Prove $(y-x^2)$ is a prime ideal in $\mathbb{R}[x,y]$, but not maximal.

Question

My guess is to use the fact that when we take the quotient, $\mathbb{R}[x,y]/(y-x^2)$, this will become an integral domain but not a field.

I am not sure how to take the quotient, though. I am also unfamiliar with the ring of polynomials of two variables. As a set, can I write $\mathbb{R}[x,y] = \{a + bx + cy + dxy + ex^2+fy^2+...| a,b,c,d,e,f,...\in \mathbb{R}\}$?

And would it be correct if I assume that in the quotient, $y = x^2$?

If my above two guesses are correct, then I would assume that the quotient becomes $\mathbb{R}[x]$, since all the $y$ terms can be turned into $x^2$. But isn't this ring an integral domain, since there are no zero divisors, but not a field, since not every real polynomial has an inverse?

Answer 1

Yes, the quotient is $\Bbb R[x]$. One way to see this is to consider the homomorphism $\phi:\Bbb R[x,y]\to\Bbb R[x]$ with $\phi(f(x,y))=f(x,x^2)$. This homomorphism is surjective, and its kernel is the ideal $(y-x^2)$, so by the First Isomorphism Theorem for rings, $\Bbb R[x,y]/(y-x^2)\cong \Bbb R[x]$.

As $\Bbb R[x]$ is not a field, then $(y-x^2)$ is maximal. But ine can write down lots of maximal ideals containing this ideal, for example $(x-a,y-a^2)$ for any $a\in\Bbb R$.

Answer 2

Indeed, $x\mapsto x$, $y\mapsto x^2$ gives us a homomorphism $\Bbb R[x,y]\to\Bbb R[x]$. Show that the kernel is $(x^2-y)$ and you are done as that shows $\Bbb R[x,y]/(x^2-y)\cong \Bbb R[x]$, which is an integral domain, but not a field.