Finding the kernel of ring homomorphisms from rings of multivariate polynomials I am trying to find the kernels of the following ring homomorphisms:
$$
f:\Bbb C[x,y]\rightarrow\Bbb C[t];\ f(a)=a\ (a\in\Bbb C),f(x)=t^2,f(y)=t^5.
$$
$$
g:\Bbb C[x,y,z]\rightarrow\Bbb C[t,s];\ g(a) = a\ (a\in\Bbb C), g(x)=t^2,g(y)=ts,g(z)=s^2.
$$
$$
h:\Bbb C[x,y,z]\rightarrow\Bbb C[t];\ h(a)=a\ (a\in\Bbb C), h(x)=t^2, h(y)=t^3, h(z)=t^4.
$$
I want to write them as ideals generated by as few elements as possible.
I am not used to this sort of question and I do not know what to do. For example, in case of $f$, if we let $a_{ij}$ be the coefficient of $x^iy^j$ in $p\in \Bbb C[x,y]$, then the coefficient of $t^k$ in $f(p)$ is
$$
\sum_{i,j\ge 0;\ 2i+5j=k} a_{ij}.
$$
Thus $p$ is in the kernel iff this is 0 for all $k>0$ and $a_{00}=0$.  However, I do not know how to write this condition in terms of (fewest possible) generators of ideals.  
I would appreciate your help.
 A: The base ring $k$ can be arbitrary.


*

*We have $x^5-y^2 \in \ker(f)$, hence $f$ lifts to a homomorphism of $k$-algebras $\overline{f} : k[x,y]/(x^5-y^2) \to k[t]$. The purpose of this factorization is that the quotient ring is much simpler: Every element of $k[x,y]/(x^5-y^2)$ has the form $p+qy$ for some $p,q \in k[x]$ (which are in fact unique). In order to show $\ker(f)=(x^5-y^2)$, we have to show that $\overline{f}$ is injective. So assume $p+qy$ is in the kernel, then $p(t^2) + q(t^2) t^5 = 0$ in $k[t]$. In the first summand there are only monomials with an even power of $t$, in the second summand we have odd powers. Hence, both summands vanish. But if $p(t^2)=0$, then $p=0$, and similarily $q=0$.

*We have $xz-y^2 \in \ker(g)$, hence $g$ lifts to $\overline{g} : k[x,y,z]/(xz-y^2) \to k[t]$. An element in the quotient ring has the form $p+qy$ for $p,q \in k[x,z]$. If it lies in the kernel, we have $p(t^2,s^2) + q(t^2,s^2) ts=0$ and exactly the same reason as above shows that both summands vanish, but then also $p,q$. This proves $\ker(g)=(xz-y^2)$.

*We have $x^2-z,\, x^3-y^2 \in \ker(h)$, so $h$ lifts to $\overline{h} : k[x,y,z]/(x^2-z,x^3-y^2) \to k[t]$. This is isomorphic to $k[x,y]/(x^3-y^2) \to k[t]$ mapping $x \mapsto t^2$, $y \mapsto t^3$. An elements of the quotient has the form $p+qy$ for $p,q \in k[x]$. If it lies in the kernel, we have $p(t^2) + q(t^2) t^3=0$ and again we can conclude $p=q=0$ by the same reason as above. This proves $\ker(h)=(x^2-z, x^3-y^2)$.
A: A solution for $f$:
Evidently, $y^2-x^5\in \ker f$. Let $p(x,y)\in \ker f$.
Divide as polynomials of $y$: $p(x,y)=q(x,y)(y^2-x^5)+r(x)y+s(x)$. Then  $r(x)y+s(x)\in \ker f$, i.e. $r(t^2)t^5+s(t^2)=0$. But the first summand contains only odd powers, while the second one  contains only even powers. Hence $r=s=0$ (since $f$ does not change coefficients) and $\ker f=(y^2-x^5)$.
As to $g$ and $h$, one can use the Buchberger's algorithm.
Addition: Similarly for $g$ divide on $y^2-xz$ as polynomials of $x$, you will get  $\ker f=(y^2-xz)$.
For $h$ the ideal  $\ker h$ is not principal, it is generated (at least) by $y^2-x^3$ and $z-x^2$, so you can divide at first  on $z-x^2$ as polynomials of $z$ and then ...
