Calculating the kernel of a homomorphism

Question

Let $R := k[x, y]$ be a polynomial ring over field $k$. Consider the homomorphism $\lambda : k[x, y, z] \to R \times R$, defined by $\lambda(x) := (x, x)$, $\lambda(y) := (y, y)$ and $\lambda(z) := (0, x)$. Why $\operatorname{Ker}(\lambda) = (z x − z^2)k[x, y, z]$?

Answer 1

Any ring map to a product of rings $\varphi : R \to S \times S'$ is equivalent to two ring maps $\varphi_1 : R \to S$ and $\varphi_2 : R \to S'$, via $\varphi(r) = (\varphi_1(r), \varphi_2(r))$. In particular $\ker \varphi = \ker \varphi_1 \cap \ker \varphi_2$.

Now if $\varphi_1, \varphi_2 : k[x,y,z] \to k[x,y,z]$ are given by $x \mapsto x, y \mapsto y, z \mapsto 0$ and $x \mapsto x, y \mapsto y, z \mapsto x$, then $\ker \varphi_1 = (z), \ker \varphi_2 = (z - x)$ (check this!). Furthermore if $f \in (z) \cap (z-x)$, then $f = gz = h(z-x)$ for some $g, h$, but $z$ is a prime element and $z \nmid z-x$, so $z \mid h(z-x) \implies z \mid h \implies f = (\frac{h}{z})z(z-x) \in z(z-x)$. Thus $\ker \lambda = \ker \varphi_1 \cap \ker \varphi_2 = (z) \cap (z - x) = (z^2-zx)$.

Answer 2

Hint:

As is mentioned by @Hagen von Eitzen, $\langle g\rangle\triangleq \langle zx-z^2\rangle$, the ideal generated by $g=zx-z^2$ is in $\ker(\lambda)$. For any polynomial $f\in\ker(\lambda)$, use division algorithm (see for example here): $$ f=(xz-z^2)q+r $$ where the remainder $r$ is not divisible by the leading term $xz$. Since $\lambda(f)=0$, we also have $\lambda(r)=0$. The only thing you need is to show that $r=0$ from the fact $\lambda(r)=0$. The monomials in $r$ are $x^iy^j, y^kz^l$.