# Maximal ideals of polynomial ring with two variables that contains ideal generated by irreducible polynomial

Consider the polynomial ring $R=F[x,y]$, where $F$ is a field. Let $p(x)\in F[x]$ be an irreducible polynomial.

What are all the maximal ideals of $R$ that contain the principal ideal $p(x)R$?

My attempt:

By correspondence theorem for rings, this is equivalent to finding all maximal ideals of $R/p(x)R$. $p(x)R$ is a prime ideal of $R$, so $R/p(x)R$ is an integral domain. I am kind of stuck here.

This is a continuation, but not an exact duplicate of my previous question (Maximal Ideals in $K[y]$).

Thanks for any help.

Update: I have an attempted solution using $R/p(x)R\cong K[y]$, where $K=F[x]/(p(x))$.

Then let $\pi:R\to K[y]$ be the quotient map. Then since the maximal ideals in $K[y]$ are $g(y)K[y]$, by correspondence theorem, the maximal ideals of $R$ containing $p(x)R$ are $$\pi^{-1}(g(y)K[y])$$ where $g(y)\in K[y]$ is an irreducible polynomial.

I am a little dissatisfied with my answer (mainly because of the $\pi^{-1}$), is there any more explicit representation of the maximal ideals $\pi^{-1}(g(y)K[y])$?

Usually $k[x,y]/(p(x))$ and $k[y]$ are not isomorphic. For example, if $k = \mathbb{Q}$ or $\mathbb{R}$ and $p(x)$ is an irreducible polynomial of degree at least $2$, then these rings are different -- say, if $p(x) = x^2+1$, then here $k[x,y]/(p(x))$ is $k[y][i]$, where $i^2 = -1$, not just $k[y]$. There is a natural map involving $k[y]$ here, but it is the inclusion map $k[y] \to k[x,y]/(p(x))$ giving $k[x,y]/(p(x))$ as a finitely generated module over $k[y]$ (specifically, if $p(x)$ has degree $d$, then $1, x, x^2, ..., x^{d-1}$ is a module basis). In other words, if $L = k[x]/(p(x))$ is the field extension of $k$ by a root of $p(x)$, then $k[x,y]/(p(x))$ is $L[y]$, not $k[y]$.
With this change, your answer is correct: if $\mathbb{m}$ is a maximal ideal of $k[x, y]/(p(x)),$ then the ideal $(p(x), \mathbb{m})$ of $k[x, y]$ is maximal. The maximal ideals $\mathbb{m}$ of $k[x,y]/(p(x))$, however, aren't just the principal ideals generated by irreducible polynomials $g(y) \in k[y]$, because those polynomials may factor over the field $L = k[x]/(p(x))$. In the same example as before, $g(y) = y^2+1$ is irreducible in $k[y]$, but factors as $(y+x)(y-x) = y^2 - x^2 = y^2 + 1$ in $k[x,y]/(x^2+1)$. Instead, if $q(x,y)$ is an irreducible factor of $g(y)$ in $L[y] = (k[x]/(p(x)))[y] = k[x,y]/(p(x))$, then the ideal $(p(x), q(x,y))$ is maximal in $k[x,y]$.
• +1 Thanks for your comments. Sorry, I believe my notation is confusing. What I meant to claim is $F[x,y]/(p(x))\cong (F[x]/(p(x))[y]$ instead. I have made some edits in the question. – yoyostein Aug 25 '16 at 3:13
• Also, in my solution, is it correct to claim $\pi^{-1}(g(y)K[y])=g(y)R+p(x)R$? – yoyostein Aug 25 '16 at 3:24
• Ah, I see. Yes, that's correct, with the understanding that the co-efficients of $g(y)$ are polynomials in $x$ (because $g$ is a polynomial over the field extension $K = F[x]/p(x)$), i.e. I think you should write it as $g(x,y)$, a polynomial in $F[x,y]$. – anon Aug 25 '16 at 5:32