Prime Ideal of a Polynomial Ring I am wondering how to show that the ideal (x,y) is prime and maximal in $\mathbb{Q}[x,y]$.
 A: 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).
A: 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 $0$)

A: 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.
