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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?

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    $\begingroup$ If I am not misunderstanding you, the answer is yes. This follows from Zariski's lemma. $\endgroup$ – Alex Youcis Sep 19 '13 at 11:39
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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.

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