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Let $A$ be a ring $\neq 0$, and $\mathcal{m}$ a maximal ideal of $A$. Then both of the field $A / m$ and $A^{n}$ (n-tuple direct sum) are $A$-modules.

Question: How to show that $(A / m) \otimes_A A^{n}$ is a $A / m$-vector space of dimension $n$.

It seems that we should establish an $A / m$-isomorphism from $(A / m) \otimes_A A^{n}$ to $(A / m)^n$. But I don't know how to write this precisely. Can any one write this precisely? Thanks in advance.

Lelong Wang
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1 Answers1

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You can use the well known basic properties of tensor products:

$M\otimes_A (N\oplus P)\cong (M\otimes_A N)\oplus (M\otimes_A P)$

$M\otimes_A A\cong M$

The isomorphisms are very natural. So in your case we have:

$(A/m)\otimes_A A^n=(A/m)\otimes (\bigoplus\limits_{i=1}^n A)\cong\bigoplus\limits_{i=1}^n ((A/m)\otimes_A A)\cong\bigoplus\limits_{i=1}^n (A/m)=(A/m)^n$

These are all isomorphisms of $A$-modules. But now just check that they are are also homomorphisms over $A/m$. (you already know they are closed under addition and are invertible, so you just have to check scalar multiplication)

Mark
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