Show that $f(y)$ is not in the ideal $\langle xy, x-yz \rangle \triangleleft k[x,y,z]$

Question

I thought I conceptually understood this as soon as I saw it. But, looking into it some more I am getting very confused. We are given that the variety of the ideal is $$V(I) = \lbrace (x,y,z)\in k[x,y,z]:x=y=0\,\,\text{or}\,\, x=z=0\rbrace$$ where $k$ is a field. And from here apparently there is the obvious conclusion that there is no polynomial just dependant on $y$ in the ideal. Is there a proof by contradiction showing this?

Answer 1

We can observe that $V(f(y))$ is a finite union of parallel planes: named $H_y=\{y=0\}$ and $p_i$ the point $(0, y_i, 0)$ where $f(y_i)=0$, then: $$ V(f(y)) = \bigcup_{i=1}^k \left(p_i + H_y\right) $$ By contradiction, if there exist a polynomial $f(y)\in I$, then $$ V(f(y))\supseteq V(I) $$ But the line $\{x=z=0\}\subset V(I)$ is in general not contained in $V(f(y))$: if $k$ is an infinite field, just take the a ponint $P=(0,\alpha,0)$, where $\alpha\neq y_i$ for all $i=1,...,k$; $P\in V(I)$ but $P\notin V(f(y))$.

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If the field $k$ is finite, I think we can use the same argument is a similar way in order to make some remarks: in order to have the line $\{x=z=0\}$ contained in $V(f(y))$ we need that $f(y_i)=0$ for all $y_i\in k$. This is possible now: take $k=\mathbb F_{p^n}$, and the polynomial $f(y)=y^{p^n} - y$. Then $$ V(f(y)) = \bigcup_{y_i\in \mathbb F_{p^n}} \left((0,y_i,0) + H_y\right) = \mathbb F_{p^n}^3\supset V(I) $$ Hence $y^{p^n} - y$ is a good candidate. Maybe it works?