Why does $M\otimes k(\mathfrak{m})=M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m}$? (From Matsumura, proof of Theorem 4.8.)

Question

Matsumura's Commutative Ring Theory, proof of Theorem 4.8, page 27, says:

Let $A$ be a ring, $M$ a finite $A$-module, and $\mathfrak{m}$ a maximal ideal. If $k(\mathfrak{m})=A_\mathfrak{m}/\mathfrak{m}A_\mathfrak{m}$ is the residue field of $A_\mathfrak{m}$, then
$$ M\otimes k(\mathfrak{m})=M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m}. $$

I know $A_m/mA_m\cong (A/m)_\bar{m}$, but even then $$ M\otimes k(m)\cong M\otimes (A/m)_\bar{m}=\cdots? $$

Even if $(A/m)_\bar{m}\cong A/m$, then doesn't $$ M\otimes k(m)\cong M\otimes A/m\cong M/mM $$ but that's not the same as $M_m/mM_m$, is it?

Answer 1

If $\mathfrak{p}$ is any prime ideal of $A$, we have

$$ A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} \cong \text{Frac}(A/\mathfrak{p}).$$

Let's assume this for now. In general, we have

$$M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p} = (A_\mathfrak{p} \otimes_A M) / (\mathfrak{p} A_\mathfrak{p} \otimes_A M) = (A_\mathfrak{p} / \mathfrak{p}A_\mathfrak{p})\otimes_A M,$$

where the second equality is by right-exactness of $ -\otimes_A M$. So by our assumption we have

$$M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p} \cong \text{Frac}(A/\mathfrak{p})\otimes_A M.$$

If $\mathfrak{p}=\mathfrak{m}$ is maximal, then $\text{Frac}(A/\mathfrak{m})=\text{Frac}(k(\mathfrak{m})) = k(\mathfrak{m})$, so that indeed

$$ M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m} = k(\mathfrak{m})\otimes_A M.$$

Now to show $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} \cong \text{Frac}(A/\mathfrak{p})$.

Since every non-zero element in $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is invertible, by the universal property of localization the map $A/\mathfrak{\mathfrak{p}} \rightarrow A_\mathfrak{p}/\mathfrak{p}A_p$ factors through an induced homomorphism $\text{Frac}(A/\mathfrak{p})\rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$, which is necessarily an injection (of fields). Now any element in $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is represented by an element $\frac{a}{b}$ in $A_\mathfrak{p}$, with $a\in \mathfrak{p}$ and $b\notin \mathfrak{p}$. Since $b\notin \mathfrak{p}$, it is non-zero in $\text{Frac}(A/\mathfrak{p})$, so $\frac{a}{b} \in \text{Frac}(A/\mathfrak{p})$, which maps to $\frac{a}{b}$ in $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$. This shows $\text{Frac}(A/\mathfrak{p}) \rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is also surjective.

Answer 2

This holds for any $A$-module:

$$ M\otimes_A k(\mathfrak{m})\simeq M\otimes_A (A_{\mathfrak m}\otimes_{A_{\mathfrak m}}k(\mathfrak{m}))\simeq (M\otimes_A A_{\mathfrak m})\otimes_{A_{\mathfrak m}}k(\mathfrak{m})\simeq$$ $$M_{\mathfrak m}\otimes_{A_{\mathfrak m}}k(\mathfrak{m})\simeq M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m} $$

The first isomorphism follows from $B⊗_BX≃X$, where $X$ is a $B$-module (see Matsumura, Formula 4, page 267), the second follows from the associativity of tensor product (or see Matsumura, Formula 10, page 268), while the third uses that $M_S\simeq M\otimes_AA_S$ (see Matsumura, Theorem 4.4, page 26).