Describe prime ideals and Krull dimension of $\overline{\mathbb{Q}} \otimes_{\mathbb{Q}} \overline{\mathbb{Q}}$ 
I want to describe the prime ideals of $\overline{\mathbb{Q}} \otimes_{\mathbb{Q}} \overline{\mathbb{Q}}$, where $\overline{\mathbb{Q}}$ denotes the integral closure of $\mathbb{Q}$ in $\mathbb{C}$, and then find $\dim(\overline{\mathbb{Q}} \otimes_{\mathbb{Q}} \overline{\mathbb{Q}})$.

I claim that $\overline{\mathbb{Q}} \otimes_{\mathbb{Q}} \overline{\mathbb{Q}}$ is an integral extension of $\overline{\mathbb{Q}}$, which is an integral extension of $\mathbb{Q}$. Therefore 
$\dim(\overline{\mathbb{Q}} \otimes_{\mathbb{Q}} \overline{\mathbb{Q}}) = \dim(\mathbb{Q}) = 0$.
I'm not sure if this is the right idea. But what I'm really struggling with is describing the prime ideals in the tensor product and showing that the tensor product is an integral extension.
 A: The following perspective on prime ideals is often helpful: a prime ideal in a ring $R$ is an ideal which is the kernel of a homomorphism from $R$ to a domain.
For a ring like $R=\overline{\mathbb{Q}}\otimes\overline{\mathbb{Q}}$ this is very useful, because while elements of $R$ (let alone ideals of $R$) are hard to understand, homomorphisms out of $R$ are easy by the universal property of the tensor product.  Namely, a homomorphism from $R$ to a (commutative) ring $S$ just corresponds to a pair of homomorphisms $f,g:\overline{\mathbb{Q}}\to S$ (specifically, it is the unique homomorphisms which sends $a\otimes b$ to $f(a)g(b)$).  If $S$ is nonzero, these homomorphisms are automatically injective, so you just have a ring $S$ with two different embeddings of $\overline{\mathbb{Q}}$.
Now if $S$ is a domain, it has at most one subring that is isomorphic to $\overline{\mathbb{Q}}$, namely the subfield of its field of fractions consisting of elements that are algebraic over $\mathbb{Q}$.  So our two embeddings $f$ and $g$ have the same image, and then the induced homomorphism $R\to S$ also has the same image.  This means that the image of our homomorphism $R\to S$ is just a subfield of $S$ isomorphic to $\overline{\mathbb{Q}}$, so we may assume that $S$ is actually just $\overline{\mathbb{Q}}$ itself.  Moreover, we can choose our identification of this subring of $S$ with $\overline{\mathbb{Q}}$ such that our first homomorphism $f:\overline{\mathbb{Q}}\to S$ becomes just the identity map $\overline{\mathbb{Q}}\to \overline{\mathbb{Q}}$.
So to sum up: every prime ideal of $R$ is the kernel of a homomorphism $\varphi_g:R\to \overline{\mathbb{Q}}$ of the form $a\otimes b\mapsto ag(b)$, for some homomorphism (or equivalently, automorphism) $g:\overline{\mathbb{Q}}\to\overline{\mathbb{Q}}$.  In particular, this makes it clear that every prime ideal of $R$ is maximal, since $\overline{\mathbb{Q}}$ is a field and these homomorphisms are surjective.  To describe the prime ideal $\ker(\varphi_g)$ associated to an automorphism $g$ a bit more explicitly, you can say it is generated by all elements of the form $1\otimes b-g(b)\otimes 1$ for $b\in\overline{\mathbb{Q}}$.  Clearly these elements are all in $\ker(\varphi_g)$, and conversely if you mod out all these elements, then the quotient map will factor through $\varphi_g$ since $a\otimes b$ will be identified with $ag(b)\otimes 1$.  This also shows that $g$ is uniquely determined by $\ker(\varphi_g)$, since $g$ can be recovered as the map sending each $b\in\overline{\mathbb{Q}}$ to the unique $c\in\overline{\mathbb{Q}}$ such that $1\otimes b-c\otimes 1\in\ker(\varphi_g)$.  So, prime ideals in $R$ are in bijection with automorphisms of $\overline{\mathbb{Q}}$.
(None of this discussion was special to $\mathbb{Q}$, and more generally a similar description holds for prime ideals in $\overline{K}\otimes_K\overline{K}$ for any field $K$.  Even more generally, if $L$ is an algebraic extension of $K$, similar arguments show that prime ideals in $\overline{K}\otimes_K L$ are all maximal and are in bijection with embeddings of $L$ in $\overline{K}$.)
