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

Question

Let $F$ be a field, and $F[x,y]$ be a 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 $y^2-x$|$f*g$,can we claim that $y^2-x$|$f$ or $y^2-x$|$g$?

Answer 1

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.

Answer 2

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 $