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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.

user38764
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    It is not obvious that the polynomial $xy-zw$ generates the kernel of the homomorphism you consider. – Hagen Knaf Nov 20 '12 at 08:36
  • @user38764:$x-z \not\in\langle xy-zw\rangle$ but $x-z$ belongs to the kernel. – P.. Nov 20 '12 at 09:24
  • Yes. But the homomorphism is not zero and the ideal generated by $xy-zw$ is contained in the kernel, hence there is a well defined homomorphism into $C$ – user38764 Nov 20 '12 at 10:00

3 Answers3

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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.

Community
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Marc van Leeuwen
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    From $xy-zw=(px+q)b$ (seen as an equality of polynomials in $x$ with coefficients in $C[y,z,w]$) we can conclude that $y=pb$ and $-zw=qb$ and the proof is almost done. – user26857 Aug 24 '14 at 22:22
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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.

Fredrik Meyer
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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.

baharampuri
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  • You can use $\langle \ldots \rangle$ for $\langle \ldots\rangle$. – martini Nov 20 '12 at 11:23
  • Can you show that $t_1t_2t_3\in\mathbb{C}[t_1,t_1t_2,t_1t_3]$? Even if you change $\mathbb{C}[t_1,t_1t_2,t_1t_3]$ with $\mathbb{C}[t_1,t_1t_2,t_1t_3,t_1t_2t_3]$ you are far from over. To conclude: I consider your answer as a hint which leads to nowhere. –  Sep 22 '13 at 08:59
  • I have corrected the typo why is this answer getting down voted? – baharampuri May 30 '17 at 12:55