maximal spectrum of $\mathbb{F}_2[x,y]$ What are the maximal ideals in $\mathbb{F}_2[x,y]$? In particular, I'm trying to answer the following question from my commutative algebra class: What is the number of maximal ideals in $\mathbb{F}_2[x,y]$ with quotient ring of order 8?
So far, the only maximal ideals I have found are $(x,y)$ and $(x+1,y+1)$.Since $\mathbb{F}_2$ is not algebraically closed (as $x^2+x+1$ has no root in $\mathbb{F}_2$) , I don't think Hilbert's Nullstellensatz applies. 
For the quotient ring to have order $8$, it must be generated by some elements $1,a,b$ where $a^2,b^2,ab$ are in the maximal ideal $m$, or $1,a,a^2$ where $a^3 \in m$. But I'm not sure how this helps me find $m$. 
 A: Warning
Despite appearances this is a quite subtle and non trivial question!
Solution
Clearly to any maximal ideal $\mathfrak m \subset \mathbb F_2[x,y])$ with residue field $\kappa(\mathfrak m)$ of order $8$ we can associate an $\mathbb F_2$-algebra morphism $\mathbb F_2[x,y]\to \mathbb F_8$.
Since there are $64=8^2$ such morphisms (send $x,y$ arbitrarily to $F_8$) the answer to our problem  is  $64$, right?
Wrong! We have to dismiss the points with residue field of order $2$ (corresponding to morphisms $\mathbb F_2[x,y]\to \mathbb F_2$) of which we have $4$.
OK, so the required number is $64-4=60$, right?
Wrong again! The subtle point is that for each $\mathfrak m$ with residue field of order $8$ we have three morphisms $\kappa(\mathfrak m)\to \mathbb F_8$ : they are  obtained by composing one of them with the $3$ elements of $\operatorname {Gal} (\mathbb F_8/\mathbb F_2)$.  Dividing by $3$ we obtain the required result:  
There are 20 maximal ideals in $\mathbb F_2[x,y]$ with residue field of order 8. 
