# Prime Ideal of a Polynomial Ring

I am wondering how to show that the ideal (x,y) is prime and maximal in $\mathbb{Q}[x,y]$.

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Look at the ring homomorphism $\varphi: \mathbb{Q}[x,y] \rightarrow \mathbb{Q}$ such that $\varphi(f)=f(0,0)$. What is the kernel of this map? –  Prometheus Jan 27 '11 at 12:13
@Prometheus: the kernel is exactly (x,y) and the image of the homomorphism is Q, and then using the first isomorphism to conclude that (x,y) is maximal because Q[x,y]/(x,y) is a field –  user6308 Jan 27 '11 at 12:30

Look at the quotient ring $\mathbb{Q}[x,y]/(x,y)$. Then use that an ideal $I \subset R$ is maximal iff $R/I$ is a field. Using the fact that $I$ is prime iff $R/I$ is an integral domain, you will also be able to show that $(x,y)$ is prime (and all maximal ideals are).

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Here's a hint extracted from one of my old posts (which shows even more)

Even easier: the ideals $\rm\ \ (x)\:\ <\ (x,y)\ <\ 1\$ are distinct primes$\ \ \$ (or $1$)
since the residue rings $\rm\ \mathbb Q[y]\ >\ \ \ \mathbb Q\ \ \ \ >\ 0\$ are distinct domains (or $0$)

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Why is this easier: is it not exactly equivalentto what was proposed by Sebastian and by Prometheus in his comment? –  Mariano Suárez-Alvarez Jan 28 '11 at 18:58
@Mariano: You misunderstand: the "easier" refers to the post from where it was excerpted. –  Bill Dubuque Jan 28 '11 at 19:11

Let $I$ be an ideal of $\mathbb{Q}[x,y]$ such that $(x,y) \subsetneq I \subseteq \mathbb{Q}[x,y]$. Choose $a \in I$ with $a \not \in (x,y)$. Then $a \in \mathbb{Q}$ and we conclude $I = \mathbb{Q}[x,y]$ (why?) and $(x,y)$ is maximal.

Now let $F, G \in \mathbb{Q}[x,y]$ with $FG \in (x,y)$. Note $F$ and $G$ cannot both have a constant term, so one must be in $(x,y)$. So $(x,y)$ is prime.

We could have deferred to the fact that all maximal ideals are prime. Also, Sebastian's answer is probably more practical (it is surely how I would approach this problem). However, sometimes I like seeing definitions work and thought others might as well.

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