Let $L/K$ be an algebraic extension of fields. Let $B = L[X,Y]$ and $A = K[X,Y]$. Suppose $a$, $b \in L$ and $m = (X-a,Y-b)$ is an ideal of $B$. Show that $m$ and $m \cap A$ are maximal ideals of $B$ and $A$ respectively.
I can't think of a way to show $m \cap A$ is a maximal ideal of $A$. I know that contraction of an ideal is an ideal, but it seems that's not the case with maximal ideals.
You have inclusions of integral domains $K \hookrightarrow A/(\mathfrak{m} \cap A) \hookrightarrow B/\mathfrak{m} \cong L$. Here the composition $K \to L$ is the given inclusion $K \hookrightarrow L$ (because the isomorphism $B/\mathfrak{m} \cong L$ is just by evaluation at $X \mapsto a$ and $Y \mapsto b$). $L/K$ is algebraic, and since $K$ is a field $L/K$ is an integral extension. So $\overline{X}, \overline{Y} \in A/(\mathfrak{m} \cap A) = K[X,Y]/(\mathfrak{m} \cap K[X,Y])$ are integral elements over $K$, so $A/(\mathfrak{m} \cap A)$ is integral over a field, $K$. So $A/(\mathfrak{m} \cap A)$ is a field and $\mathfrak{m} \cap A$ is a maximal ideal of $A$.