# Equivalent determinantal hypersurfaces

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?

• 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. – Jason Starr May 13 '16 at 11:29
• What is the order of the matrices? What are the degrees of the entries? – Zach Teitler May 13 '16 at 17:45
• 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 $A$? – user46071 May 17 '16 at 14:47

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