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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.

KReiser
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HumbleStudent
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  • 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. – KReiser Sep 25 '18 at 01:02

1 Answers1

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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.

To actually show irreducibility, I would not try to compute these quotients explicitly. Instead, I'd apply the generalized Eisenstein's criterion.

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.

Stahl
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