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well I'm having a hard time understanding the tensor product. Here's a problem from Atiyah and Macdonald's book:

Let $A$ be a non-trivial ring and let $m,n$ be positive integers. Let $f: A^{n} \rightarrow A^{m}$ be an isomorphism of $A$-modules. Show this implies that $n=m$.

Well the solution is at follows:

Let $m$ be a maximal ideal of $A$. Then we have an induced isomorphism:

$(A/m) \otimes_{A} A^{n} \rightarrow (A/m) \otimes_{A} A^{m}$.

Now it says that this is an isomorphism between vector spaces of dimension $n$ and $m$.

My questions are:

1) How do we know that $(A/m) \otimes_{A} A^{n}$ is a vector space over $A/m$ ? 2) How do we know it has exactly dimension $n$?

Is there some "standard" theorem that tells us this? Can you please explain this in detail?

Thanks

user6495
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    By the way, Atiyah has written many books. I suspect the book you're talking about is the (unique) one written by M.F. Atiyah and I.G. Macdonald. It is not very respectful to refer to a book by only one of the authors' names (put yourself in Macdonald's shoes). – Pete L. Clark Mar 19 '11 at 01:05
  • @Pete L. Clark: you're right, sorry. – user6495 Mar 19 '11 at 01:12
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    Also, it never hurts to mention that your rings are commutative... – Mariano Suárez-Álvarez Mar 19 '11 at 01:40
  • You can find a simple proof avoiding tensor products in Corollary IV.2.12 p. 186 of Hungerford's Algebra. – Bill Dubuque Mar 19 '11 at 19:15
  • I hope not to break any rule about that but instead of an answer I'm posting a (even strictly related) question...

    Does the former claim remain true if the rings involved are noncommutative? I mean, in the non commutative case there are lots of rings having no IBN. But what happens for modules?

     – fosco Mar 19 '11 at 09:36
  • @tetra: as you say, this is not an answer so should have been left as a comment or further question. (Well, no big deal. Now you know...) Anyway, your comment suggests that you may not have completely grasped the concept of IBN: indeed it is precisely the property that the rank of a finitely generated free module is well-defined. So, as you say, there are definitely non-commutative rings in which this question has a negative answer. From the perspective of the proofs given here, the issue is that a non-commutative ring need not admit a quotient which is a division ring. – Pete L. Clark Mar 19 '11 at 18:54

2 Answers2

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1) Let $f: A \rightarrow B$ be any homomorphism of commutative rings and let $M$ be any $A$-module. Then one may view $B \otimes_A M$ as a $B$-module via $b \cdot (\sum_i b_i \otimes x_i) := \sum_i b b_i \otimes x_i$. This process of starting with an $A$-module and getting a $B$-module is called base change.

In your particular case the map is $f: A \rightarrow A/\mathfrak{m}$, so $(A/\mathfrak{m}) \otimes_A M$ is an $A/\mathfrak{m}$-module. But since $\mathfrak{m}$ is maximal, $A/\mathfrak{m}$ is a field, and a module over a field is simply a vector space over that field. This answers your first question.

2) You should establish the following two properties of tensor products:

(i) For any $A$-module $M$, there is a canonical isomorphism $M \otimes_A A \stackrel{\sim}{\rightarrow} M$. (Thus "changing the base from $A$ to $A$" has no effect.)

(ii) For any family $M_i$ of $A$-modules indexed by a set $I$ and any $A$-module $M$, there is a canonical isomorphism $(\bigoplus_{i \in I} M_i) \otimes_A N \stackrel{\sim}{\rightarrow} \bigoplus_{i \in I} (M_i \otimes_A N)$.

What you want follows easily from these facts (which are themselves ubiquitously useful). Try it yourself and ask again if you have further questions.

Pete L. Clark
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  • OK thanks 1) is clear. Now for 2) can we simply say $A^{n} \otimes A/m \cong (A \otimes A/m)^{n}$. Now observe $A/m$ is an A-module so by (i) we have $(A \otimes A/m)^{n} \cong (A/m)^{n}$ which is a vector space of dimension $n$. ? – user6495 Mar 19 '11 at 01:11
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    @user6495: Well, you can "simply" say it once you show (ii) in what Pete writes, because $(A\otimes A/\mathfrak{m})^n = \oplus_{i=1}^n(A\otimes A/\mathfrak{m})$. So if you have shown the tensor distributes over direct sums, then the isomorphism you assert follows. – Arturo Magidin Mar 19 '11 at 02:12
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The original question you are trying to solve (that $A^m \cong A^n$ as $A$-modules implies $m = n$) does not need tensor products. If $M$ and $N$ are isomorphic $A$-modules then, for any ideal $I$ in $A$, $M/IM$ and $N/IN$ are isomorphic $A/I$-modules. More precisely, if $f \colon M \rightarrow N$ is an $A$-module isomorphism with inverse $g$, then the induced map $M/IM \rightarrow N/IN$ given by $x \bmod IM \mapsto f(x) \bmod IN$ is well-defined and $A/I$-linear, and it's an $A/I$-module isomorphism because the map $y \bmod IN \mapsto g(y) \bmod IM$ going the other way is an inverse (just check on each element that these two maps undo each other). Taking $M = A^m$, $N = A^n$, and $I = \mathfrak m$ to be a maximal ideal in $A$, we get from an $A$-module isom. of $A^m$ with $A^n$ that the $A/\mathfrak m$-vector spaces $A^m/{\mathfrak m}A^m$ and $A^n/{\mathfrak m}A^n$ are isomorphic. Now it remains to show $A^m/IA^m \cong (A/IA)^m$ for any ideal $I$, and this is left to you. No tensor products are needed. I think this is useful because it makes this result available if you are teaching an abstract algebra class without having to introduce tensor products already. Tensor products are of course very important, but they need not be used here. In fact, later on in the course when you have tensor products you can come back and reprove this result on isomorphisms with them and observe that the argument is basically the same as the one that didn't use tensor products.

After learning about tensor products, and then exterior powers, you can prove finer properties of $A$-linear maps $A^m \rightarrow A^n$. See Corollary 5.11 in http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf, which incidentally provides a third way to handle the original question via exterior powers without using maximal ideals in $A$.

KCd
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