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I would like to know why $\mathbb{C}[x,y]$ is not isomorphic to $\mathbb{C}[x] \otimes _{\mathbb{Z}} \mathbb{C}[y]$ as rings.

Thank you! 1

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Do you have a specific map that you thought should be an isomorphism? –  Alex B. Dec 6 '10 at 5:28
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@user4465: Essentially, you only have bilinearity over $\mathbb{Z}$ in the latter. –  Arturo Magidin Dec 6 '10 at 5:41
    
No, Alex. I have no specific map. –  user4465 Dec 6 '10 at 5:44
    
The second one isn't even an integral domain. Note: $\mathbb{C}\otimes_{\mathbb{Z}}\mathbb{C}$ has 4 square roots of 2. –  George Lowther Dec 6 '10 at 12:46
    
...in fact, there's infinitely many –  George Lowther Dec 6 '10 at 13:00

3 Answers 3

Because $\mathbb C \otimes_{\mathbb Z} \mathbb C$ is not isomorphic to $\mathbb C$. (There is a natural map from the first to the second, with an absolutely enormous kernel.)

Added: In light of the discussion below, it might help to add the remark that this kernel is full of non-trivial zero-divisors (see e.g. Robin's answer).

One way to think of this is that $\mathbb C \otimes_{\mathbb Z} \mathbb C = \mathbb C \otimes_{\mathbb Q} \mathbb C$ contains $\overline{\mathbb Q}\otimes_{\mathbb Q} \overline{\mathbb Q}$, and this latter ring already is full of zero divisors. (Robin's answer gives one explicit example. More generally, $\overline{\mathbb Q}$ is the union of all finite Galois extensions $L$ of $\mathbb Q$ lying in $\mathbb C$, and for any such $L$, the tensor product $L \otimes_{\mathbb Q} L$ is isomorphic to a product of $[L:\mathbb Q]$ copies of $L$.)

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I should have specified that I meant not isomorphic as rings. How do we know that just because the natural map is not an isomorphism, that there is not some other map which is an isomorphism? –  user4465 Dec 6 '10 at 5:42
    
Perhaps the issue is that i am not sure that $R[x]\cong S[x]$ implies $R\cong S$. –  user4465 Dec 6 '10 at 5:50
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The point of Matt E's answer is that if R is a ring which admits a non-injective homomorphism to some other nonzero ring S, then the kernel of that homomorphism is a nonzero, proper ideal in R, so R is not a field. –  Pete L. Clark Dec 6 '10 at 6:34
    
I see, that makes a lot of sense. I'll have to think about why that implies the non-isomorphism of the polynomial rings. –  user4465 Dec 6 '10 at 6:49
    
I see that $1 \otimes 1 + i \otimes i \mapsto 0$ by the natural map, for example, and it's obvious that that element is nonzero in the tensor product, but I'm not sure how to actually prove it. –  user4465 Dec 6 '10 at 6:55

To show that $1\otimes 1+i\otimes i$ is nonzero in $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$ note that it maps to $1\otimes 1+i\otimes i$ in $\mathbb{C}[X]\otimes_{\mathbb{R}}\mathbb{C}[X]$. This is a tensor product over a field, so a basis as an $\mathbb{R}$-vector space is got by tensoring togther bases on each side. Now as the set of elements of the forms $X^n$ and $iX^n$ are bases of $\mathbb{C}[X]$ over $\mathbb{R}$ then $1\otimes 1$ and $i\otimes i$ are linearly independent over $\mathbb{R}$.

As $$(1\otimes 1+i\otimes i)(1\otimes 1-i\otimes i)=0$$ then $\mathbb{C}[X]\otimes_{\mathbb{Z}}\mathbb{C}[X]$ isn't an integral domain, unlike $\mathbb{C}[X]\otimes_{\mathbb{C}}\mathbb{C}[X]$.

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Note that the two are not even isomorphic as $\mathbb{C}$-vector spaces, since the LHS has countable dimension, while for the RHS $\{1\otimes z|\;\bar{z}\in \mathbb{C}/\mathbb{Z}\}$ is an uncountable set that is linearly independent over $\mathbb{C}$.


Edit: another way to see that the rings are not isomorphic, which is just an addition to Matt's answer, is e.g. to note that there is a different number of elements of finite multiplicative order $n$ for any $n>2$. E.g. on the LHS, an element must clearly be a scalar to be of finite multiplicative order, so the only elements of order 4 are $i$ and $-i$. On the other hand, on the RHS, $i\otimes 1$, $-i\otimes 1$, $1\otimes i$ and $-1\otimes i = 1\otimes -i$ are all distinct elements of order 4.

To prove that two elements of the tensor product are really distinct (this applies to any tensor products), you have to construct a $\mathbb{Z}$-bilinear map from $\mathbb{C}[x]\times\mathbb{C}[y]$ to some other $\mathbb{Z}$-module that takes the two corresponding elements (in this case ($i$,1) and (1,$i$), say) to different images. To understand, why that is a necessary and sufficient condition for them to be distinct in the tensor product, go back to the definition of tensor product via a universal property.

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I am regarding them as rings, so there is no $\mathbb{C}$-vector space structure. Could there then be another $\mathbb{C}$-vector space structure where they are isomorphic? –  user4465 Dec 6 '10 at 5:58
    
@user4465 I am confused by your comments. In a comment to Matt's answer, you wrote that you are not talking about an isomorphism of rings, which seems to contradict the above comment. Moreover, $\mathbb{C}$ naturally sits inside both of them, so there is a $\mathbb{C}$-vector space structure and it is subsumed by and strictly weaker than their ring structure. Accordingly, the last question in your comment doesn't make sense, since the vector space structure is forced upon you by the ring structure. –  Alex B. Dec 6 '10 at 6:30
    
No, there can't be since they have different dimension. Dimension is an invariant of vector spaces. –  Sean Tilson Dec 6 '10 at 6:32
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In the previous comment, I meant to emphasize that I mean "not isomorphic as rings". Also, I see that there is a $\mathbb{C}$-vector space structure, but why should the ring isomorphism preserve it? –  user4465 Dec 6 '10 at 6:47
    
The comment to Matt's answer parses as "I meant (not(isomorphic as rings))" = "I meant nonisomorphism in the category of rings", rather than "I (meant-not) (isomorphic as rings)" which has the opposite interpretation. –  T.. Dec 6 '10 at 13:03

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