Is $(x^2 - x + yz)$ a prime ideal in $\mathbb{C}[x,y,z]$?
I have been trying to prove that it is.
My first approach was to try to show that $f(x,y,z) = x^2 - x + yz$ is irreducible but I wasn't able to do it.
I also tried to show that $\frac{\mathbb{C}[x,y,z]}{(x^2 - x + yz)}$ is an integral domain, but I don't know what the quotient ring looks like so that didn't help either.
Suppose towards a contradiction that $f=x^2-x+yz$ is reducible. Then there exist nonconstant $g,h\in\Bbb{C}[x,y,z]$ such that $f=gh$. Comparing degrees shows that $$\deg_zg+\deg_zh=\deg_zgh=\deg_zf=1,$$ so without loss of generality $g=uz+v$ with $u,v\in\Bbb{C}[x,y]$ and $h\in\Bbb{C}[x,y]$. We see that $$x^2-x+yz=f=gh=uhz+vh,$$ which shows that $uh=y$ and $vh=x^2-x$. In particular $h$ divides both $y$ and $x^2-x$, contradicting our assumption that $h$ is nonconstant. Hence $f$ is irreducible.
Analogous arguments work when comparing the degrees in $x$ or $y$, or the total degrees; try them all out.