I want to prove that this ideal: $I=(y^3-xz, xy^2-z^2, x^2-yz)$ is prime in $K[x,y,z]$. I think it would be a good idea to prove that the factor ring $K[x,y,z]/I$ has no zero divisors.
In this factor ring the following relations would be satisfied:
$y^3 = xz$
$xy^2=z^2$
$x^2=yz$
So, for example $y^4=x^3$ and $y^5=z^3$
Does that in some way help me prove that the factor ring is an integral domain?
Could you help me a bit?
Polynomials in $K[x,y,z]$ are linear combinations over $K$ of $\{x^ay^bz^c\mid a,b,c\geq 0\}$. Prove that, modulo $I$, every polynomial in $K[x,y,z]$ are linear combinations over $K$ of $\{y^a,xy^a,zy^a\mid a\geq 0\}$. (This can be done, it is tricky to write. Start with say $x^ay^b$, then decrease its $x$-degree by $x^2=yz$; after maybe $xz=y^3$, you will get term $y^cz^d$, where $d<a$. So, now use $z^2=xy^2$ and ... Consider other cases in a similar way.)
$\phi$ maps the set $\{y^a,xy^a,zy^a\mid a\geq 0\}$ to $\{t^{3a},t^{3a+4},t^{3a+5}\mid a\geq 0\}$, which is linearly independent over $K$. Therefore, for every linear combination $q$ of $\{y^a,xy^a,zy^a\mid a\geq 0\}$ over $K$, if $\phi(q)=0$, then $q=0$.
Now, take $p(x,y,z)\in\ker(\phi)$; write $p(x,y,z)$ modulo $I$ as a linear combination $q$ of $\{y^a,xy^a,zy^a\mid a\geq 0\}$ over $K$. Since $p=q\mod I$, then $\phi(q)=0$, hence $q=0$, and therefore $p\in I$.