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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.

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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 '14 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.

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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 $F[x,y]$

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