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Given an infinite field $K$, one can prove that any maximal ideal of $K[X,Y]$ can't be principal. In fact, every non-principal prime ideal is a maximal ideal, and can be generated by two polynomials.

I am wondering whether the same result holds in $\mathbb F_q[X,Y]$. Can we find a principal ideal $I = (P(X,Y))$ for some irreducible polynomial $P$ that is a maximal ideal ?

Such a polynomial $P$ must have positive degrees in both $X$ and $Y$. Indeed, given an irreducible polynomial $Q(X)$ in only one variable, the quotient $$\mathbb F_q[X,Y]/(Q(X))\cong (\mathbb F_q[X]/(Q(X)))[Y]$$ is a ring of polynomials over a field and as such, can never be a field.

Moreover, such a polynomial $P$ must have the form $$P(X,Y) = \sum_{i=0}^n P_i(X)Y^i$$ where $d\geq 1$, the $P_i$'s are polynomials in $X$ and $P_n$ is a nonzero polynomial that vanishes identically on $\mathbb F_q$, thus $P_n$ must be divisible by $\prod_{\alpha \in \mathbb F_q} (X-\alpha)$.
Indeed, if there were some $\alpha \in \mathbb F_q$ such that $P_n(\alpha) \not = 0$, then the ideal $I:=(P,X-\alpha)$ would contain $(P)$ strictly. If $(P)$ were to be maximal, $I$ would be the whole of $\mathbb F_q[X,Y]$. Writing down the fact that $1\in I$ and evaluating $X=\alpha$ would leads us to the conclusion $n=\deg_Y(P)=0$, which is absurd.

This is all I could infer so far. With respect to the above, I tried looking at $P(X,Y) = (X^{p-1}-1)XY - 1$ or $P(X,Y) = (X^{p-1}-1)XY - X - 1$ in $\mathbb F_p[X,Y]$ for some prime number $p$ but I have trouble determining whether the quotient is a field or not.

Would somebody know the answer of the problem, and according to it, give a proof or a counter-example ? Thank you very much in advance.

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Suzet
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    For any field for any $f \in k[X,Y] - k$ let $t = X$ if $f \not \in k[X]$, $t=Y$ otherwise.

    Then $k[X,Y]$ is a finitely generated $k[f,t]$ module and $k[f,t]/(f) \cong k[t]$ and $k[x,y]/(f)$ is a finitely generated $k[t]$ module thus it doesn't contain $k(t) = k[t,{ \frac{1}{p(t)}, p \in k[t]_{Irr}}]$ since it is not finitely generated as a $k$-algebra.

     – reuns Apr 05 '19 at 19:49
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    Read the comment under my answer in the first thread (and the answer if you want to see an elementary proof). In my opinion Nullstellensatz is a big gun for such an elementary result. – user26857 Apr 05 '19 at 19:57
  • Thanks a lot to both of you, I accept both your nice answers as well. They absolutely solve my problem. – Suzet Apr 06 '19 at 06:39

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Here's my try. It seems correct to me :

If $I=(P)$ is maximal, then by the Nullstellensatz, $\mathbb{F}_q[X,Y]/(P)$ is a finite extension of $\mathbb{F}_q$, that is a $\mathbb{F}_{q^n}$ for some $n$.

Then, in $\mathbb{F}_q[X,Y]/(P)$, $\bar{X}^{q^n}=\bar{X}$ ($\bar{X}$ being the class of $X$ in $\mathbb{F}_q[X,Y]/(P)$), so $X^{q^n}-X \in (P)$ , so $P$ divides $X^{q^n}-X$, so $P \in \mathbb{F}_q[X]$, but you show it is not possible.