# Intersection of Algebraic Varieties.

Let $\mathbb K$ be an algebraically closed field. Consider the set $M_n(\mathbb K)$ of all matrices of order $n$. Identify the set $M_n(\mathbb K)$ with the affine space $\mathbb A^{n^2}_{\mathbb K}$.

The set $V_1=\{A\in M_n(\mathbb K):A=-A^T\}$ of anti-symetric matrices is an algebraic variety of dimension $\frac{n^2-n}{2}$.

The set $V_2=\{A\in M_n(\mathbb K):\det(A)=0\}$ of singular matrices is an algebraic variety of dimension $n^2-1$.

My question is:

The set $V_1\cap V_2$ is algebraic variety, that is, it is irreducible as an algebraic set? If the answer is yes, what is its dimension?

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If $n$ is odd, $V_1\cap V_2=V_1$, so the answer is certainly yes in that case. –  Matthew Pressland Apr 24 '13 at 18:23
One thing that might be useful is that every component of the intersection not contained in V_2 has the expected dimension, namely $\frac12 n(n-1)-1$. This is a basic fact about intersecting with hypersurfaces: it is in Hartshorne Ch.1 or Shafarevich. Sadly, I don't know a good way to check irreducibility. –  Asal Beag Dubh Apr 24 '13 at 19:35
You probably want to assume $\mathrm{char}(K) \neq 2$? –  Martin Brandenburg Apr 24 '13 at 20:10
Yes, I assume $char(K)\neq 2$. Thank Martin –  zacarias Apr 24 '13 at 23:28

For odd $n$ and $A\in V_1$, we have $$\det(A)=\det(-A^T)=(-1)^n\det(A^T)=-\det(A),$$ so $V_2\subseteq V_1$ and the intersection is just $V_1$.
In the following we assume that $n$ is even.
The variety $V_1$ is cut out by $X_{ij}+X_{ji}$ and has coordinate ring $$\mathbb K[X_{ij}]/(X_{ij}+X_{ji}\mid i,j\in[n])\cong\mathbb K[X_{ij}\mid i>j].$$ Hence, for $r:=(n^2-n)$, $V_1\cong\mathbb A^r_{\mathbb K}$ is an affine subspace of $\mathbb A^{n^2}_{\mathbb K}$. The restriction map sends the determinant polynomial to the square of the pfaffian polynomial $\mathrm{pf}$. There's Exercise 13 in Chapter 2.9 of Matrices: Theory and Applications by Denis Serre, which says that the Pfaffian is irreducible. I am pretty sure it will work a little something like this proof that the determinant is irreducible. Hence, $V_1\cap V_2=Z(\mathrm{pf}^2)\subseteq \mathbb A^r_{\mathbb K}$ is an irreducible hypersurface in $\mathbb A^r_{\mathbb K}$, therefore irreducible in $\mathbb A^{n^2}_{\mathbb K}$ as well. The dimension is $r-1$ as Asal Beag Dubh already pointed out.
Nice answer. One should add the hypothesis $n$ is even. –  user18119 Apr 26 '13 at 0:13