I am reading Algebra by Artin and am facing a problem in the following example:
Let $f : R[x,y]$ to $R[t]$ be a ring homomorphism that sends $x$ to $t^2$ and $y$ to $t^3$ i.e., it sends $g(x,y)$ to $g(t^2,t^3)$. Then the polynomial $h = y^2-x^3$ lies in the kernel of $f$ .So $<y^2-x^3>$ belongs to ker($f$).
Then he says that ker($f$) = $<y^2-x^3>$ . One side's implication is clear and trivial. How do I prove the implication of the other side, namely that $ker(f)$ is a subset of $<y^2-x^3>$ ? Thanks in advance for any help.
Suppose $g(x,y) \in \operatorname{ker}(f)$, then we can apply to division algorithm to $g$ to obtain:
$$g(x,y)=h(x,y)(y^2-x^3)+r(x,y)$$
Where the degree of $y$ in $r(x,y)$ is one. In other words, $r(x,y)=p(x)y+q(x)$ for some polynomials $p,q\in \mathbb R[x]$
Then since $g$ and $(y^2-x^3)$ is in the kernel, $r$ must also be in $\operatorname{ker}(f)$. i.e:
$$p(t^2)t^3+q(t^2)=0$$
But $p(t^2)t^3$ consists only of odd powers of $t$, and $q(t^2)$ consists only of even powers of $t$, so $p$ and $q$ must both be zero respectively.
In other words, $g(x,y)=h(x,y)(y^2-x^3)$, and thus $g(x,y)\in (x^2-y^3)$.