1
$\begingroup$

I have two matrices $A$ and $B$ (of the same order) whose entries are homogeneous polynomial of the same degree. I have that $\det A=0$ and $\det B=0$ define the same hypersurface of $\mathbb{P}^n$ (not two equivalent ones but just the same) but I have proved that $A$ cannot be written as $A=MBN$ with $M$ and $N$ invertible matrices. How is that possible? How $A$ and $B$ should be related?

$\endgroup$
  • 1
    $\begingroup$ You did not ask this, but anyway, this phenomenon is responsible for several "descent" results, i.e., certain incidence divisors, etc., descend from a "big" space such as the Hilbert scheme to a "small" space such as the Chow variety precisely because different matrices can give the same determinant. $\endgroup$ – Jason Starr May 13 '16 at 11:29
  • $\begingroup$ What is the order of the matrices? What are the degrees of the entries? $\endgroup$ – Zach Teitler May 13 '16 at 17:45
  • $\begingroup$ I have a square $A$ matrix of order 3, the entries are homogeneous polynomials of degree 3 in 9 variables. I know that $\det A=0$ defines a reducible hypersurface consisting of 3 hypersurfaces of degree 3. What can I say about $A$? $\endgroup$ – user46071 May 17 '16 at 14:47
4
$\begingroup$

An example would be $A = \text{diag}(X_0X_1, X_0X_1)$ and $B = \text{diag}(X_0^2, X_1^2)$. In this case you can look at the structure of the cokernels $Q_A$, $Q_B$ of the corresponding endomorphisms $k[X_0, \ldots, X_n]^{\oplus 2}$ to tell them apart. For $A$ you get that $Q_A$ is annihilated by $X_0$ after you invert $X_1$, but for $Q_A$ this is not the case. In your example, is the hypersurface irreducible? (That would be more interesting.)

$\endgroup$
  • $\begingroup$ It is reducible. Actually, $B$ is a diagonal matrix, $A$ is not. $\endgroup$ – user46071 May 13 '16 at 12:07
  • $\begingroup$ If $B$ is diagonal, its determinant is the product of the diagonal entries. How can that be irreducible unless its size is one? $\endgroup$ – Mohan May 13 '16 at 12:54
  • $\begingroup$ I said that it is reducible indeed. But $A$ is not a diagonal matrix. $\endgroup$ – user46071 May 13 '16 at 12:59

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.