$(x^2-yz,z-1)=(x^2-y,z-1)$ in $k[x,y,z]$

Question

Why is $(x^2-yz,z-1)=(x^2-y,z-1)$ in $k[x,y,z]$?

I tried to write $x^2-yz$ in terms of $x^2-y$ and $z-1$ and to write $x^2-y$ in terms of $x^2-yz$ and $z-1$.

Answer 1

Let $I = (x^2 - yz, z-1)$, and $J = (x^2-y,z-1)$.

Then $z - 1 \in I \implies z \equiv1 \pmod{I}$, hence

\begin{align*} x^2 - y &= x^2 - (y)(1)\\[3pt] &\equiv x^2 - (y)(z) \pmod{I}\\[3pt] &\equiv 0 \pmod{I}\\[3pt] \implies x^2 - y &\in I \end{align*}

It follows that $J \subseteq I$.

Similarly, $z - 1 \in J \implies z \equiv1 \pmod{J}$, hence

\begin{align*} x^2 - yz &\equiv x^2 - (y)(1) \pmod{J}\\[3pt] &\equiv 0 \pmod{J}\\[3pt] \implies x^2 - yz &\in J \end{align*}

It follows that $I \subseteq J$.

Therefore $I=J$.

Answer 2

To show the two ideals are equal, you want to write $x^2-yz$ as a linear combination of $x^2-y$ and $z-1$ with coefficients in $k[x,y,z]$ and similarly with $x^2-y$. You see they differ by $y$ and $yz$ terms, so try combinations where you multiply $z-1$ by $y$.

Answer 3

For example:

$$x^2-yz=-y(z-1)+(x^2-y)\in\left(\,x^2-y,\,z-1\,\right)$$