If $K / k$ is a finite seperable field extension and $L / k$ is some other field extension, show that $K \otimes_k L$ is a semisimple $k-algebra.
Since $K$ is a finite dimensional vector space over $k$, I know it's a free $k$-module. So $K \cong k^s$ for $s<\infty$. So I can write $K \otimes_k L \cong k^s \otimes L \cong L^s$. Wouldn't this imply that $K \otimes_k L$ is a semisimple $k$-algebra?
If this is not correct (which I doubt it is since I didn't use the seperability of K), could I be steered in the right direct please.
By the primitive element theorem, $K=k(\alpha)$ for some $\alpha$.
If $p(x)$ is the minimal polynomial of $\alpha$ over $k$, then $k(\alpha)\cong k[x]/p(x)$. this is because the $k$-algebra homomorphism $k[x]\to k(\alpha)$ given by evaluating polynomials at $\alpha$ is surjective with kernel $p(x)$; if any element of the kernel were not a multiple of $p(x)$, its remainder upon division by $p(x)$ would $\alpha$ as a root and smaller degree than $p(x)$, contra minimality.
Thus, $K\otimes_k L=k[x]/p(x)\otimes_k L$. This is isomorphic, as an algebra, to $L[x]/p(x)$. This is extension of scalars - a term usually used for modules, but applies here for algebras.
To see that there is an ismorphism (of $k$-algebras), first write down what the isomorphism is, see that it's invertible, and see that it preserves addition and multiplication.
Let's apply this with $k[x]\otimes L\cong L[x]$. The map $k[x]\otimes L\to L[x]$ is given by linearly extending the rule $x^n\otimes a\mapsto ax^n$, so that $p_1(x)\otimes a_1+\cdots+p_k(x)\otimes a_k\mapsto a_1p_1(x)+\cdots+a_kp_k(x)$. You simply turn the $\otimes$ symbol into actual multiplication in $L[x]$, since both $L$ and $k[x]$ are subalgebras! The inverse is given by $b_mx^m+\cdots+b_1x+b_0\mapsto x^m\otimes b_m+\cdots+x\otimes b_1+b_0$. Can you show this really is an inverse? (Remember you can slide scalars from $k$ across the $\otimes$ symbol, though not other scalars from $L$.) Can you show both maps are algebra homomorphisms?
The same applies with $k[x]/p(x)\otimes L\cong L[x]/p(x)$.
Now suppose $p(x)$ factors into irreducibles $p_1(x)\cdots p_r(x)$ over $L$. Note these factors must all be distinct, since otherwise $p(x)$ would have repeated roots in the algebraic closure of $k$ and then $K=k[x]/p(x)$ would not be separable, a contradiction. Therefore, the $p_i(x)$s are coprime and Chinese remainder theorem applies, which says $L[x]/p(x)\cong L[x]/p_1(x)\oplus \cdots L[x]/p_r(x)\cong L_1\oplus\cdots\oplus L_r$ which we can check is semisimple, i.e. has no nilpotent elements, given $L_i:=L[x]/p_i(x)$s are field extensions of $L$.
If you need help showing the explicit maps $k[x]\otimes L\leftrightarrow L[x]$ I gave are inverses of each other, or that they're algebra homomorphisms, or that the same applies to $k[x]/p(x)\otimes_k L\leftrightarrow L[x]/p(x)$, or how to show $\bigoplus L_i$ is semisimple, be sure to ask. Remember the trick to the homomorphism out of the tensor product is to turn $\otimes$ into plain-old multiplication, and the trick to the inverse is to separate out the $L$-coefficients using the $\otimes$ symbol - these are key to seeing why the maps are inverses and algebra homomorphisms.