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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?

1 Answers1

3

About question I

  1. For $n=1$ all nonzero quadratic forms are nondegenerate, smooth and reducible.
  2. For all $n\geq 1$ the properties "nondegenerate" and "smooth" are equivalent.
    Quadratic forms satisfying these conditions are often called regular in the literature.
  3. For $n\geq3$ these properties "nondegenerate" and "smooth" imply "irreducible" .
  4. 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.
  5. However for $n=2$ "irreducible" $\implies$ "nondegenerate".
  6. 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.
  7. 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.

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    I'm sure that (most of) these results are contained inside The algebraic and geometric theory of quadratic forms. – Eoin Sep 22 '20 at 19:39
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    @Eoin. No, they are definitely not. I challenge you to post the 7 page numbers corresponding to my 7 points. Anyway, I think a vague comment that ("most" ?) results are somewhere inside a 450 page book is not especially helpful. – Georges Elencwajg Sep 22 '20 at 19:47
  • 1903 answers and one question. good ratio. You might enjoy Classical Groups and Geometic Algebra by Larry C. Grove. everything relevant is surely in Lam, Quadratic Forms over Fields, but I don't remember answering any question on MSE by quoting from Lam. Also, as to my interest, he does not do spinor genus. He does do the spinor norm as a group homomorphism as Theorem 1.13 on page 108, in the chapter on Clifford algebras. Does not define genus, I guess that is what happens when the coefficients are in a field..... – Will Jagy Sep 22 '20 at 20:07
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    Thanks for your comment, @Will Jagy. I checked in Lam (whom I greatly admire) but could find nothing related to my question. Could you please give a precise reference? – Georges Elencwajg Sep 22 '20 at 20:32
  • Proposition 7.19 on p. 43 (nondegenerate), Remark 7.20 on p. 44 (regular), this is a stretch but Corollary 7.30 on p. 45 and Notation 7.3 on p. 40 (reducible)
  • Proposition 22.1 on p. 93
  • Page 94
  • If n=2, then nondegenerate does not imply irreducible (an isotropic nondegenerate form of dimension 2 is a hyperbolic plane, Proposition 7.13, e.g. xy Example 7.5).
  • Page 93 (an irreducible quadric of dimension 0 is a point)
  • 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
  • Georges, now that I have printed out the three past questions, I think my contribution to the discussion is my question http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr and my answer. Evidently the method I settled on is taught in some countries in college. I went as far as writing a C++ program that begins with inputting a symmetric matrix $H$ of (rational) integers. It produces rational matrices $P,Q,D$ of determinant $\pm 1$ with $PQ=QP = I$ and $D$ diagonal, then $P^TH P = D$ and $Q^T DQ = H.$ – Will Jagy Sep 22 '20 at 21:46
  • $$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \
    • 3 & 0 & 1 \
    • 1 & 1 & - \frac{ 1 }{ 3 } \

    \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
  • $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \ 2 & \frac{ 1 }{ 3 } & 1 \ 3 & 1 & 0 \ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \ 0 & - 3 & 0 \ 0 & 0 & \frac{ 1 }{ 3 } \ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 3 \ 0 & \frac{ 1 }{ 3 } & 1 \ 0 & 1 & 0 \ \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 3 \ 2 & 4 & 5 \ 3 & 5 & 6 \ \end{array} \right) $$ – Will Jagy Sep 22 '20 at 21:47
  • @Eoin. Of course everything I wrote can be deduced from the extremely advanced book you mention. Every question on this site, which is not a research journal and contains no original result, has its answer in some book. My aim was to have a concise summing up of the main relations between the properties I examine. You got 23 upvotes for proving that the image of a pathwise connected space under a continuous map is pathwise connected. Is that stated in no book? Did somebody comment "I'm sure that this is contained in Bourbaki's Topology" ? – Georges Elencwajg Sep 22 '20 at 23:19
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    @Georges No, of course the post of mine that you cite is in a number of books or notes on topology. I also don't know what information the OP has to ask this question; I was only providing a reference that the OP could look to if they were trying to find more about the subject and your examples. Lam's book was another that I thought could be a good reference. (Both of these though are really just comments). – Eoin Sep 22 '20 at 23:19
  • @Eoin. Ok, I think we can agree on your last comment. – Georges Elencwajg Sep 22 '20 at 23:21