# Irreducible degree two homogeneous polynomials in polynomial ring in 4 variables

I was wondering if polynomials $$xw-y^2$$, $$yz-w^{2}$$ and $$xz-yw$$ are irreducible in polynomial ring $$R = K[x,y,z,w]$$, where $$K$$ is a field of characteristic zero.

Since the polynomial ring is a UFD, these polynomials are irreducible if and only if they are prime. So one way to show this would be if $$R/(p)$$ is a domain for $$p$$ any of the polynomials above.

If $$I = \langle xz-yw\rangle$$, I have a guess that $$R/I$$ would just be polynomial ring $$K[x,z]$$ which is clearly a domain. But I don't know if that is true. Also I don't have much of an idea for other two polynomials, although I note that argument is going to be same for both of them.

• I've edited your question to make use of MathJax, the typesetting system for math on this website. Please use it in the future! It makes questions easier to read and understand. Commented Sep 25, 2018 at 1:02

First: your claim about $$k[x,y,z,w]/(xz - yw)$$ being $$k[x,z]$$ is not quite correct. We would have $$k[x,y,z,w]/(y,w)\cong k[x,z],$$ or even that $$k[x,y,z,w]/(x - y, z - w)\cong k[x,z].$$ However, when forming $$k[x,y,z,w]/(xz - yw),$$ you have the relation that $$xz = yw,$$ which you can't split up into the two relations $$x = y$$ and $$z = w.$$ In fact, there cannot be any abstract isomorphism between $$k[x,y,z,w]/(xz - yw)$$ and $$k[x,z]:$$ $$k[x,z]$$ is a UFD, but $$k[x,y,z,w]/(xz - yw)$$ is not!
Now, as you noted, irreducibility of your first two polynomials is equivalent: we can define an automorphism of $$k[x,y,z,w]$$ by \begin{align*} \varphi : k[x,y,z,w]&\to k[y,w,x,z] = k[x,y,z,w]\\ x&\mapsto y,\\ y&\mapsto w,\\ w&\mapsto z,\\ z&\mapsto x. \end{align*} Applying $$\varphi$$ to $$xw - y^2$$ gives $$\varphi(xw - y^2) = \varphi(x)\varphi(w) - \varphi(y)^2 = yz - w^2.$$ Therefore, the first polynomial is irreducible if and only if the second is.
For $$xw - y^2,$$ we may write $$k[x,y,z,w] = k[x,z,w][y].$$ Then if $$\mathfrak{p} = (x),$$ it is clear that $$\mathfrak{p}$$ is a prime ideal in $$k[x,z,w],$$ and moreover that $$xw\in\mathfrak{p},$$ but $$xw\not\in\mathfrak{p}^2 = (x^2).$$ Thus, $$-y^2 + xw$$ satisfies Eisenstein's criterion for $$\mathfrak{p} = (x),$$ and is irreducible.
You can play the same game with $$xz - yw:$$ again write $$k[x,y,z,w] = k[x,z,w][y],$$ and again let $$\mathfrak{p} = (x).$$ $$\mathfrak{p}$$ is still prime, and $$xz\not\in\mathfrak{p}^2 = (x^2).$$ Thus, $$xz - yw$$ is irreducible.
Notice that nowhere did I assume anything about the characteristic of $$k$$! The above holds for all fields.