# Problem Set 2 :::{.problem title="AM 2.1"} Show that \[ m\ZZ + n \ZZ = 1 \implies C_m \tensor_\ZZ C_n = 0 ,\] and more generally \[ C_m\tensor_\ZZ C_n \cong C_d, \qquad d = \gcd(m, n) .\] ::: :::{.solution} To fix notation, set $C_m \da \gens{x \st x^m} = \ts{1=x^0, x, x^2,\cdots, x^{m-1}}$, written multiplicatively. The $n$th power map $x\mapsto x^n$ induces a SES \[ 0 \to \ZZ \mapsvia{(\wait)^n} \ZZ \to C_n \to 0 \quad \in \zmod .\] Apply the right-exact functor $(\wait)\tensor_\ZZ C_m$ and use that $\ZZ\tensor_\ZZ (\wait) \homotopic \id$ to obtain \[ \cdots \to C_m \mapsvia{(\wait)^n} C_m \to \coker((\wait)^n) \cong C_n\tensor_\ZZ C_m \to 0 ,\] so it suffices to show surjectivity, that every element in $C_m$ has an $n$th root -- i.e that if $y\in C_m$ then $y=z^n$ for some $z\in C_m$. This immediately reduces to finding $n$th roots of the generator $x$, since if $y=z^n \in C_m$, writing $y=x^k$ for some $k$, we have \[ y=x^k = z^n \implies z = x^{k\over n} = (x^{1\over n})^k ,\] and thus $z$ can be expressed as a power of an $n$th root of $x$. That such a root can always be found follows from Bezout's identity: since $m, n$ are coprime, there are solutions $(a, b)$ to $1 = am + bn$, so \[ x = x^1 = x^{am + bn} = x^{am} x^{bn} = (x^b)^n ,\] using that $(\wait)^m$ annihilates every element in $C_m$, making $x^b$ an $n$th root of $x$. More generally, using the same resolution and tensoring with any $A\in \zmod$ yields \[ \cdots \to A \mapsvia{(\wait)^n} A \to \coker( (\wait)^n ) \cong {A\over nA} \to 0 ,\] the submodule of $n\dash$divisible elements, and take $A = C_m$ to get \[ {A\over nA} = {C_m \over n C_m} \cong {\ZZ \over m\ZZ + n \ZZ }\cong {\ZZ\over d\ZZ} \cong C_d .\] ::: :::{.remark} Note that similarly applying $\Hom_{\zmod}(C_n, \wait)$ yields \[ \Hom_{\zmod}(C_n, C_m) \cong \ker( (\wait)^n ) \cong C_d .\] ::: :::{.problem title="AM 2.2"} Let $A\in \CRing, \mfa \in \Id(A), M\in \mods{A}$. Show that $(A / \mfa) \tensor_{A} M \cong M / \mfa M \in \mods{A}$. > Tensor the exact sequence $0 \rightarrow \mathfrak{a} \rightarrow A \rightarrow A / \mathfrak{a} \rightarrow 0$ with $M$. ::: :::{.solution} Applying the hint yields the following: \begin{tikzcd} 0 && \mfa && A && {A/\mfa} && 0 \\ \\ \cdots && {\mfa \tensor_A M} && {A\tensor_A M \cong M} && {A/\mfa \tensor_A M} && 0 \arrow[from=1-1, to=1-3] \arrow[""{name=0, anchor=center, inner sep=0}, "\iota", hook, from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-1, to=3-3] \arrow[""{name=1, anchor=center, inner sep=0}, "{\iota_*}", hook, from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow["{(\wait)\tensor_A M}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Thus it suffices to show $\im \iota_* \cong \mfa M$. This is clear since $\mfa \injects A$ is an inclusion, and the natural map $A\tensor_A M \iso M$ is given by $(a, m) \mapsto am$. ::: :::{.remark} Note that there is a map \[ f: \mfa \times M &\to \mfa M \\ (a, m) &\mapsto am ,\] which is clearly surjective and bilinear, lifting to map out of the tensor product by the universal property. However, it is *not* always an isomorphism, and it being an isomorphism for all ideals is equivalent to $M$ being flat as an $A\dash$module. In other words, \[ \mfa \tensor_A M \cong \mfa M \quad \forall \mfa\in\Id(A) \iff M\in \mods{A}^\flat .\] An easy counterexample: - $A = k[\eps]/\gens{\eps^2}$ - $\mfa = \gens{\eps} \in \Id(A)$ - $M = \mfa$ Then $\mfa\tensorpower{A}{2} \to \mfa^2$ is not injective. Note that $\mfa^2=0$ in $A$, so it STS $\mfa\tensorpower{A}{2}\neq 0$. The claim is that \[ \mfa\tensorpower{A}{2} \cong \mfa\tensorpower{A/\mfa}{2} \cong \mfa\tensorpower{k}{2} ,\] which is the tensor product of two 1-dimensional $k\dash$vector spaces, and is thus 1-dimensional over $k$. ::: :::{.problem title="AM 2.9"} Let \[ 0\to A \mapsvia{d_1} B \mapsvia{d_2} C \to 0 \in \mods{A} \] with $A, C\in \mods{A}^\fg$, and show $B\in \mods{A}^\fg$ (i.e. $B$ is finitely generated as an $A\dash$module). ::: :::{.solution} Let $\mcc, \mca$ be generators for $C$ and $A$ respectively, and consider \[ \mcb \da \ts{ d_1(a) \st a\in \mca }\union \ts{\tilde c \in d_2\inv(c) \st c\in \mcc} ,\] where the $\tilde c$ are arbitrarily chosen lifts of the generators $c\in \mcc$. Then $\mcb$ is a finite set, and the claim is that it generates $B$ as an $A\dash$module. ::: :::{.problem title="AM 2.11"} Let $A$ be a ring $\neq 0$. - Show that $A^{m} \cong A^{n} \Rightarrow m=n$. - If $\phi: A^{m} \rightarrow A^{n}$ is surjective, then $m \geq n .$ - If $\phi: A^{m} \rightarrow A^{n}$ is injective, is it always the case that $m \leq n$? > Hint: Let $m$ be a maximal ideal of $A$ and let $\phi: A^{m} \rightarrow A^{n}$ be an isomorphism. Then $1 \otimes \phi:(A / m) \otimes A^{m} \rightarrow(A / m) \otimes A^{n}$ is an isomorphism between vector spaces of dimensions $m$ and $n$ over the field $k=A / m .$ Hence $m=n .$ (Cf. Chapter 3, Exercise 15.) ::: :::{.problem title="AM 2.14"} A partially ordered set $I$ is said to be a directed set if for each pair $i, j$ in $I$ there exists $k \in I$ such that $i \leq k$ and $j \leq k$. Let $A$ be a ring, let $I$ be a directed set and let $\left(M_{i}\right)_{i \in I}$ be a family of $A$-modules indexed by $\cdot I$. For each pair $i, j$ in $I$ such that $i \leq j$, let $\mu_{i j}: M_{i} \rightarrow M$, be an $A$-homomorphism, and suppose that the following axioms are satisfied: (1) $\mu_{l t}$ is the identity mapping of $M_{i}$, for all $i \in I$; (2) $\mu_{i k}=\mu_{j k} \circ \mu_{l y}$ whenever $i \leq j \leq k$. Then the modules $M_{1}$ and homomorphisms $\mu_{i f}$ are said to form a direct system $\mathbf{M}=\left(M_{t}, \mu_{t j}\right)$ over the directed set $I$. We shall construct an $A$-module $M$ called the direct limit of the direct system M. Let $C$ be the direct sum of the $M_{1}$, and identify each module $M_{i}$ with its canonical image in $C$. Let $D$ be the submodule of $C$ generated by all elements of the form $x_{i}-\mu_{t j}\left(x_{i}\right)$ where $i \leq j$ and $x_{i} \in M_{i}$. Let $M=C / D$, let $\mu: C \rightarrow M$ be the projection and let $\mu_{i}$ be the restriction of $\mu$ to $M_{i}$. The module $M$, or more correctly the pair consisting of $M$ and the family of homomorphisms $\mu_{i}: M_{1} \rightarrow M$, is called the direct limit of the direct system $M$, and is written $\colim_i M_{i}$. From the construction it is clear that $\mu_{t}=\mu_{j} \circ \mu_{i f}$ whenever $i \leq j$. ::: :::{.problem title="AM 2.16"} Show that the direct limit is characterized (up to isomorphism) by the following property. Let $N$ be an $A$-module and for each $i \in I$ let $\alpha_{i}: M_{t} \rightarrow N$ be an $A-$ module homomorphism such that $\alpha_{t}=\alpha_{j} \circ \mu_{t s}$ whenever $i \leq j$. Then there exists a unique homomorphism $\alpha: M \rightarrow N$ such that $\alpha_{1}=\alpha \circ \mu_{t}$ for all $i \in I$. ::: :::{.problem title="AM 2.20"} Keeping the same notation as in Exercise 14 , let $N$ be any $A$-module. Then $\left(M_{1} \otimes N, \mu_{i j} \otimes 1\right)$ is a direct system; let $P=\colim_i \left(M_{i} \otimes N\right)$ be its direct limit. For each $i \in I$ we have a homomorphism \[ \mu_{i} \otimes 1: M_{i} \otimes N \rightarrow M \otimes N ,\] hence by Exercise 16 a homomorphism $\psi: P \rightarrow M \otimes N$. Show that $\psi$ is an isomorphism, so that \[ \colim_i \left(M_{i} \otimes N\right) \cong\left(\colim_{i} M_{i}\right) \otimes N . \] > Hint: For each $i \in I$, let \[ g_{i}: M_{t} \times N \rightarrow M_{t} \otimes N \] be the canonical bilinear mapping. Passing to the limit we obtain a mapping $g: M \times N \rightarrow P$. Show that $g$ is $A$-bilinear and hence define a homomorphism $\phi: M \otimes N \rightarrow P$. Verify that $\phi \circ \psi$ and $\psi \circ \phi$ are identity mappings. :::