# If F is a field, then $F[x,y]$ is a Principal Ideal Domain?

Let $F$ be a field, and $F[x,y]$ be a ring of polynomials in two variables. Is $F[x,y]$ a Principal Ideal Domain?

Also show that $F[x,y]/(y^2-x)$ and $F[x,y]/(y^2-x^2)$ are not isomorphic for any field $F$.

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No; for example, $(x,y)$ is a maximal ideal which is not principal in $F[x,y]$.
And also, $F[x,y]/(x^2-y^2)$ is not an integral domain since $(x-y)(x+y)=x^2-y^2$. On the other hand, the polynomial $y^2-x$ is irreducible and hence $F[x,y]/(x-y^2)$ is an integral domain.
What does the notation $(x,y)$ stand for? –  Manos Jul 28 '11 at 16:18
The ideal generated by $x$ and $y$! –  Bruno Joyal Jul 28 '11 at 16:29
why $F[x,y]/(y^2-x)$ is an integral domain? I try to prove it, let f(x,y),g(x,y) belong to 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 another question, is $(y^2-x)$ a prime ideal in F[x,y]? (we know F[x,y] is not PID, even if $y^2-x$ is irreducible, we still don't know whether it is prime$) – Youli Aug 2 '11 at 21:10 One way to see that the ideal generated by$y^2-x$is prime in$F[x,y]$is to note that it's prime in$F(x)[y]$. – Bruno Joyal Aug 3 '11 at 17:03 Another way to see that$F[x,y]/(x-y^2)$is an integral domain is to observe that dividing by the ideal$(x-y^2)$amounts to identifying$x$with$y^2$, so the quotient ring is isomorphic to$F[y]$. – Andreas Blass Aug 7 at 20:45 If$A$is a commutative ring, a classical result states that the polynomial ring$A[x]$is a PID if and only if$A$is a field. It is a good exercise. In your case, as$F[x]$isn't a field,$F[x,y] \simeq (F[x])[y]$cannot be a PID. (I'm not claiming it's the best proof). For the second question, Bruno's answer will be hard to improve upon. - One more easy way to see is,$(y^2-x)|f(x,y)g(x,y) \implies$exist$f_0(x),g_0(x)\in F[x]$such that$y^2-x|yf_0(x)+g_0(x)$(if none of$f,g$are zero) which is clearly absurd. So,$(y^2-x)$is prime ideal of$F[x,y]\$