$\mathbb{C}[x,y,z]/(xy-z^2)$ is not a field I know that the ring  $R=\mathbb{C}[x,y,z]/(xy-z^2)$ is not a field but an integral domain as you can easily show with Eisenstein criterion that $(xy-z^2)$ is an irreducible polynomial. Now I don't understand, why the irreducibility does not imply that $R$ is a field, because in other quotient rings I've seen before it was enough to show that " $I$ is irreducible then $R/I$ is a field ". Why here not? How can I show that $R$ is not a field here? Do I have to find a bigger Ideal than $(xy-z^2)$ in order to show that $(xy-z^2)$ is not maximal? I don't see any simpler way.
 A: If $R=A/I$, where $A$ is a commutative ring with $1\ne 0$, and $I$ is an ideal of $A$, then


*

*$R$ is an integral domain if and only if $I$ is a prime ideal.$\\[4pt]$

*$R$ is a field if and only if $I$ is a maximal ideal.


In the context of the current problem, let


*

*$A=k[x,y,z]$, where $k$ is a field.$\\[4pt]$

*$I=(xy-z^2)$.$\\[4pt]$

*$R=A/I$.


Since $A$ is a UFD and $xy-z^2$ is irreducible in $A$, it follows that $I$ is a prime ideal, hence $R$ is an integral domain.

To show that $R$ is not a field, it suffices to show that $I$ is not a maximal ideal.

But the ideal $M=(x,y,z)$ is clearly maximal, and we have $I\subseteq M$, hence it suffices to show $I\ne M$.

As an example, we have $z\in M$, but since $z$ is not a multiple of $xy-z^2$, it follows that $z\not\in I$.

Therefore $I\ne M$, so $I$ is not maximal, hence $R$ is not a field.

Notes:

Rings such as $\mathbb{Z}$ and $k[x]\;$(where $k$ is a field) are principal ideal domains (PIDs), i.e., integral domains such that every ideal is principal.

Since a PID is a UFD, every irreducible element is a prime element, and every prime element is irreducible.

In a PID, the maximal ideals are precisely the nonzero prime ideals, and an ideal is maximal if and only if it's a principal prime ideal (a principal ideal generated by a prime element).

But not all UFDs are PIDs. In particular, if $n > 1$, the ring $k[x_1,...,x_n]$ (where $k$ is a field) is a not a PID, since for example, the ideal $(x_1,...,x_n)$ is not principal.
A: For an element $\;p\in\Bbb C[x,y,z]\;$ , let us denotes by $\;\overline p\;$ its image in the quotient ring $\;R:=\Bbb C[x,y,z]/\langle xy-z^2\rangle\;$.
Then for example, $\;\overline{xy}\neq0\in R\;$ , yet there is no $\;q\in \Bbb C[x,y,z]\;$ s.t. $\;\overline{xy}\cdot \overline q=\overline1\;$ , since that'd mean that
$$xyq-1\in\langle xy-z^2\rangle\subset\Bbb C[x,y,z]\iff xyq+(xy-z^2)h=1\,,\,\,\text{for some}\;h\in\Bbb C[x,y,z]\implies\;$$
substituting $\;z=0,\,y=1\;$ , that'd mean $\;x(q-h)=1\;$ . End now the argument.
Another way: show that $\;\langle xy-z^2\rangle\subsetneq \langle xy-z^2,\,x\rangle\;$
A: $\mathbf C[x,y,z]$ is a U.F.D. (of Krull dimension $3$). As in all U.F.D.s, irreducible elements generate a prime ideal of height $1$, thus the quotient $\mathbf C[x,y,z]/(xy-z^2)$ is an integral domain of dimension $2$, and field has dimension $0$.
