Spec of tensor product of fields 
Suppose $K/k$ is a finite separable extension of degree $n$. How to show that there exists a finite separable extension $k'/k$ such that $\operatorname{Spec}(K \otimes_k  k') $ consists of $n$ points?

(It is in Gortz and Wedhorn, Algebraic Geometry I, Exercise 4.16)
 A: Probably it is wrong (I'm not familiar with field theory) but I think the following observations could be useful: 


*

*A finite separable extension of $k$ has a primitive element. Call $p(x)$ its minimal polynomial: it has degree $n$ and it is separable.

*$K=\frac{k[x]}{(p(x))}$

*Consider $k'$ as the splitting field of $p(x)$, it is a separable extension of $k$, because $p(x)$ is separable and, if $p(x)= (x-\alpha_1)\cdots (x-\alpha_n)$, we have the following equalities: $$K \otimes_k k'=\frac{k[x]}{(p(x))}\otimes_k k'= \frac{k'[x]}{(p(x))}= \frac{k'[x]}{(x-\alpha_1)}\times \dots \times \frac{k'[x]}{(x-\alpha_n)}=(k')^n$$

A: Since $K/k$ is separable, for any extension $k^\prime$ of $k$, $K\otimes_kk^\prime$ is a product of finite separable extensions of $k^\prime$. This is seen by writing $K=k[t]/(f(t))$ for some monic irreducible separable $f(t)\in k[t]$ (of degree $n$), in which case
$K\otimes_kk^\prime=k[t]/(f(t))\otimes_kk^\prime\cong k^\prime[t]/(f(t))
=\prod_{i=1}^r k^\prime[t]/(p_i(t))$
where the $p_i(t)$ are the irreducible factors of $f(t)$ in $k^\prime[t]$ (in algebro-geometric language one says that $\mathrm{Spec}(K)$ is geometrically reduced over $k$). In particular, if $k^\prime$ is taken to be a separable extension of $k$ over which $f(t)$ splits, say a splitting field of $f(t)$ over $k$, then all the $p_i$ have degree one, so the isomorphism becomes $K\otimes_kk^\prime\cong \prod_{i=1}^n k^\prime$, where $n=\deg(f)$. Then $\mathrm{Spec}(K\otimes_kk^\prime)=\coprod_{i=1}^n\mathrm{Spec}(k^\prime)$ is a disjoint union of $n$ points.
