Proving that $\mathbb C[x,y,z,w]/(xw-yz)$ is an integral domain using Eisenstein's criterion. Here is Einsteins's criterion for general rings I am using:



Let's write $\mathbb C[x,y,z,w] \cong \mathbb C[x,y,z][w]$.
Now, the ideal $(z)$ is prime in $\mathbb C[x,y,z]$ since $\mathbb C[x,y,z]/(z) \cong \mathbb C[x,y]$ is an integral domain.
Next, consider the polynomial $xw-yz$ as a polynomial in $w$ over $\mathbb C[x,y,z]$, i.e., $xw-yz \in \mathbb C[x,y,z][w]$.
Now by Eisenstein:
$\bullet$ $x \notin (z)$
$\bullet$ $-yz \in (z)$
$\bullet$ $-yz \notin (z^2)$,
and so $xw-yz$ is not the product of polynomials of degree $<1$ in $\mathbb C[x,y,z][w]$?
Therefore, $xw-yz$ is irreducible in $\mathbb C[x,y,z][w]$. Since $\mathbb C[x,y,z][w]$ is a UFD, then $xw-yz$ is prime and thus $(xw-yz)$ is a prime ideal. Hence $\mathbb C[x,y,z,w]/(xw-yz)$ is an integral domain.
Have I made correct use of Eisenstein's criterion? Is this sufficient?
 A: Yes, your use of Eisenstein's criterion is perfectly correct. The only problem is that you conclude that

...and so $−$ is not the product of polynomials of degree $<1$ in $ℂ[,,][]$.
Therefore, $−$ is irreducible in $ℂ[,,][]$.

But the above does not immediately imply that $xw-yz$ is irreducible. In fact the first sentence is already clear without Eisenstein's criterion; polynomials of degree $<1$ are constant, and so their product is also constant, so certainly not equal to $xw-yz$.
In fact, as stated the theorem does not seem directly useful for proving that $xw-yz$ is irreducible. On the other hand, since this polynomial is linear in $w$, if it is reducible then it must factor as
$$xw-yz=c\cdot q(w),$$
where $c\in\Bbb{C}[x,y,z]$ and $\deg q=1$. But then $c$ divides both $x$ and $yz$, so it is a unit in $\Bbb{C}[x,y,z]$ and hence in $\Bbb{C}[x,y,z,w]$, which shows that $xw-yz$ is irreducible.
On the other hand, Eisenstein's criterion in its slightly more general form, with a very similar proof, tells you that the polynomial is irreducible if  the $a_i$ together generate the whole ring $R$.
