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Let $A$ be a local ring with maximal ideal $\mathfrak{m}$, and let $M$ be an $A$-module.

I want to turn the following object into an $A/\mathfrak{m}$-module: $$A/\mathfrak{m} \otimes_A M$$

I attempt to multiply in this way:

$$\left( [a],\sum_{i = 1}^n [b_i] \otimes_A m_i \right) \mapsto \sum_{i=1}^n [ab_i] \otimes_A m_i$$

(Here, $[a]$ is the class of $a \in A$ modulo $\mathfrak{m}$.)

Is this idea correct, and why or why not?

I think the fact that $\mathfrak{m} = \text{Ann}(A/\mathfrak{m})$ has something to do with it.

Thanks!

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    $\begingroup$ It might be easier to see that $A/\mathfrak m\otimes_A M\cong M/\mathfrak mM$ as an $A$-module. $\endgroup$ Feb 5, 2015 at 1:38
  • $\begingroup$ Is the isomorphism $\sum_{i=1}^n [b_i] \otimes_A m_i \mapsto \sum_{i=1}^n [b_i]m_i$? $\endgroup$ Feb 5, 2015 at 1:47
  • $\begingroup$ That's the map. It's easier than that. Do you know the universal definition of tensor product. You have a bi-linear map: $A/\mathfrak{m}\times M\to M/\mathfrak{m}M$, defined as $([a],m)\mapsto [am]$. You need to show this is well-defined and bilinear. $\endgroup$ Feb 5, 2015 at 1:52
  • $\begingroup$ So by the universal property I get an $A$-map from $A/\mathfrak{m} \otimes_A M$ to $M/\mathfrak{m}M$ which I can then show is bijective. $\endgroup$ Feb 5, 2015 at 1:58
  • $\begingroup$ Or you can show that $M/\mathfrak{m}M$ satisfies the universal property. $\endgroup$ Feb 5, 2015 at 2:01

2 Answers 2

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I believe this is better understood when non (necessarily) commutative rings are involved.

If $R$ and $S$ are rings, an $R$-$S$-bimodule $_RM_S$ is an abelian group $M$ which is also a left $R$-module and a right $S$-module, so that $$ (rx)s=r(xs) $$ for all $r\in R$, $x\in M$ and $s\in S$.

If $M_S$ is a right $S$-module and $N$ is a left $S$-module, then $M\otimes_SN$ has nothing more than the structure of abelian group, generally speaking. However, if $_RM_S$ is a bimodule, then $M\otimes_SN$ can be regarded as a left $R$-module by defining $$ r(x\otimes y)=(rx)\otimes y $$ extended by linearity (only with respect to sums). This is well defined, as the map $$ l_r\colon M\times N\to M\otimes_SN \qquad l_r(x,y)=(rx)\otimes y $$ is balanced, so it induces a group homomorphism $\lambda_r\colon M\otimes_SN\to M\otimes_SN$ such that $\lambda_r(x\otimes y)=(rx)\otimes y$. It is routine to verify that we get a left $R$-module structure on $M\otimes_SN$ by defining $r(x\otimes y)=(rx)\otimes y$ (extended by additive linearity).

When a ring $A$ is commutative, any $A$-module can be regarded as an $A$-$A$-bimodule in an obvious way; it's this feature that allows to regard $M\otimes_AN$ as an $A$-module, using the above setup.

In your case, $A/\mathfrak{m}$ is an $A/\mathfrak{m}$-$A$-bimodule, so $A/\mathfrak{m}\otimes_AM$ becomes an $A/\mathfrak{m}$-module by $$ [a]([x]\otimes y)=[ax]\otimes y $$ extended by additive linearity.

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Whenever you have an algebra $A \to B$ and an $A$-module $M$ the tensor product of $A$-modules $B \otimes_A M$ has the structure of a $B$-module. You can construct this cleanly via the $A$-trilinear map \begin{align*} B \times B \times M &\to B \otimes_A M\\ (b, x, m) &\mapsto (bx) \otimes m. \end{align*} But your comment about annihilators is good too and ties into this construction: if $I$ is an ideal of $A$ then $(A/I) \otimes_A M \simeq M/(IM)$ and if $I$ is contained in the annihilator of $M$ then $IM = 0$, so $(A/I) \otimes_A M \simeq M$. But maybe it's better to show this without tensor products.

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