Does an ideal in $k[x_1, \dots, x_n]$, $k$ a field, contain only finitely many irreducible elements?

Question

Let $k$ be a field and consider a nonunital ideal $I$ in $k[x_1, \dots, x_n]$. Since $k$ is Noetherian, we may write $I = (f_1, \dots, f_k)$ where each $f_i$ is nonconstant irreducible.

Is it true that $I$ can contain only finitely many irreducible elements? If so, is there an elementary proof (not using too much commutative algebra)? What if $I$ is assumed to be a maximal ideal?

I don't know how to answer when a prime polynomial can be a linear combination of other primes.

Answer 1

This is not true for $n \ge 2$, if $k$ is an infinite field. For example, in the ideal $(x_1, x_2)$ there are infinitely many irreducibles $\{ x_1 + ax_2: a \in k\}$.