Maximal ideals of polynomials in two variables I want to show that any maximal ideal $M$ of $k[X,Y]$ is of the form $(f,g)$, where $f\in k[X]$ is monic and irreducible, and $g\in k[X,Y]$ whose reduction modulo $f$ is irreducible in the ring $(k[X]/(f))[Y]$. 
My attempt: Let $M$ be any maximal ideal of $k[X,Y]$, and $M' = k[X]\cap M$. It is clear that $M'$ is an ideal of $k[X]$. I want to prove that $M'$ is a maximal ideal of $k[X]$, since then I can use the fact that $k[X]$ is a PID, which implies that $M' = (f)$ for some irreducible polynomial $f\in k[X]$. One can easily see that $k[X]/M'\cong k[\overline{x}]$, where $\overline{x} = X \bmod M$. So it is enough to show that $k[\overline{x}]$ is a field. 
My Question: How can one show that $k[\overline{x}]$ is a field?.
Note: I also need to show that $M'\neq (0)$, since otherwise $k[X]\cong k[\overline{x}]$, and $k[X]$ clearly isn't a field. And maybe it is easier to show that $k[\overline{x}]$ is an integral domain, which would imply that $M'$ is a prime ideal, and since $k[X]$ is a PID, $M'$ is also a maximal ideal. 
 A: *

*If $M$ is a maximal ideal of $k[X,Y]$ then $M \cap k[X]$ is a prime ideal of $k[X]$ which (thanks to the $\gcd$ algorithm) is a PID thus $M \cap k[X]= (f(X))$ 
prime ideal implies $f(X)\in k[X]$ is irreducible,  $k[X]/(f(X))$ is an integral domain and a finite dimensional $k$-vector space thus it is a field,
$k[X]/(f(X))[Y]$ is a PID and $M \bmod (f(X))$ is a prime ideal of it thus $M \bmod (f(X)) = (g(X,Y))$ with $g(X,Y)\in k[X]/(f(X))[Y]$ irreducible ie. $$M = (f(X),g(X,Y))$$

*Conversely if $f(X)\in k[X]$ is irreducible then $(f(X))$ is a prime ideal ( because $f(X)|u(X)v(X)$ means $f(X) |\gcd(u(X), f(X))\gcd(v(X), f(X))$), 
thus $k[X]/(f(X))$ is a field, if $g(X,Y) \in k[X]/(f(X))[Y]$ is irreducible then $ k[X]/(f(X))[Y]/(g(X,Y)$ is a field and hence $(f(X),g(X,Y))$ is a maximal ideal.

It is not so obvious how to keep things elementary to show $M$ maximal implies $f\ne 0$ : $(f)$ is the kernel of $K[X]\to K[X,Y]/M$, if $f=0$ then $K[X,Y]/M$ contains $K[X]$ thus $Frac(K[X])$, if $Y$ is not algebraic over $Frac(K[X])$ then $M = (0)$ is not a maximal ideal, thus $Y$ is algebraic over $Frac(K[X])$, let $p(X,Y)=\sum_{j=0}^n g_j(X)Y^j\in  K[X][Y]$ be its minimal polynomial, take $\alpha \in \overline{K}$ such that $g_n(\alpha) \ne 0$, let $h(X)\in K[X]_{monic}$ be $\alpha$'s minimal polynomial, let $P(Y) = p(\alpha,Y)\in K(\alpha)[Y]$ then $K[X,Y]/(h(X),M) = K(\alpha)[Y]/(P(Y))$ so $(h(X),M)$ is a larger proper ideal contradicting the maximality of $M$
