Let $K$ be a separable $F$-algebra, and $E$ be a extension field of $F$. We see that $E\otimes _FK$ is a separable $E$-algebra. Then $E\otimes _FK$ is a separable extension field of $E$.
$E\otimes _FK$, as a field, is a simple ring.
However, $E\otimes _FK$, as a separable $E$-algebra, is a direct sum of some simple $E$-algebras, which we denote by $E_i$. Then $E\otimes _FK$ direct sum of separable extension fields $E_i$ of $E$ (It is well known that a separable $E$-algebra is a separable extension field over $E$). Then $E\otimes _FK=E_1\oplus\cdots \oplus E_r$, which have nonzero ideals $E_1,\cdots,E_r)$...
It may be looks like a contraction.
Thanks everyone!
It is simply incorrect that $E\otimes_F K$ is a field. This is true sometimes (in the case that you end up having $r=1$), but not in all cases. For instance, $\mathbb{C}\otimes_\mathbb{R}\mathbb{C}\cong\mathbb{C}\times\mathbb{C}$, which is not a field. See https://mathoverflow.net/questions/82083/when-is-the-tensor-product-of-two-fields-a-field for a discussion of when the tensor product of two fields is a field.