Let $q(x_1,x_2,...,x_n) =\sum a_{ij}x_ix_j\in k[x_1,x_2,...,x_n] \; (a_{ij}=a_{ji})$ be a quadratic form in $n$ variables over a field $k$ of characteristic $\neq 2$, but not assumed algebraically closed nor anything else.
Question I
Are there any implications between the following properties?
a) The polynomial $q$ is irreducible i.e. it is not the product of two linear forms in $k[x_1,x_2,...,x_n]$.
b) The quadratic form $q$ is nondegenerate i.e. its rank is $n$ [the rank of $q$ being the rank of the matrix $(a_{ij})$] .
c) The quadric $V(q)\subset \mathbb P^{n-1}$ given by the equation $q=0$ is smooth.
Concretely c) means that the linear system of $n$ equations $\frac {\partial q} {\partial x_i}(a)=0$ has no solution $a\in k^n$ besides the solution $a=0$.
Question II
Is there a book in which these implications are spelled out?
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Georges Elencwajg
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About question I
- For $n=1$ all nonzero quadratic forms are nondegenerate, smooth and reducible.
- For all $n\geq 1$ the properties "nondegenerate" and "smooth" are equivalent.
Quadratic forms satisfying these conditions are often called regular in the literature. - For $n\geq3$ these properties "nondegenerate" and "smooth" imply "irreducible" .
- For $n=2$ the property "nondegenerate" (or "smooth") does not imply "irreducible".
For example the form $x_1x_2\in k[x_1,x_2]$ is non degenerate but of course the polynomial $x_1x_2$ is reducible. - However for $n=2$ "irreducible" $\implies$ "nondegenerate".
- For $n=3$ the implication "irreducible $\implies$ "nondegenerate" is false in general.
For example the quadratic polynomial $x_1^2+x_2^2\in \mathbb R[x_1,x_2,x_3]$ is irreducible but the corresponding quadratic form is degenerate.
On the other hand the implication "irreducible $\implies$ "nondegenerate" is true if $k$ is algebraically closed. - For all $n\geq 4$ and all fields $k$ the implication "irreducible" $\implies$ "nondegenerate" is FALSE!
It suffices to consider the quadratic polynomial $x_1^2+x_2^2+x_3^2 \in k[x_1,x_2,..., x_n]$ : that polynomial is irreducible over any field of characteristic not $2$, but the corresponding quadratic form is of course degenerate as soon as $n\geq 4$.
About question II
No, I don't know any book nor any other document in which these (easy) results are explicitly spelled out. Does some user know one?
Since questions of this type often pop up (see here, here or there ), I decided to try and give a reasonably complete answer.

Georges Elencwajg
- 150,790
I don't know if I can find 5 or 6 but, I didn't claim to be able to either.
– Eoin Sep 22 '20 at 21:07\end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 3 \ 2 & 4 & 5 \ 3 & 5 & 6 \ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & - 1 \ 0 & 0 & 1 \ 0 & 1 & - \frac{ 1 }{ 3 } \ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \ 0 & - 3 & 0 \ 0 & 0 & \frac{ 1 }{ 3 } \ \end{array} \right) $$ Output is in Latex code suitable for MSE.
– Will Jagy Sep 22 '20 at 21:46