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I know that if $R$ is a commutative ring, then $R^n\simeq R^m$ as $R$-modules is equivalent to $n=m$.---(1)

But there is some confusing example.

Let $A=\prod_{n\in \mathbb{N}}R$. Then $A$ is a commutative ring and $$A^2=\prod_{n\in \mathbb{N}}R\times \prod_{n\in \mathbb{N}}R=\prod_{n\in \mathbb{N}}R=A.$$ Thus $A^2$ and $A$ should be isomorphic as $R$-modules. But it contradicts (1) above. What's wrong?

I also have one more question.

If $M$ is free $R$-module and have a finite basis $S$, is it possible to have another infinite basis?

I know if $R$ is commutative ring with unity, the number of basis should be same. But if it is not, I think it is not true. Is there a counter example?

Thanks in advance for your help.

user26857
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Andrew
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  • I don't think your statement (1) is correct as your (counter)example seems legit. – Quang Hoang Oct 12 '15 at 10:27
  • I think that (1) works, because we can choose a maximal ideal of $R$. Then $R/m$ is a field and taking a tensor product $R^n\simeq R^m$ with $(R/m)$, we get $R/m$-module isomorphism between them. Then by the invariance property of dimension of vector sapce over $R/m$, we get to know $n=m$. – Andrew Oct 12 '15 at 10:57
  • Does such an ideal $m$ exist? – Quang Hoang Oct 12 '15 at 13:20
  • Sure, by Krull theorem. – Andrew Oct 12 '15 at 13:46
  • @Andrew - this proof is missing something as $R/m$ is probably not flat over $R$, thus we do not know right away that the tensored map is an isomorphism. (It will be surjective, as tensor product is right exact, but maybe not injective.) However, this gives us $m\geq n$, and then symmetry finishes the proof. Alternatively, see here: http://math.stackexchange.com/questions/106786/am-hookrightarrow-an-implies-m-leq-n-for-a-ring-a-neq-0. – Ben Blum-Smith Oct 12 '15 at 14:27
  • @Ben Blum-Smith, Thank you for your caution. But I just followed the proof of exercise 2.11 of Attiyah & Macdonald. In the proof, it seems that they overlook the flatness of $R/m$ and explains that $R/m \otimes_R R^n \simeq R/m \otimes_R R^m$ from $R^n\simeq R^m$ in an innocuous way. Are they also missing the flatness of $R/m$? – Andrew Oct 12 '15 at 17:03
  • @BenBlum-Smith $M\simeq N\Rightarrow M\otimes_RX\simeq N\otimes_RX$, no matter what $X$ is. – user26857 Oct 13 '15 at 07:13
  • @Andrew and user26857, point taken. I actually had ex. 2.11 of Atiyah & MacDonald in mind; in particular I was recalling that proving that injectivity of $R^m\rightarrow R^n$ implies $m\leq n$ (which is true) requires more than just tensoring with $R/m$ for some maximal $m$. But as user26857 points out, of $M,N$ are actually isomorphic, then of course they are still isomorphic after tensoring by something, flat or not. – Ben Blum-Smith Oct 13 '15 at 17:17

2 Answers2

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In your example $A\simeq A^2$ as rings, but not as $A$-modules. (As you already know, this is not possible.)

A free module can't have a finite and an infinite basis. Suppose it has a finite basis $x_1,\dots,x_n$ and an infinite basis $(y_i)_{i\in I}$. Then each $x_j$ is a (finite) linear combination of the $y_i$'s. Let $J\subset I$ be a finite subset which contains all indices of the $y_i$'s which appear in a linear combination as mentioned above. Then they form a system of generators (actually a basis) for our module, so any other $y_j$ is a linear combination of them, a contradiction.

user26857
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  • Thank you very much. I thought that the map $f:A^2\to A$ defined by $(x_1,x_2,\cdots)\times(y_1,y_2,\cdots)->(x_1,y_1,x_2,y_2,\cdots)$ is $A$-module isomorphism but it turned out only a ring isomorphism. But I am also curious that how can we know they are not isomorphic as $A$-module? I am sure it should be, but proving this is a different one. – Andrew Oct 12 '15 at 16:46
  • A is commutative, so A and AxA can not be isomorphic as A-modules. – user26857 Oct 12 '15 at 16:50
  • I see. It follows from (1). – Andrew Oct 12 '15 at 17:06
  • To see it concretely, consider that your map $f$ is not compatible with the $A$-module multiplication. E.g. take $R=\mathbb{Z}$. Then multiplication of $((x_1,x_2,\dots),(y_1,y_2,\dots))\in A^2$ (seen as an $A$-module) by, say, $(1,2,3,\dots)$, yields $((x_1,2x_2,\dots),(y_1,2y_2,\dots))$, and the image of this under $f$ is $(x_1,y_1,2x_2,2y_2,\dots)$. On the other hand, if you take the image under $f$ and then multiply, you get $(x_1,2y_1,3x_2,4y_2,\dots)$, which is different. – Ben Blum-Smith Oct 13 '15 at 17:26
  • @Andrew - I forget if the software pings you automatically but that last comment was intended for you. – Ben Blum-Smith Oct 13 '15 at 20:46
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an $A$-module is a commutative group $M$ together with a multiplication map $\mu : A \times M \to M$.

In your counter-example, you have got two $A$-modules $(M_1 = A^2, \mu_1)$ and $(M_2 = A,\mu_2)$ and you have argued that they are isomorphic as $R$-modules (there is a group isomorphism between them that is also compatible with the unrelated multiplication maps $\nu_i : R \times M_i \to M_i$).

But this isomorphism (that you didn't completely describe) is not compatible with the maps $\mu_i$.

For example, the image of $\mu_1(1,0,0,\ldots)$ is an $R$-module of rank $2$, while the image of $\mu_2(1,0,0,\ldots)$ is an $R$-module of rank $1$.

mercio
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  • My claim is $A$ and $A^2$ is isomorphic as $A$ modules. (here the $A$ multiplication on these $A$ modules arises in the digonal way) – Andrew Oct 12 '15 at 13:50
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    yes but your "proof" is not a proof that they're isomorphic as $A$-modules, but is is only a proof that they're isomorphic as $R$-modules – mercio Oct 12 '15 at 13:58
  • Sorry, your are right. I edited my claim that they are ring isomorphic not as $A$-modules. – Andrew Oct 12 '15 at 17:00