# Is $(y^2-x)$ a prime ideal in $F[x,y]$?

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 $F[x, y]/(y^2 - x)$? – Qiaochu Yuan Aug 3 '11 at 2:09
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]? – Youli Aug 3 '11 at 2:13
In a UFD irreducible implies prime. – jspecter Aug 3 '11 at 2:13
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. – Omar Antolín-Camarena Aug 3 '11 at 2:13
While F[x,y] is not a PID, it is a UFD. Do you know how that helps? – hardmath Aug 3 '11 at 2:14

In the quotient ring $F[x,y]/(y^2-x)$, you have the relation $x=y^2$, which means that $F[x,y]/(y^2-x)$ is isomorphic to $F[t]$ under $x \mapsto t^2$, $y \mapsto t$. Since $F[t]$ is a domain, $(y^2-x)$ is prime.

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HINT $\$ Since $\rm\:R[x]/(x-r)\ \cong\: R\:,\:$ we infer $\rm\ (x-r)\$ is prime in $\rm\:R[x]\iff R\:$ is a domain.

But in your case $\rm\: r = y^2\:$ and $\rm\ R = F[y]\:$ is a domain.

REMARK $\$ It is instructive to look at this equivalence a bit more explicitly

$\rm\qquad x-r\ \ prime\ \ \iff\ \ x-r\ |\ f(x)\ g(x)\ \Rightarrow\ x-r\ |\ f(x)\ \ or\ \ x-r\ |\ g(x)$

$\rm\qquad \phantom{\x-r\ \ prime} \iff\ \ \ \ \ \ f(r)\ g(r) = 0\ \ \ \Rightarrow\ \ \ \ f(r) = 0 \ \ \ \ or\ \ \ \ g(r) = 0$

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