonzero invariant factors.

To see that this is the case, we can apply the structure theorem for finitely-generated modules over a PID. This gives us \begin{align*} M \cong F \oplus \bigoplus R/(r_i) \end{align*}

where \(F\) is free of finite rank, \(R/(r_i)\) is cyclic torsion, and \(r_i \divides r_{i+1} \divides \cdots\) are the invariant factors of \(M\).

We thus have \begin{align*} M \cong R^n/\operatorname{im}A \cong F \oplus \bigoplus R/(r_i), \end{align*}

which will be pure torsion if and only if \(F = 0\).

But if we compute the smith normal for of \(A\), we obtain \begin{align*} SNF(A) = \left[ \begin{array}{rrrrrr} d_1 & 0 & \cdots & 0 & \cdots & 0 \\ 0 & d_2 & \cdot & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots & \cdots & 0\\ 0 & 0 & \cdots & d_n & \cdots & 0 \end{array}\right] \end{align*}

where \(d_1 \divides d_2 \divides \cdots \divides d_n\), and thus \begin{align*} \begin{align*} \operatorname{im}A \cong \operatorname{im}SNF(A) &\cong d_1 R \oplus d_2 R \oplus \cdots \oplus d_n R \\ \\ \implies M = R^n/\operatorname{im}A &\cong \frac{R^n}{d_1 R \oplus d_2 R \oplus \cdots d_n R} \\ \\ &\cong R/(d_1) \oplus R/(d_2) \cdots \oplus R/(d_n) \end{align*} \end{align*}

where \(R/(d_i)\) is a cyclic torsion module precisely when \(d_i \neq 0\). If instead some \(d_i = 0\), we then have \(R/(d_i) \cong R\), which is a free \(R{\hbox{-}}\)module, yielding non-torsion elements in \(M\).

But \(\operatorname{det}(A) = \operatorname{det}(SNF(A)) = \prod_{i=1}^n d_i\), and so if \(d_i=0\) for some \(i\) iff \(\operatorname{det}A = 0\) iff \(\operatorname{rank}A < n\).

Part 2

Identifying \begin{align*} R \times F = F[x] \oplus F \cong F[x] \oplus \frac{F[x]}{(f)} \end{align*} where \(f\) is any degree 1 polynomial in \(F[x]\), by the structure theorem we can pick a matrix \(A \in M_2(F[x])\) with invariant factors \(d_1 = 0, d_2 = f\). Then by the same argument given in part 1, we would have \begin{align*} (F[x])^2/\operatorname{im}A \cong \frac{ F[x] }{(d_1)} \oplus \frac{F[x]}{(d_2)} = F[x] \oplus \frac{F[x]}{(f)} \end{align*}

So we can choose \(n=2\), and say \(f(x) = x+1\), and then just pick a matrix that is already in Smith normal form: \begin{align*} \begin{align*} A = \left[ \begin{array}{cc} x+1 & 0 \\ 0 & 0 \end{array}\right]. \end{align*} \end{align*}

Problem 9

Part 1

Let \(M\) be a finitely generated module over \(R\) a PID.

Then \begin{align*} M \cong F \oplus \bigoplus_{i=1}^n R/(d_i) \end{align*}

where \(F\) is free of finite rank and \(R/(d_i)\) are cyclic torsion modules (the invariant factors) satisfying \(d_1 \divides d_2 \divides \cdots \divides d_n\).

Equivalently,

\begin{align*} M \cong F \oplus \bigoplus_{i=1}^n R/(p_i^{s_i}) \end{align*}

where \(F\) is free of finite rank, \(p^i \in R\) are (not necessarily distinct) prime elements (the elementary divisors), and \(s_i \in {\mathbf{Z}}^{\geq 1}\).

Part 2

Since \({\mathbf{Z}}^4\) is a finitely generated module over the PID \({\mathbf{Z}}\), the structure theorem applies, and we can write \(M \cong {\mathbf{Z}}^k \oplus \bigoplus {\mathbf{Z}}/(r_i)\) for some \(k\leq 4\) and some collection \(r_i\) of invariant factors.

If we write \(M \cong {\mathbf{Z}}^4/N\) where \(N\) is the submodule generated by the prescribed relations, then we can construct a homomorphism of \({\mathbf{Z}}{\hbox{-}}\)modules \(L: {\mathbf{Z}}^4 \to N\) which is given by the matrix \begin{align*} A_L = \left(\begin{array}{rrrr} 3 & 12 & 3 & 6 \\ 0 & 6 & 0 & 0 \\ -3 & 6 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \end{align*}

Then \(\operatorname{im}A_L \cong N\), and we can compute the Smith normal form, \begin{align*} SNF(A_L) = \left(\begin{array}{rrrr} 3 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \end{align*}

which shows that the invariant factors are \(3, 6, 6, 0\). We can thus write \(\operatorname{im}A_L \cong 3{\mathbf{Z}}\oplus 6{\mathbf{Z}}\oplus 6{\mathbf{Z}}\), and so \begin{align*} M \cong \frac{{\mathbf{Z}}^4}{3{\mathbf{Z}}\oplus 6{\mathbf{Z}}\oplus 6{\mathbf{Z}}} \cong {\mathbf{Z}}\oplus {\mathbf{Z}}/(3) \oplus {\mathbf{Z}}/(6) \oplus {\mathbf{Z}}/(6). \end{align*}

Problem 10

Part 1

An element \(x\in M\) is torsion iff there exists some nonzero \(r\in R\) such that \(rm = 0\), or equivalently \(\mathrm{Ann}(x) \neq 0\).

Part 2

Let \(R = {\mathbf{C}}[x],~ M = {\mathbf{C}}^2\), and \begin{align*} A = \left[\begin{array}{ccc} 1 & 2 \\ 2 & 1 \end{array}\right] \in M_2({\mathbf{C}}). \end{align*}

Then \({\mathbf{C}}^2\) is a module over \({\mathbf{C}}[x]\) with action given by \begin{align*} p(x) \curvearrowright\mathbf{v} \coloneqq p(A)\mathbf{v} \end{align*}

Then \(M\) is cyclic as an \(R{\hbox{-}}\)module and generated by the basis vector \([1, 0]^2 \in {\mathbf{C}}^2\), since \begin{align*} \begin{align*} (tA + s) \left[\begin{array}{cc} 1 \\ 0 \end{array}\right] &= \left[\begin{array}{cc} x \\ y \end{array}\right] \\ \implies \left[\begin{array}{cc} t & 2t \\ 2t & t \end{array}\right] \left[\begin{array}{cc} 1 \\ 0 \end{array}\right] + \left[\begin{array}{cc} s \\ 0 \end{array}\right] &= \left[\begin{array}{cc} x \\ y \end{array}\right] \\ \implies \left[\begin{array}{cc} t + s \\ 2t \end{array}\right] &= \left[\begin{array}{cc} x \\ y \end{array}\right] \\ \implies \left[\begin{array}{cc} 1 & 1 \\ 2 & 0 \end{array}\right] \left[\begin{array}{cc} t \\ s \end{array}\right] &= \left[\begin{array}{cc} x \\ y \end{array}\right] \\ ,\end{align*} \end{align*}

which is a linear system of equations represented by an invertible matrix, which always has a solution. So every \(\mathbf{v}\in {\mathbf{C}}^2\) is the image of some polynomial in \(A\).

It is then easy to see that \({\mathbf{C}}^2\) is torsion as a module over \({\mathbf{C}}[x]\), since by Cayley-Hamilton we have \(\mathrm{Ann}(A) = (\mathrm{minpoly}(A)) = (x^2-2x-3)\), and so letting \(p(x) = x^2-2x-3\), we find that \begin{align*} \forall~ \mathbf{v}\in {\mathbf{C}}^2 \quad p(A) \curvearrowright\mathbf{v} = 0 \curvearrowright\mathbf{v} = 0. \end{align*}

Part 3

Suppose \(R\) is a domain, \(M\) an \(R{\hbox{-}}\)module, and let \begin{align*} T(M) = \left\{{m \in M {~\mathrel{\Big\vert}~}rm = 0 \text{ for some $r\neq 0 \in R$}}\right\}. \end{align*}

Then \(T(R)\) is a submodule iff for all \(r\in R\) and all \(m,n\in T(M)\) we have \(rm + n \in T(M)\).

So pick annihilators \(a_m, a_n \neq 0 \in R\) where \(a_m m = 0\) and \(a_n n = 0\).

Since \(a_m \neq 0\) and \(a_n \neq 0\), the product \(a_m a_n \neq 0\) because \(R\) is a domain.

Since \(0 \in T(M)\), we can suppose \(rm+n \neq 0\) (otherwise this is in \(T(M)\) trivially). Then

\begin{align*} \begin{align*} a_m a_n (rm + n) &= a_m a_n r m + a_m a_n n \\ &=r a_n (a_m m) + a_m (a_n n) \\ &= r a_n 0 + a_m 0 \\ &= 0 .\end{align*} \end{align*}

where the commutativity of \(r, a_n, a_m\) follows from the fact that these are all elements of \(R\), which is a domain, and in particular is commutative. \(\hfill\blacksquare\)