Prime ideals in $L \otimes_K \overline{K}$ Let $L|K$ be a separable finite field extension and $\overline{K}$ an algebraic closure of $K$, then we have
$$ L\otimes_K \overline{K} \cong \prod_{\sigma \in \text{Hom}_K (L, \overline{K})} \overline{K}. $$
Now, it is stated here (with $L = k(x), K=k(y)$) that

So, we would want to say that $L|K$ is unramified if and only if $\text{Spec} (L\otimes_{K}\overline{K})$ has $n = [L:K]$ points. But this happens precisely when $L|K$ is separable.

Now, here is my confusion: The ideals in the product above are of the form $\prod I_i$, where $I_i$ are ideals in $\overline{K}$ and in particular, I would think that all expect $(1) \times ... \times(1)$ are prime ideals. But this would give $2^n - 1$ prime ideals.
I think I am missing something about the algebra-structure here, where am I wrong?
 A: First, recall the standard fact that for a finite seperable extension like $L/K$ there are precisely $n = [L : K]$ embeddings of $L$ into $\overline K$ which fix $K$. Hence, this tensor product is $L \otimes_K \overline K \cong (\overline K)^n$. It seems to me that you already used this fact, but I just want to explicitly mention it.
Anyways, let's focus on the alleged $2^n - 1$ prime ideals of the form $(1) \times \dots \times (1)$. It seems like you're saying that for every nonempty subset $S$ of $\{1, \dots, n\}$ we associate a prime ideal by taking a product of $(1)$ in the slots corresponding to the elements of $S$. More formally, we associate to this subset $S$ the product $I_1 \times \dots \times I_n$ such that $I_j = (1)$ if $j \in S$ and $I_j = (0)$ otherwise. Now, this construction will indeed get you all $2^{n - 1}$ nonzero ideals of $(\overline K)^n$ but they will not all be prime. Indeed, say we had $(1) \times (0) \times (0) \subseteq (\overline K)^3$ corresponding to $S = \{1\}$. Then the quotient $(\overline K)^3 / (1) \times (0) \times (0)$ is $(\overline K / (1)) \times (\overline K / (0)) \times (\overline K / (0)) \cong (\overline K)^2$. This is not a domain, so the ideal $(1) \times (0) \times (0)$ is not prime.
Indeed, not all subsets $S \subseteq \{1, \dots, n\}$ yield prime ideals in this fashion. So which ones do? We can compute the quotient $(\overline K)^n / \prod I_j$ explicitly as $\prod \overline K / I_j$. Now, if all of these factors are trivial we get the trivial ring, which is not a domain. That is, $S = \{1, \dots, n\}$ corresponds to the unit ideal which is not prime. If at least two of these factors are nontrivial then we get a product of at least two nontrivial rings, which is not a domain. Hence, any subsets $|S| \leq n - 2$ fail to yield prime ideals. We are left precisely with those subsets $|S| = n - 1$, corresponding to the ideals of the form $(1) \times \dots \times (1) \times (0) \times (1) \times \dots \times (1)$ which are the unit ideal in all but one slot. The quotient by this ideal is precisely $\overline K$, so it is in fact maximal.
Hence, $Spec(L \otimes_K \overline K)$ corresponds to subsets of $\{1, \dots, n\}$ of cardinality $n - 1$. There are $\binom{n}{n - 1} = n$ of these, as desired.
As a completely unnecessary aside, there is an interesting generalization of this computation to the prime spectrum of an arbitrary product of fields $Spec(\prod_{i \in I} F_i)$. These turn out to be in bijection to the ultrafilters on the power set of $I$, and the set of all ideals corresponds to the filters on the power set of $I$.
