If an $F$-algebra $A$ satisfies a multilinear identity $f$ then for any extension field $K$ of $F$, $A\otimes_F K$ also satisfies $f$. The following problem is from Herstein's Noncommutative rings, the chapter about "A theorem of Kaplanksy".
If an $F$-algebra $A$ satisfies a multilinear identity $f$ then for any extension field $K$ of $F$, $A\otimes_F K$ also satisfies $f$.
I tried putting elementary tensors $a_i\otimes_F k_i$ as arguments of the polynomial $f$ and using multiniearity, but didn't get anywhere.
Now, I have prety much zero knowledge of tensors apart from the definition, so I'd also ask you to recommend me a book on tensor products with more examples and problems.

Every element of $A\otimes_F K$ is a $K$-linear combination of elements of $A$ (where we identify $A$ with a subalgebra of $A\otimes_F K$ in the canonical way via $a\mapsto a\otimes 1$).  Explicitly, if you have a tensor $\sum_i a_i\otimes k_i$, then it is just the linear combination $\sum_i k_i a_i$ (where here $a_i$ is identified with the tensor $a_i\otimes 1\in A\otimes_F K$).  Since $f$ is linear in each input, this means that if you evaluate $f$ on a bunch of elements of $A\otimes_F K$, you just get a big $K$-linear combination of $f$ evaluated on elements of $A$.  (For instance, if $f$ has two inputs, then $f(\sum_i k_ia_i,\sum_j k_j'a_j')=\sum_{i,j}k_ik_j'f(a_i,a_j')$.)  So if $f$ always vanishes when evaluated on elements of $A$, it will also vanish when evaluated on your elements of $A\otimes_F K$.
