Ideals of $\mathbb{Q}(\sqrt[3]{2}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt[3]{2})$ 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 $\mathbb{Q}(\sqrt[3]{2}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt[3]{2})$?
 A: 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 $0$).
