Calculate $\mathcal{I}(V)$ where $V=\{(x,y,z) \in k^3 : y=x^2, z=xy\}$.

Question

Let $V=\{(x,y,z) \in k^3 : y=x^2, z=xy\}$.

In trying to show that $\mathcal{I}(V)=(y-x^2,z-xy)$. I have a few questions.

1. $V=\mathcal{V}(y-x^2,z-xy)$, so at least I know (by the Nullstellensatz) that $\mathcal{I}(V)=\text{rad}(y-x^2,z-xy)$. How do I show that $(y-x^2,z-xy)$ is radical?

2. I know that $(y-x^2,z-xy)$ is radical iff $S:=k[x,y,z]/(y-x^2,z-xy)$ is reduced, so I tried to show that $S \simeq k[x]$. (Since $k[x]$ is obviously reduced, once I have this then the item above follows.) I can at least see that the obvious homomorphism $S \rightarrow k[x]$ given by identity on $k[x]$ and sending $y \mapsto x^2$, $z \mapsto x^3$ is surjective, but showing that it is injective boils down to showing that if $f(x,x^2,x^3)=0$ then $f \in (y-x^2,z-xy)$. How would I show this?

3. I had another idea for how to calculate $S:=k[x,y,z]/(y-x^2,z-xy)$. Is it correct? I worry that it's too informal:

   $S:=k[x,y,z]/(y-x^2,z-xy)=k[x,x^2,x^3]=k[x]$.

Answer 1

Lemma: let $R$ be a commutative ring with unity and $F,G\in R[X]$. Assume $G$ is almost monic (leading coefficient is a unit). Then $\exists Q,H\in R[X]$ such that $F=GQ+H$ with $\deg(H)<\deg(G)$

Proof: We use induction on $deg(F)$ . Say $F(X) = \sum _{i=0} ^n a_i X^i $ and $G(X)= \sum_{j=0} ^m b_j X^j$. Note that $b_m$ is a unit in $R$. If $ deg(F)<deg(G)$ take $Q=0, H=F$. Otherwise consider $F_1 := F- a_n b_m^{-1} X^{n-m}G$ and observe that $deg (F_1)< deg (F)$. Now use induction hypothesis and we are done.

Suppose $f\in k[X,Y,Z]$ such that $f(t,t^2,t^3)=0\ \forall t\in k$. Then using the division algorithm you can write $f=p(X,Y,Z)(Z-XY)+q(X,Y)(Y-X^2)+r(X)$ where $p,q,r$ are polynomials with coefficients in $k$. Then we get $r(t)=0\ \forall \ t \ \in k $ and hence $r(X)=0$ as $k$ is infinite. Thus $f=p(X,Y,Z)(Z-XY)+q(X,Y)(Y-X^2)$ as desired

Answer 2

Following the second approach. Given any polynomial in $f(x,y,z)$, changing all the $z$ that may appear in the polynomial by $xy$, you get a new polynomial $f_1(x,y)$ such that
$f(x,y,z)\in f_1(x,y)+(z-xy)$. Now changing all the $y$ that can appear in $f_1(x,y)$ by $x^2$, you end up with a polynomial $f_2(x)$ such that $f_1(x,y)\in f_2(x)+(y-x^2)$. Hence, $f(x,y,z)\in f_2(x)+(z-xy,y-x^2)$. Now, if $f(x,y,z)+(z-xy,y-x^2)=f_2(x)+(z-xy,y-x^2)$ is in the kernel of your morphism, this means that its image, which is $f_2(x)$ has to be $0$. So, $f(x,y,z)\in (z-xy,y-x^2)$.