$xy-zw$ is an irrreducible element in $\mathbb C[x,y,z,w]$

Question

Prove that $xy-zw$ is an irreducible element in the polynomial ring $\mathbb C[x,y,z,w]$.

My attempt was:

Consider the homomorphism from $\mathbb C[x,y,z,w]$ to $\mathbb C$ induced by the map $x,y,z,w$ onto $1,2,1,2$ respectively.

Since $\mathbb C$ is an integral domain the ideal generated by $xy-zw$ is prime and hence irreducible as $\mathbb C[x,y,z,w]$ is an integral domain.

Is this correct?

Thanks.

Answer 1

Note that $\mathbb{C}[x,y,z,w]/(xy-zw) \cong \mathbb{C}[x,\frac{zw}{x},z,w]$. (Try to construct an explicit isomorphism.) The right hand side is an integral domain, hence the ideal on the left is irreducible.

Answer 2

See this question for a general proof that the generic determinant is an irreducible polynomial. I will adapt the proof to this special case.

I will suppose $C$ is an integral domain (maybe you meant $\Bbb C$; that is a particular integral domain). Suppose $xy-zw=ab$ with $a,b\in C[x,y,z,w]$. Since the degree of $xy-zw$ in $x$ is $1$, one of $a,b$ must be of degree $1$ in $x$ and the other of degree $0$ (it does not contain $x$); by symmetry we may suppose $a$ is of degree $1$, say $a=px+q$ with $p,q\in C[y,z,w]$ and $p\neq0$. Now $b$ cannot contain either $z$ or $w$, lest the leading term in that variable produce a term in the product $pxb$ divisible by $xz$ respectively by $xw$ and which cannot be canceled by any other terms in the product $ab$, contradicting the fact that $ab=xy-zw$ contains no such terms. So $a$ is of degree $1$ in $z$ and $w$ as well as in $x$. Repeating the argument with $z$ in the place of $x$, we see that $b$ cannot contain any $y$ either (as $xy-zw$ has no terms divisible by $yz$), so it is a constant in $C$. But that constant divides the coefficients of all terms in $ab$, which means it is invertible, so the decomposition $ab$ is trivial. As a consequence $xy-zw$ is an irreducible polynomial in $C[y,z,w]$.

I used the fact that $C$ is an integral domain in the argument that the leading term of $b$ produces a (nonzero) term of $pxb$. I haven't tried to see if one can construct a ring with zero divisors over which $xy-zw$ factors non-trivially. But in any case question about factorisation usually suppose integral domains.

Answer 3

Consider $\phi: \mathbb{C}[x,y,z,w] \longrightarrow \mathbb{C}[t_1,t_2,t_3]$ where $\phi(x)=t_1$, $\phi(y)=t_1t_2t_3$, $\phi(z)=t_1t_3$, $\phi(w)=t_1t_2$. Now show that $\phi$ is surjective onto $\mathbb{C}[t_1,t_1t_2,t_1t_3, t_1t_2t_3]$ with the ideal $\langle xy-zw\rangle$ as the kernel.