TITLE: Is the ideal $I:=\langle xy-z,x^5-z^3 \rangle $ prime in $\Bbb{C}[x,y,z]$?
QUESTION [6 upvotes]: Let us take the ideal $I:= \langle xy-z,x^5-z^3 \rangle $ of the ring
  $\Bbb{C}[x,y,z].$ We want to find if this ideal is prime.

My thoughts: We define $f:=xy-z,\ g:=x^5-z^3 \in \Bbb{C}[x,y,z].$ The first thought is to prove that  the quotient ring $\Bbb{C}[x,y,z]/\langle xy-z,x^5-z^3\rangle$ is an integral domain.
We observe that $f:=-z+xy\in \Bbb{C}[x,y][z]$ is irreducible in $\Bbb{C}[x,y][z]=\Bbb{C}[x,y,z]$ since $f$ has degree $1$ and $-1\in U(\Bbb{C}[x,y])=\Bbb{C^*}$ . And, from the general Eisenstein Criterion, we can take that $g$  is irreducible too.
1) Is it true that (since $f,g$ are irreducible ) $\gcd (f,g)=1\implies 1\in I \iff I=\Bbb{C}[x,y,z]?$ And if the answer is no, why?
2) I found in this post that although in $K[x]$  an irreducible polynomial generates a prime (maximal) ideal, now this is not always true, so we can not claim from irreducibility of $f,g$, that $I$ is prime. 
Could you please give me a help?
Thank you in advance.

REPLY [3 votes]: Here is a (possibly) simpler and more intuitive way to understand what happens when you take the quotient of $R=\mathbb C[x,y,z]$ by the ideal $I=\left< xy-z , x^5 - z^3 \right>$.
When we find a quotient of a ring by an ideal, we are essentially setting everything in the ideal equal to $0$.  That amounts to introducing "replacement rules" that allow us to rewrite polynomials in $R$ in a different (hopefully simpler) form.
In this particular case, for example, the element $xy - z \in I$ tells us that in $R/I$, we have $xy-z=0$, so $xy=z$.  That means that any polynomial in three variables can be rewritten as an equivalent polynomial in just two variables by replacing every occurrence of $z$ with $xy$.  So, for example, the polynomials $x^2+y+z^3$ and $x^2+y+(xy)^3$ are different elements of $R$, but are identified in $R/I$.  The practical upshot of this is that $\mathbb C[x,y,z]/\left<xy-z\right> \cong \mathbb C[x,y]$.
Now, what about the second generator of $I$, namely $x^5 - z^3$?  Once we have already taken account of the replacement rule in the previous paragraph, this tells us that in $R/I$ the polynomial $x^5 - (xy)^3$ is set equal to $0$.  Equivalently, we have $x^3(x^2-y^3) = 0$ as an identity in $R/I$.
But this clearly shows that $R/I$ is not an integral domain, because in $R/I$ we have $x^3 \ne 0$ and $x^2-y^3\ne 0$, but their product is $0$.  If you want you can also express this in terms of the original ideal $I$:  we have $x^3\notin I$ and $x^2-y^3\notin I$, but $x^3(x^2-y^3) \in I$.  So $I$ is not prime.