HW5

Text-Ex. 2.2. Let $A$ be a ring, $a$ an ideal, $M$ an $A$-module. Show that $(A/a){\otimes }_{A}M$ is isomorphic to $M/aM$. (Tensor the exact sequence $0\to a\to A\to A/a\to 0$ with $M$.)

\begin{proof} Note that $0\to a\to A\to A/a\to 0$ is exact, then $a\otimes M\to A\otimes M\to A/a\otimes M\to 0$ is exact. Define $\pi :A\to A/a$, then $\tilde{\pi }:A\otimes M\to A/a\otimes M$ is defined by $\tilde{\pi }=\pi \otimes \mathrm{id}$, and $\mathrm{im}\pi =A\otimes M/\ker \pi \simeq A/a\otimes M$. It suffices to show $A\otimes M/\ker \pi \simeq M/aM$. Recall that $\phi :A\otimes M\to M,b\otimes m\mapsto bm$ is an isomorphism.

If $b\otimes m\in \ker \tilde{\pi }$ with $b\in A$, then $\overline{b}\otimes m=0$, which deduces that $\overline{b}m={\overline{c}}^{\prime }{m}^{\prime }$ for some ${\overline{c}}^{\prime }\in a$ and so $\phi (\ker \tilde{\pi })\subseteq aM$. Conversely, each element $t\in aM$ has a preimage $c\otimes m$ with $c\in a$, $\phi (c\otimes m)=t$ and $\tilde{\pi }(c\otimes m)=0$. Therefore, $\phi (\ker \tilde{\pi })=aM$ and so $\ker (\tilde{\pi }\circ \phi )=aM$. Since $\tilde{\pi }\circ \phi :M\to (A/a){\otimes }_{A}M$ is surjective, we have $M/aM\simeq (A/a){\otimes }_{A}M$. \end{proof}

Text-Ex. 2.3. Let $A$ be a local ring, $M$ and $N$ finitely generated $A$-modules. Prove that if $M\otimes N=0$, then $M=0$ or $N=0$. (Let m be the maximal ideal, $k=A/\mathrm{m}$ the residue field. Let $M_k=k{\otimes }_{A}M\simeq$ $M/\mathrm{m}M$ by Exercise 2. By Nakayama’s lemma, $M_k=0\Rightarrow M=0$. But $M{\otimes }_{A}N=0\Rightarrow {\left(M{\otimes }_{A}N\right)}_k=0\Rightarrow M_k{\otimes }_k{N}_k=0\Rightarrow M_k=0$ or $N_k=0$, since $M_k,N_k$ are vector spaces over a field.)

\begin{proof} Let $m$ be the unique maximal ideal of $A$, and let $k:=A/m$. Then $k$ is a field. By Exercise 2, $M_k=k{\otimes }_{A}M\simeq M/mM$ and $N_k=k{\otimes }_{A}N\simeq N/mN$. Since $M{\otimes }_{A}N=0$, by Exercise 2.15 one have

$$0=k{\otimes }_A(M{\otimes }_{A}N){\otimes }_{A}k=(k{\otimes }_{A}M){\otimes }_A(N{\otimes }_{A}k)=M_k{\otimes }_A{N}_k.$$

Furthermore, $M_k{\otimes }_A{N}_k\simeq M_k{\otimes }_k{N}_k$ as $M_k,N_k$ are $k$-modules.

Assume that $M={⟨m_1,\cdots ,m_s⟩}_A$ and $N={⟨n_1,\cdots ,n_t⟩}_A$. Since $M_k=k{\otimes }_{A}M\simeq M/mM$, $M_k$ is a vector space over $k$, where $\{1\otimes m_1,\cdots ,1\otimes m_s\}$ is a set of basis. Similarly, $N_k$ is also a $k$-vector space, where $\{1\otimes n_1,\cdots ,1\otimes n_t\}$ is a set of basis. Then $M_k{\otimes }_k{N}_k$ is a $k$-vector space, whose basis is $\{(1\otimes m_i)\otimes (1\otimes n_j):i=1,\cdots ,s;j=1,\cdots ,t\}$ and $\dim M_k{\otimes }_k{N}_k=st$.

However, $M_k{\otimes }_k{N}_k\simeq M_k{\otimes }_A{N}_k=0$ yields that $st=0$ and one of $s,t$ is $0$. That is, one of $M_k,N_k$ is trivial. WLOG we assume that $M_k=0$. By Nakayama’s lemma, as $M=mM$, that is $M_k=0$, we have $M=0$. \end{proof}

4. Let $M_i(i\in I)$ be any family of $A$-modules, and let $M$ be their direct sum. Prove that $M$ is flat $\iff$ each $M_i$ is flat. Here we assume $|I|<\infty$.

\begin{proof} For any exact sequence $0\to N\stackrel{\phi }{\to }K\stackrel{\psi }{\to }L\to 0$, it suffices to show that the sequence

$$0\to \left({\oplus }_{i\in I}{M}_i\right)\otimes N\stackrel{\mathrm{id}\otimes \phi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes K\stackrel{\mathrm{id}\otimes \psi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes L\to 0$$

is exact iff $0\to M_i\otimes N\to M_i\otimes K\to M_i\otimes L\to 0$ is exact for all $i\in I$.

Note that there is an isomorphism ${\theta }_N$ such that $\left({\oplus }_{i\in I}{M}_i\right)\otimes N\stackrel{{\theta }_N}{\simeq }{\oplus }_{i\in I}{M}_i\otimes N$, and we can define ${\theta }_K,{\theta }_L$ similarly. Since $\mathrm{id}\otimes \phi$ is injective iff

$${\phi }^{\ast }={\theta }_K\circ (\mathrm{id}\otimes \phi )\circ {\theta }_N^{-1}:{\oplus }_{i\in I}(M_i\otimes N)\to {\oplus }_{i\in I}(M_i\otimes K)$$

is injective iff ${\phi }^{\ast }{|}_{M_i\otimes N}$ is injective for all $i$ iff $M_i\otimes N\to M_i\otimes K$ is injective for all $i$, we know $0\to \left({\oplus }_{i\in I}{M}_i\right)\otimes N\stackrel{\mathrm{id}\otimes \phi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes K$ is exact iff $0\to M_i\otimes N\to M_i\otimes K$ is exact for all $i$. Similarly we can prove $\left({\oplus }_{i\in I}{M}_i\right)\otimes K\stackrel{\mathrm{id}\otimes \psi }{\to }\left({\oplus }_{i\in I}{M}_i\right)\otimes L\to 0$ is exact iff $M_i\otimes K\to M_i\otimes L\to 0$ is exact for all $i$. \end{proof}

Text-Ex 2.7. Let $p$ be a prime ideal in $A$. Show that $p[x]$ is a prime ideal in $A[x]$. If $m$ is a maximal ideal in $A$, is $m[x]$ a maximal ideal in $A[x]$ ?

\begin{proof} To show $p[x]$ is a prime ideal in $A[x]$, it suffices to show $A[x]/p[x]$ is an integral domain. Note that $A[x]/p[x]\simeq (A/p)[x]$. Assume that $f(x)={\sum }_{i=0}^n{a}_i{x}^i$ and $g(x)={\sum }_{j=0}^m{b}_j{x}^j$ are elements of $(A/p)[x]$ with $f(x)g(x)=0$, then $a_n{b}_m\in p$ and so $a_n{b}_m=0$ with $a_n,b_m\ne 0$, which is impossible as $p$ is a prime ideal of $A$.

If $m$ is a maximal ideal in $A$, $m[x]$ is not necessarily a maximal ideal in $A[x]$. For example, $2\mathbb{Z}\subseteq \mathbb{Z}$ is a maximal ideal, but $\mathbb{Z}[x]/2\mathbb{Z}[x]\simeq F_2[x]$ is not a field, as $x\ne 0$ is not a unit. \end{proof}

Text-Ex. 2.8.

• i) If $M$ and $N$ are flat $A$-modules, then so is $M{\otimes }_{A}N$.
• ii) If $B$ is a flat $A$-algebra and $N$ is a flat $B$-module, then $N$ is flat as an $A$-module.

\begin{proof} i) It suffices to show for any injective map $f:L^{\prime }\hookrightarrow L$, $f\otimes 1:L^{\prime }\otimes (M\otimes N)\to L\otimes (M\otimes N)$ is flat. Since $M$ is flat, the map $g:L^{\prime }\otimes M\to L\otimes M$ is injective. Since $N$ is flat, $k:(L^{\prime }\otimes M)\otimes N\to (L\otimes M)\otimes N$ is injective. Recall that there exists unique isomorphism $(M_0\otimes N_0)\otimes P_0\to M_0\otimes (N_0\otimes P_0)$, then $k$ injective yields that $f\otimes 1$ injective. Therefore, $M{\otimes }_{A}N$ is flat.

ii) For any injective map $h:L^{\prime }\hookrightarrow L$ with $A$-modules $L,L^{\prime }$, we aim to show $h\otimes 1_N:L^{\prime }{\otimes }_{A}N\to L{\otimes }_{A}N$ is injective. Since $B$ is a flat $A$-algebra, $L^{\prime }{\otimes }_{A}B\to L{\otimes }_{A}B$ is injective. Since $N$ is a flat $B$-module and $L^{\prime }{\otimes }_{A}B,L{\otimes }_{A}B$ are $B$-modules, $(L^{\prime }{\otimes }_{A}B){\otimes }_{B}N\to (L{\otimes }_{A}N){\otimes }_{B}N$ is injective. By Exercise 15, $(L{\otimes }_{A}B){\otimes }_{B}N\simeq L{\otimes }_A(B{\otimes }_{B}N)\simeq L{\otimes }_{A}N$. Therefore, $h\otimes 1_N$ is injective and so $N$ is flat as an $A$-module. \end{proof}