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

ogerard
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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? – Ofir Jan 27 '11 at 12:13
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    @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 –  Jan 27 '11 at 12:30

3 Answers3

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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).

Sebastian
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Hint $ $ (excerpted from one of my old posts - which shows even more)

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

Bill Dubuque
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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.

Hans Parshall
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