Specifically, I'm trying to solve the following problem:
Let $R$ be an integral domain and let $x$, $y$ and $t$ be indeterminates. Let $R[x,y]$ denote the ring of polynomials in $x$ and $y$ over $R$, and $R[t]$ the ring of polynomials in $t$ over $R$.
It can be verified (you need not do so) that the map $\Phi:R[x.y]\rightarrow R[t]$ by $\Phi(p(x,y))=p(t^2,t^3)$ is a ring homomorphism.
(a) Prove that the kernel of $\Phi$ is the ideal generated by $x^3-y^2$.
(Hint: it may be helpful to express $p(x,y)$ as
$$p(x,y)=(x^3-y^2)q(x,y)+x^2a(y)+xb(y)+c(y),$$
where $q(x,y)\in R[x,y]$, and $a(y)$, $b(y)$, $c(y)\in R[y]$.)(b) Prove that the ideal generated by $x^3-y^2$ is a prime ideal of $R[x,y]$.
For part (a), it is clear that $(x^3-y^2)\subseteq\ker(\Phi)$, but I'm struggling to show that $\ker(\Phi)\subseteq(x^3-y^2)$. It's not even entirely apparent to me that any $p(x,y)\in R[x,y]$ can be expressed in the way suggested in the hint, but assuming that is can, I suppose the idea is to show that whenever
$$\Phi(p(x,y))=\Phi((x^3-y^2)q(x,y)+x^2a(y)+xb(y)+c(y))=0+t^4a(t^3)+t^2b(t^3)+c(t^3)$$
is equal to $0$, it must be the case that $p(x,y)\in(x^3-y^2)$.
For part (b), it seems like it's easier to show that if $a,b\notin(x^3-y^2)$, then $ab\notin(x^3-y^2)$ than it is to show that if $ab\in(x^3-y^2)$ then either $a\in(x^3-y^2)$ or $b\in(x^3-y^2)$, but I'm not sure how to approach either of these directions.
Best Answer
Suppose $p \in \operatorname{Ker}$, then $\Phi(p)=0$, so $t^4\alpha(t^3)+t^2b(t^3)+c(t^3)$ is the zero polynomial in $R[t].$ If any of $\alpha, b, c$ were not the zero polynomials in $R[t]$, then we have a contradiction, as then the polynomial $f(t)=t^4\alpha(t^3)+t^2b(t^3)+c(t^3)$ is non-zero. Therefore, $p(x,y)= q(x,y)(x^3-y^2) \in (x^3-y^3)$. Finally, by the first isomorphism , we have $R[x,y]/(x^3-y^2) \cong R[t^2,t^3]$ which is an integral domain, so $x^3-y^2$ is a prime ideal. ($\Phi$ is not surjective in this case, ie $t$ is not in the image.)