Why isn't this ideal equal to $0$? Let $I$ be a maximal ideal in $K[X,Y]$, where $K[X,Y]$ is the polynomial ring in two indeterminates over a field $K$. I would like to prove that at least one of $I \cap K[X]$ and $I \cap K[Y]$ is not equal to the zero ideal.
I think that if $I \cap K[X]=I \cap K[Y]=0$, then this assumption causes a contradiction with $I$'s maximality. However, I can't prove it.
Would you have any hints or ideas?
Thank you.
 A: [Updated due to comment below from Lukas Hager.]
This follows from the Zariski Lemma. Zariski says that any finitely-generated field extension is a finite(-dimensional) extension.
You actually get that $I\cap K[x]\neq 0$ and $I\cap K[y]\neq 0.$
Even more generally, for any non-constant $f(x,y)\in K[x,y],$ $I\cap K[f(x,y)]\neq 0.$
Since $I$ is maximal, $k=K[x,y]/I$ is a finitely generated field extension of $K,$ and thus finite-dimensional.
If $n$ is the dimension, and $\overline x$ is the image of $x$ in $k,$ then $1,\overline x,\overline x^2,\dots,\overline x^n$ are not linearly independent over $K,$ so for some non-zero $q(x)\in K[x],$ $q(\overline x)=0$ But that means $q(x)\in I.$
This proves the same for maximal ideals of $K[x_1,x_2,\dots,x_m]$ for any $m.$ Indeed, Zariski's Lemma is equivalent to this general case.
Of course, if this were an exercise before knowing Zariski, this answer would be inadequate without also proving Zariski...

My Original Post
Not an answer, but we can show that if $I\cap K[x]\neq 0$ then $I\cap K[y]\neq 0.$
Let $k=K[x,y]/I$ be the quotient. Since $I$ is maximal, $k$ is a field extension of $K.$ Let $\overline x,\overline y$ be the images of $x,y$ in $K.$
Since $I\cap K[x]$ is an ideal in $K[x],$ if it is non-zero, let there is an irreducible $q(x)\in K[x]$ such that $q(\overline x)=0$ in $k.$ So $\overline x$ is algebraic over $K.$
If $I= \langle q(x)\rangle,$ then $I$ is not maximal, because we can find a bigger ideal, $\langle q(x),y\rangle.$
So some $f(x,y)\in I$ is not divisible by $q(x).$
Write $f$ as a polynomial in $K[x][y],$ so $$f(x,y)=\sum_{i=0}^n a_{i}(x) y^i.$$
Then not all the $a_i(x)$ are divisible by $q(x),$ so $g(y)=f(\overline x,y)$ is a non-zero polynomial in $K[\overline x][y],$ and $g(\overline y)=f(\overline x,\overline y)=0.$
So $\overline y$ is algebraic over $K(\overline x)$ and $\overline x$ is algebraic over $K,$ so $\overline y$ is algebraic over $K,$ and thus some non-zero $p(y)\in I.$

This means that your statement is equivalent to the statement:

If $I$ is a maximal ideal in $K[x,y]$ then $K[x,y]/I$ is a finite  extension of $K.$

A: Let $\mathfrak{m}$ be maximal such that $\mathfrak{m} \cap K[X] =0$, then $x$ (which I'll use as notation for the class of $X$ in $K[X,Y]/\mathfrak{m}$, same for $y$ and $Y$) is transcendental over $K$.
Case 1: $y$ is transcendental over $K(x)$, then we have that $K[X,Y]/\mathfrak{m}$ contains both $K(x)$ and $K(y)$. But $K[X,Y]/\mathfrak{m}$ is generated as an algebra by $X$ and $y$, so we get that $K[X,Y]/\mathfrak{m}=K(x)(y)=K(x,y)\cong K(X,Y)$
Case 2: $y$ is algebraic over $K(x)$, so $K[X,Y]/\mathfrak{m}$ is a finite extension of $K(x)$.
I claim that in both cases, $K[X,Y]/\mathfrak{m}$ is not finitely generated as a $K$-algebra, which is a contradiction. In the first case, this is easy: there are infinitely many irreducible polynomials in $K[X,Y]$, so considering the collection $\{\frac{1}{p}\}$, where $p$ runs over the irreducble polynomials gives the result.
In the second case: Apply the Artin-Tate lemma (which has a short and elementary proof, see the link), to see that $K(x) \cong K(X)$ is a finitely generated algebra. As in the first case, there are infinitely many irreducible polynomials in $K[X]$, so the same argument gives a contradiction.
