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Let $Q(x_0, \cdots, x_n)$ be a homogeneous (square free) polynomial of degree two in $k[x_0, \cdots, x_n]$ ($k$ algebraically closed, $\mathrm{char}(k)\neq 2$). In other words $Q$ determines a quadric in $\mathbb{P}^n$.

We can always write this polynomial as a quadratic form: $Q = \mathbf{X}^T A \mathbf{X}$ where $\mathbf{X}^T=(x_0, x_1, \cdots, x_n)$. Also we might as well assume $A$ is symmetric and treat the evaluation of $Q$ as a symmetric bilinear form on $k^{n+1}$.

Is this true that $Q$ is irreducible if and only if $A$ is non-degenerate?

In fact it is pretty simple to show that if $A$ is non-degenerate then $Q$ is irreducible. It is also simple to show that if $Q$ is reducible then $A_\mathrm{red}=\frac{1}{2}(\mathbf{v}\mathbf{w}^T + \mathbf{w}\mathbf{v}^T)$ for some vectors $\mathbf{v}$ and $\mathbf{w}$, and a simple check shows that $\det A_\mathrm{red}=0$. So basically I'm asking about the things in between:

Is there any irreducible $Q$ such that $A$ is degenerate?

Hamed
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1 Answers1

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Any quadric under the hypothesis ($k$ algebraically closed, $\mathrm{char}(k)\neq 2$) can be easily transformed by a homogeneous linear change of coordinates to $x_0^2+\cdots+x_m^2$ for some $m\leq n$. The quadric is non-degenerate if and only if $m=n$, while it is irreducible if $m\geq 2$.

theHigherGeometer
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Mohan
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    But then if $-1$ is a square in $k$ (like in $\mathbb{C}$) then $x_0^2+x_1^2 = (x_0+ix_1)(x_0-ix_1)$ so this polynomial is not irreducible. – Hamed Sep 26 '15 at 18:08
  • Perhaps he is pointing out that you can write a second "if and only if". – Hoot Sep 26 '15 at 19:44
  • @Mohan Hi Mohan, do you have a reference for this? Thanks – AmorFati Jun 12 '18 at 22:06
  • @KyleBroder It probably can be found implicitly in many places, but it is fairly elementary to prove. – Mohan Jun 12 '18 at 22:41