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I was wondering if the ring $\mathbb{Q}(\sqrt[3]{2}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt[3]{2})$ is a PID. I believe that it is because I think $\mathbb{Q}(\sqrt[3]{2}) \otimes_{\mathbb{Q}}\mathbb{Q}[x]$ is a PID, which are just polynomials with coefficients in the field $\mathbb{Q}(\sqrt[3]{2})$. Also, I was wondering if someone can describe the the prime ideals of $\mathbb{Q}(\sqrt[3]{2}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt[3]{2})$?

bob
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    No, $R=\mathbb{Q}(2^{1/3}) \otimes_{\mathbb{Q}}\mathbb{Q}(2^{1/3}) $ is not a PID, not even a domain. Write $R=\mathbb{Q}(2^{1/3})[X]/(X^3-2)$ and factor $X^3-2$ in $\mathbb{Q}(2^{1/3})[X]$. – Georges Elencwajg Mar 25 '13 at 10:18

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Indeed ${\bf Q}(\sqrt[3]{2})\otimes_{\bf Q}{\bf Q}(\sqrt[3]{2})\cong{\bf Q}(\sqrt[3]{2})[x]/(x^3-2)$ is a quotient of the PID ${\bf Q}(\sqrt[3]{2})[x]$. However not every quotient of a PID is again a PID (quotienting by a prime does yield another PID). Here, we have the factorization $x^3-2=(x-\sqrt[3]{2})(x^2+\sqrt[3]{2}x+\sqrt[3]{4})$ (the latter does not even have real roots let alone roots in ${\bf Q}(\sqrt[3]{2})$ and it is quadratic so it is irreducible), hence

$$\frac{{\bf Q}(\sqrt[3]{2})[x]}{(x^3-2)}\cong\frac{{\bf Q}(\sqrt[3]{2})[x]}{(x-\sqrt[3]{2})}\times\frac{{\bf Q}(\sqrt[3]{2})[x]}{(x^2+\sqrt[3]{2}x+\sqrt[3]{4})}\cong {\bf Q}(\sqrt[3]{2})\times {\bf Q}(\sqrt[3]{2},\sqrt{-3}).$$

You should verify yourself that if $H,K$ are fields then the proper nontrivial ideals of $H\times K$ are only $H\times0$ and $K\times 0$ using the fact that a field has no proper nontrivial ideals. Thus this is in fact a principal ideal ring but it cannot be a principal ideal domain (since the direct product of two non-trivial rings will always have zero divisors, in particular elements with precisely one coordinate $0$).

Martin Brandenburg
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anon
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    Dear anon, your solution is correct, but for categorical reasons, you should write $H\times K$, not $H\oplus K$, and call this a product, not a sum, of rings. The sum (or coproduct) for rings and algebras corresponds to the tensor product. – Georges Elencwajg Mar 25 '13 at 10:25
  • @GeorgesElencwajg Is the word "sum" used for coproduct in a generalized categorical context? I have never seen it used as such, but I haven't studied category theory in and of itself. I like thinking of it as a sum since then we have distributivity $(A\oplus B)\otimes C\cong(A\otimes C)\oplus(B\otimes C)$, and this sort of thinking is inherent in arguably categorical-flavored Clebsch–Gordan problems in a large variety of representation theory contexts (namely decomposing tensor products of irreducibles into irreducibles, or equivalently, structure constants of the virtual representation ring). – anon Mar 25 '13 at 10:32
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    Dear anon, no, "sum" is not a very widespread terminology for "coproduct". However all the authors I know refer to the ring you mention as the product of the rings and use the notation $R\times S$: Mike Artin, Atiyah-Macdonald, Bourbaki, Grothendieck, Jacobson, Lang,... That said, you wrote a nice, detailed answer, and this is the main thing to emphasize: +1 – Georges Elencwajg Mar 25 '13 at 10:42
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    Unfortunately, someone down-voted this answer. This is a pity, because the answer is quite nice and I would be sad if the little terminological difference evoked in my comments above had led to this too harsh reaction. – Georges Elencwajg Mar 25 '13 at 11:20
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    Good answer. In general, if $L$ is a finite separable field extension of $K$, and $F$ is an algebraic extension of $K$, then $L \otimes_K F$ is a finite direct product of fields. – Martin Brandenburg Mar 25 '13 at 11:41