Irreducibility of a polynomial in $F[x,y]$.

Question

This is a rather easy question, but I'm not entirely confident in why I think this is true.

Let $F$ be a field. Consider $F[x,y]$. I want to show that the ring $F[x,y]/(y^{2} - x)$ is an integral domain.

My thought process is that because $F$ is a field, $F[x,y]$ is a unique factorization domain. I'm inclined to say that $y^{2} - x$ is irreducible in $F[x,y]$ (and therefore a prime ideal, so the quotient we are considering would indeed be an integral domain). But I'm not entirely confident in my judgment that $y^{2} - x$ is irreducible. Is it a stretch to say that this follows from Eisenstein's criterion?

note: $F[x,y]/(y^2-x)$ is not a polynomial (and polynomials can't be integral domains) — anon, Mar 07 '14 at 00:55
@user121097 Assuming the leading coefficient is a unit in the scalar ring, yes. (For example $2T-2$ is not irreducible in $\Bbb Z[T]$.) — anon, Mar 07 '14 at 01:12
you can also use the idea from this answer: http://math.stackexchange.com/a/487064/ — Camilo Arosemena-Serrato, Mar 07 '14 at 15:54

score 7 · Accepted Answer · answered Mar 07 '14 at 01:23

It does follow from Eisenstein's Criterion. View $y^2 - x$ as a polynomial in $y$, with coefficients in $F[x]$. Then $x \in F[x]$ is a prime element (since $F[x]/(x) \cong F$ is a domain), and $x$ divides all coefficients of $y^2 - x$ except that of the leading term $y^2$, and $x^2$ does not divide the constant term $-x$. By Eisenstein, $y^2 - x$ is irreducible in $F[x][y] = F[x,y]$.

In fact, this reasoning works even if $F$ is only a domain (not even necessarily a UFD).

Irreducibility of a polynomial in $F[x,y]$.

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