Compute $Z(x^2+yz,y)$ and $I(Z(x^2+yz,y))$ (Zariski topology)

Question

Let $R=F[x,y,z]$ where $F$ is a field. I want to compute $V=Z(x^2+yz,y)$ and $I(Z(x^2+yz,y))$.

$$(x,y,z)\in Z(x^2+yz,y)\iff \begin{cases}x^2+yz=0\\ y=0\end{cases}\iff x=y=0$$ therefore, $$Z(x^2+yz,y)=\{(0,0,z)\in F^3\mid \forall f\in (x^2+yz,y), f(x,y,z)=0\}$$ Is it correct ? can I do better ?

Then, $$I(Z(x^2+yz,y))=\{f\in F[x,y,z]\mid \forall (x,y,z)\in V, f(x,y,z)=0\},$$ since $$(x,y,z)\in V\iff x=y=0,$$ we get $$f\in I(Z(x^2+yz,y))\iff f(0,0,z)=0\iff f\in (x,y).$$ Is it correct ?

Finally, why $$(x^2+yz,y)=(x^2,y)\ \ ?$$ Is it because $$f\in (x^2+yz,y)\iff f=g(x^2+yz)+hy=gx^2+y(h+z)\in (x^2,y)\ \ ?$$

Answer 1

(1) Yes. $Z(I) = \{(0, 0, z): z \in F\}$

(2) Yes. $I(Z(I)) = (x, y)$.

Hint: to see that $f(0, 0, a) = 0$ for all $a \in F$ implies $f \in (x, y)$ use polynomial division to get $f = gx + hy + r$, where $g \in F[x, y, z]$, $h \in F[y, z]$ and $r \in F[z]$.

(3) Yes.