Let $S=\mathbb F_q[x]$ be the polynomial ring over the finite field $\mathbb F_q$. If $I=\langle p(x)\rangle$ is a maximal ideal of $S$ ($p(x)$ is irreducible), then the field $S/I$ is also a finite field. This is a basic result of field theory.

Now, consider the polynomial ring in $n$ indeterminates $R=\mathbb F_q[x_1, \dots , x_n]$. Let $J=\langle f_1(x_1, ..., x_n), \dots, f_n(x_1, ..., x_n)\rangle $ be a maximal ideal of $R$ (the polynomials $f_i$, for $i=1, ..., n$, are irreducible). It is true that in this case the field $R/J$ is also finite?

  • 2
    $\begingroup$ If I am not misunderstanding you, the answer is yes. This follows from Zariski's lemma. $\endgroup$ Sep 19, 2013 at 11:39

1 Answer 1


This follows from an equivalent to Hilbert's Nullstellensatz known as Zariski's Lemma:

Theorem (Zariski): Let $k$ be a field. If $K$ is another field which is of finite type over $k$, then $K$ is actually a finite extension of $k$

Thus, in your case the field $\mathbb{F}_q[x_1,\ldots,x_n]/J$ is a field which is finitely generated over $\mathbb{F}_q$, and thus by Zariski's lemma must be actually a finite extension of $\mathbb{F}_q$, and thus finite.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.