It is quite known fact that the determinant of arbitrary symmetric matrix is an irreducible polynomial in algebra $\mathbb C [x\_{ij}, 1\leq i,j\leq n]$ ($x\_{ij}=x\_{ji}$) (see this: atlas.mat.ub.es/personals/sombra/papers/cayley/cayley.ps ). 


Is there any geometric proof of this statement like the proof of irreducibility of determinant (from the biduality theorem in Gelfand-Zelevinsky-Kapranov book)?


Upd: I have found an algebraic proof of this fact (But I still need geometric). Since $\det (A A^{T})=\det(A)^2$ our polynomial (if not irreducible) is a square of irreducible. Since $\det diag(a\_1,\ldots ,a\_n)=a\_1\cdot\ldots\cdot a\_n$ our polynomial cannot be a square of any polynomial.



Irreducibility of determinant of symmetric matrix
I don't know a proof based on biduality, but here is a short geometric proof, which generalizes to other similar situations (all degenerate matrices, singular hypersurfaces and many other examples considered in the book by Gelfand, Kapranov and Zelevinsky).


In all these cases the discriminant variety admits a ``canonical'' resolution of singularities, which in the case of symmetric matrices is constructed as follows. Consider the space $X$ of couples (a degenerate symmetric $n\times n$ matrix, a 1-dimensional subspace in the kernel of the matrix). This projects both to the space of matrices and to $\mathbf{P}^n$. The second projection gives the structure of a vector bundle over $\mathbf{P}^n$ on $X$, so $X$ is irreducible. The image of $X$ under the first projection is the discriminant hypersurface, which is irreducible, so is given by an irreducible polynomial $f$. By the Nullstellensatz the determinant is a power of $f$ times a constant. Now consider the family of matrices with $t,1,1,\ldots, 1$ on the diagonal and zeroes elsewhere (here $t\in\mathbf{C}$). Restricting the determinant to this family we get $t$, so the determinant is precisely $f$.


