Why this tensor product is a vector space?

Question

Let $A$ be a commutative ring with unity, $\mathcal{m}\subset A$ a maximal ring of $A$. Let $M$ be an $A-$module finitely generated.

Why $(A/\mathcal{m})\otimes_{A} M$ is a finite dimensional vector space? I really have no intuition about tensor products so I can't see it.

Answer 1

If $M$ is an $A$-module (finitely generated), then it is an abelian group under addition with a multiplication from $A$. You can think of this as a vector space, but instead of having a scalar multiplication from a field, you have a scalar multiplication from a commutative ring $A$.

If $m \subset A$ is a maximal ring, then $A/m$ is a field.

In the tensor product $(A/\mathcal{m})\otimes_{A} M$, elements look like pairs $(r,x)$ for $x \in M$ and $r \in A/m$.

Now, since $M$ is a finitely generated $A$-module, all elements are linear combinations of some finite basis set with coefficients in $A$. I think that it is easy enough to check that the axioms for a finitely generated vector space hold, but I'm happy to provide more help in the comments if needs be.

Answer 2

Generally, given three commutative rings with unity, say, $A,B,C$ and two bimodules $_B N_A, _A M_C$ one can form the tensor product over $A$, $M \otimes_A N$ which is a left-module over B and a right-module over. These module structure are naturally inherited by of $M, N$. As a speacial case, you have $N=A/ \mathfrak m$ is a residue field so $(A/\mathfrak m) \otimes M$ is a finite-dimensional vector space over $A/\mathfrak m$. Moreover, $\left \{ 1 \otimes m_i \right \}$ generates it as a vector space where $\left \{m_i \right \}$ generates $M$ as a $A$ - module and this proves the tensor product is finite-dimensional.