# If for $A$ a commutative nonzero ring $A^m ≅ A^n$ as $A$-modules, then $m = n$

This is the problem I need to solve:

Let $A$ be a nonzero ring. Show that if $A^m ≅ A^n$, then $m = n$.

The book I got this problem from suggests using the following method to solve it:

Let $M$ be a maximal ideal of $A$ and let $f: A^m \rightarrow A^n$ be an isomorphism. Then $1\otimes f:(A/M) \otimes A^m \rightarrow (A/M) \otimes A^n$ is an isomorphism between vector spaces of dimensions $m$ and $n$ over the field $K = A/M$. Hence $m = n$.

The trouble is, I have no idea how to use this hint, mainly because I do not comprehend tensor products. We haven't gone over them in class due to some inclement weather closing it, and nothing I have read about them in the book or looking around on the internet, including a few answers to a question on this very website, makes sense to me. I just can't seem to get my head around what they are, how they're made from the modules they're made from, or what they're for.

Is there a way to solve this problem without using tensor products? And if there isn't, are there any proofs of other problems which use tensor products that I can read and maybe get an understanding of how to use this object in a proof?

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You can try to prove that $A/M\otimes A^n\cong(A/M)^n,$ by showing that the tensor product commutes with the direct sum, perhaps with the aid of some canonique maps? –  awllower Jan 31 at 15:27
What do you mean by "$\cong$"? Do you mean isomorphic as $A$-modules? If you mean isomorphic as rings, I do believe there are rings that satisfy $A^m \cong A^n$ for $m \neq n$. –  Hurkyl Jan 31 at 16:49
Indeed: an infinite product of copies of $F_2$ is such a ring. The problem needs to be phrased in terms of modules. –  rschwieb Jan 31 at 16:57
Okay, I clarified the title. I wasn't sure whether A^n was anything other than a module in the first place, which is why I didn't say so to begin with. –  Xindaris Jan 31 at 17:09

Show that $f(M^n) = M^m$. This is harder than it sounds, but not too hard. That means that $\bar f:A^n / M^n \to A^m / M^m : x + M^n \mapsto f(x) + M^m$ is well-defined and bijective; it is obviously $A$-linear. However $A^n / M^n \cong (A/M)^n$ so we get an isomorphism $\hat f : (A/M)^n \to (A/M)^m$ that is $A/M$-linear. Since $A/M$ is a field, and dimension is well defined for vector spaces like $(A/M)^n$ and $(A/M)^m$ we get $m=n$.

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Tensor products just let you skip that first step. –  Jack Schmidt Jan 31 at 16:04
Are there any hints on doing the first part? I've been trying to get it, but the best I can come up with is that f(Mm) is a submodule of An with the same operations as Mm, and f-1(Mn) is a submodule of Am with the same operations as Mn. I'm not really sure how to get from there to how f(Mm) is actually equal to f(Mn). –  Xindaris Feb 3 at 15:03
That's nearly it. $M^n = M \cdot A^n := \{\sum m_i \cdot v_i : m_i \in M, v_i \in A^n\}$ and $f(\sum m_i v_i) = \sum m_i f(v_i)$ so $f(M^n) = f(M \cdot A^n) = M \cdot f(A^n) = M \cdot A^m = M^m$. –  Jack Schmidt Feb 3 at 16:18

Yeah, I don't know why tesnor products are recommended here either.

Convince yourself that the existence of a module isomorphism between $R^n$ and $R^m$ is tantamount to the existence of an $n\times m$ matrix $A$ and an $m\times n$ matrix $B$ such that $AB=I_n$ and $BA=I_m$, all with entries in $R$. This is just basic linear algebra about what $Hom_R(R^n,R^m)$ looks like.

Now if you take a maximal ideal $M$ and apply the projection of $R$ onto $R/M$ to the entries of these matrices, you get matrices over a field with the same property. But we know this is not possible for fields since finite dimensional vector spaces have unique dimension.

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Just to help the OP learn tensor products: "reducing the entries of the matrix mod $M$" and "applying the projection of $R$ on $R/M$ to the entries of those matrices" are other ways of describing $1 \otimes f$. –  Jack Schmidt Jan 31 at 16:46
@JackSchmidt Very neat: that transforms the exercise into an illustration of an aspect of the tensor product. –  rschwieb Jan 31 at 16:56