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Let $F$ be a field, and $F[x,y]$ be the ring of polynomials in two variables and we know that $F[x,y]$ is integral domain but not Principal Ideal Domain. We know that $y^2-x$ is irreducible in $F[x,y]$.

How to prove that $(y^2-x)$ is a prime ideal in $F[x,y]$?

If we let $f$ and $g$ be in $F[x,y]$, such that $y^2-x\mid fg$, can we claim that $y^2-x\mid f$ or $y^2-x\mid g$?

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    Can you recognize $F[x, y]/(y^2 - x)$?2011-08-03
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    yes,I have this question just because I try to prove $F[x,y]/(y^2-x)$ is an integral domain. let f(x,y),g(x,y) be in F[x,y], suppose $[f+(y^2−x)]*[g+(y^2−x)]=0$, we will get $y^2−x$|f(x,y)∗g(x,y), then that lead to this question, is $(y^2−x)$ a prime ideal in F[x,y]?2011-08-03
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    In a UFD irreducible implies prime.2011-08-03
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    Follow Qiaochu's suggestion, or try (proving and) using this description of the ideal: it's the set of polynomials $f(x,y)$ such that $f(y^2,y)$ is identically zero.2011-08-03
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    While F[x,y] is not a PID, it is a UFD. Do you know how that helps?2011-08-03
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    In a UFD irreducible implies prime, is this a theorem?2011-08-03
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    http://en.wikipedia.org/wiki/Unique_factorization_domain see properties2011-08-03

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