Structure of the variety of $n$-tuples of $m \times m$ matrices with zero product Consider the functor sending a commutative ring $R$ to $\{(A_1,\dots,A_n) \in ( M_m(R) )^n | A_1 \dots A_n =0 \}$ which defines a scheme over $\mathbb Z$, let $X$ be its base change to $\mathbb C$.
When $m=1$, it's easy to see $X$ is not irreducible but reduced. In general, is $X$ reduced? Can we describe the irreducible components and smooth locus?
The main question is about the computation of cohomology groups. As $X$ is a cone (so is contractible topologically), we know higher etale cohomology is trivial (by comparison theorem with singular cohomology). 

What is the compactly supported cohomology group $H^i_{et, c}(X,
 \mathbb Z_l)$?

Note $X$ may not be (rationally) smooth, so Poincare duality may fail. The approach through counting over finite fields is also very complicated. 
There is a natural conjugacy action of $GL_m$ on $X$ (and action of $(\mathbb G_m)^n$ by scaling). If one wants to use this to compute the compactly supported cohomology, how to describe the orbits?
Edit: a motivation for such consideration is the nilpotent cone, which is rationally smooth and the number of points over a finite field has a very simple expression.
 A: I won't give a full answer, but a way to attack the problem of computing the dimensions of the cohomology groups with compact support in the special case $n=2$.
Let $X_k = \{ (A,B) \in \mathcal{M}_m(\mathbb{K}), \ A B = 0, \ \textrm{rank}(B) = k\}$. Then, in the case $m=2$, your $X$ is the disjoint union of the $X_k$ for $k \in [0, \ldots, m]$. I will compute the number of points over $\mathbb{F}_{q}$ of $X_k$ by describing $X_k$ as a sort of iterated fibrations.
Let $p : X_k \longrightarrow \mathrm{Gr}(k,\mathbb{F}_{q}^m)$ which sends a pair $(A,B)$ to the image of $B$. The target of $p$ has cardinality $\binom{m}{k}_q$ (where $\binom{m}{k}_q$ denotes the $q$-binomial coefficient).
The fiber of $p$ over $L$ is the variety:
$$ p^{-1}(L) = \{(A,B) \in \mathcal{M}_m(\mathbb{F}_{q})^2, \mathrm{Im}(B) = L, \ L \subset \mathrm{Ker}(A) \}.$$
It easy to see that the fibers over different $L$ are all isomorphic. Furthermore, we have a natural projection
$$q: p^{-1}(L) \longrightarrow \mathrm{Hom}(\mathbb{F}_q^m,L)_{\mathrm{rank} = k}$$ sending a pair $(A,B)$ to $B$.
Let us compute the cardinal of the target of $q$. We have a natural projection:
$$r : \mathrm{Hom}(\mathbb{F}_q^m, L)_{\mathrm{rank} = k} \longrightarrow \mathrm{Gr}(m-k,\mathbb{F}_q^m)$$
which sends an element of $\mathrm{Hom}(\mathbb{F}_q^m, L)_{\mathrm{rank} = k}$ to its kernel. The fiber of $r$ is exactly $\mathrm{GL}_k(\mathbb{F}_q)$. Since the cardinal of $\mathrm{GL}_k(\mathbb{F}_q)$ is $\prod_{j=0}^{k} (q^k-q^j)$, we deduce that:
$$ \mathrm{card}(\mathrm{Hom}(\mathbb{F}_q^m, L)_{\mathrm{rank} = k}) = \binom{m}{k}_q \prod_{j=0}^{k} (q^k-q^j).$$
In order to find the cardinal $p^{-1}(L)$, we then just have to find the cardinal of $q^{-1}(B)$ for some $B \in \mathrm{Hom}(\mathbb{F}_q^m, L)_{\mathrm{rank} = k}$ (all such $q^{-1}(B)$ are isomorphic). But we have:
$$ q^{-1}(B) = \{A \in \mathcal{M}_{m}(\mathbb{F}_q), \ L \subset \mathrm{Ker}(A) \} = \mathrm{Hom}(\mathbb{F}_q^m/L,\mathbb{F}_q^m).$$
We deduce that the cardinal of $q^{-1}(B)$ is $q^{(m-k)m}$. This gives:
$$\mathrm{Card}(p^{-1}(L) = q^{(m-k)k} \binom{m}{k}_q \prod_{j=0}^{k} (q^k-q^j).$$
This implies that:
$$ \mathrm{Card}(X_k) = q^{(m-k)k} \left(\binom{m}{k}_q \right)^2 \prod_{j=0}^{k} (q^k-q^j),$$
and finally:
$$\mathrm{Card}(X) = \sum_{k=0}^{n} q^{(m-k)k} \left(\binom{m}{k}_q \right)^2 \prod_{j=0}^{k} (q^k-q^j).$$
From this formula and the Weil conjectures (and a computer), you can probably find the dimensions of the cohomology groups with compact support of $X$ (this is the case $n=2$).
EDIT : the formula:
$$ Z(X,t) = \dfrac{P_1(X,t) \times \ldots \times P_{2d-1}(X,t)}{P_0(X,t) \times \ldots \times P_{2d}(X,t)}$$
is valid even if $X$ is not projective and singular, provided we interpret $P_i(X,t)$ as the characteristic polynomial of the Froebenius acting on compactly supported étale cohomology (see Theorem 29.8 in 
http://www.jmilne.org/math/CourseNotes/LEC.pdf). 
The computation I did dtermines entirely $Z(X,t)$. Hence, the uniqueness of the factorization of $Z(X,t)$ as a product as above guarantees that the dimensions of the compactly supported cohomology groups of $X$ can be derived from the above formula. Of course this derivation is not easy, but it can certainly done using a computer algebra system. 
EDIT EDIT As Will Sawin points it out, the uniqueness is not necessarily valid in the singular context, so it requires some more work to get the Betti numbers from the Zeta function.
