Primes containing a fixed element $f$ in a finitely generated $k$-algebra.

Question

Let $R$ be a finitely generated $k$-algebra where $k$ is a field. Let $0 \neq f \in R$. I was wondering if it is the case that there are only finitely many primes containing $f$? Any explanation or counter example is appreciated. Thank you.

Answer 1

No, this is generally far from the case. Consider for example $x \in k[x,y]$. Every ideal of the form $(x, g)$ for $g \in k[y]$ irreducible is maximal (the quotient $k[x,y]/(x,g) \simeq k[y]/(g)$ is a field, by irreducibility of $g$). Since there are infinitely many irreducible polynomials in $k[y]$, we find infinitely many maximal (hence prime) ideals over $(x)$. If $k$ is infinite, you can even restrict to the maximal ideals of the form $(x, y - a)$ for $a \in k$.

A very similar counter-example is $x \in \mathbb{Z}[x]$. For any prime $p \in \mathbb{Z}$, the ideal $(x, p) \subset \mathbb{Z}[x]$ is maximal (the quotient is $\mathbb{F}_p$). There are infinitely many primes in $\mathbb{Z}$, so once again we get infinitely many maximal ideals over $(x)$.

The first example has a nice generalization. If $f \in k[x_1, \dots, x_n]$, then $(f) \subset (x_1 - a_1, \dots,x_n - a_n)$ for $a_1, \dots, a_n \in k$ if and only if $f(a_1, \dots, a_n) = 0$. Taking $f$ to be any polynomial with infinitely many zeros then works nicely.

For a positive result in this direction, note that there are only finitely many minimal primes over any ideal $I \unlhd R$ in a noetherian ring $R$. Any finitely generated $k$-algebra is noetherian (Hilbert's basis theorem), so there are only finitely many minimal primes over your element $f$.

Answer 2

Try looking at $k[x,y]/(x^2)$ where the image of $x$ is your nonzero element for an answer to your question.