# Find the kernel of a homomorphism between polynomial rings

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.

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.)
• Oh, duh. Part (a) was obvious from the point I got to. I follow the use of the isomorphism theorem, but why does $R[t^2,t^3]$ being a integral domain imply that $(x^3-y^2)$ is prime? Commented Jan 4, 2020 at 16:34
• Because $\;R[x,y]/(x^3-y^2)\simeq R (t^2,t^3]$. Commented Jan 4, 2020 at 16:37
• There's this theorem that say $I$ is a prime ideal of a ring $R$ iff the quotient ring $R/I$ is an integral domain. Similarly, $I$ is a maximal ideal iff $R/I$ is a field. Commented Jan 4, 2020 at 16:37
• I unaccepted your answer because I'm now a bit confused. Do you mean $R[t]$ instead of $R[t^2,t^3]$? Also since $\Phi$ is not surjective, we can't directly apply the first isomorphism theorem, so maybe we need to use a subring? Commented Jan 7, 2020 at 17:44
• First iso theorem says $R[x,y]/(x^3-y-2)$ is isomorphic to the image of $\Phi$, which is $R[t^2,t^3]$. Commented Jan 7, 2020 at 18:30