Let $M$ be the projective space of nonzero $m\times n$ matrices up to scalars (in $\mathbb{K}$). In Joe Harris' Algebraic Geometry: A first course, in order to find the dimension of $M_{k}=\{A\in M:\mathrm{rk}(A)\leq k\}$, the next incidence correspondence is considered: $$ \Sigma=\{(A,\Lambda)\in M_{k}\times G(m-k,n):\Lambda\subseteq\ker A\}, $$ where $G(m-k,n)$ is the set of $m-k$-planes in the vectorial space $\mathbb{K}^{n}$. I do not get to see why this is a projective variety. I have tried to find a regular map $$ \pi: M_{k}\times G(m-k,n)\rightarrow X, $$ where $X$ is a projective variety, such that $\pi^{-1}(X)=\Sigma$. Since a regular map is continuous in the Zariski topology, $\Sigma$ would be a projective variety. The problem is that I have not find a map like that.
Another option would be to find equations for $\Sigma$. I think that the first coordinates of a point $p\in\Sigma$ must satisfy the equations of $M_{k}$, but I get lost when I try to determine equations for the rest of the coordinates.
Recall that a subset $F$ of a topological space $X$ is closed if and only if there exists an open cover $\{ X_{i} \}$ of $X$ such that $F \cap X_{i}$ is closed in $X_{i}$.
Cover the Grassmannian variety $G(m-k, n)$ with open sets $U_{i_{1}, \ldots, i_{m-k}}$ consisting of $(m-k)$-dimensional vector subspaces of $\mathbb{K}^{n}$, which we represent by a matrix $\Lambda_{(m-k) \times n}$, with non-zero minor $\Lambda_{i_{1}, \ldots, i_{m-n}}$ formed by the columns $i_{1}, \ldots, i_{m-k}$ of $\Lambda$.
Assume $(i_{1}, \ldots, i_{m-k}) = (1, \ldots, m-k)$ (for the general case, apply a permutation of coordinates). Multiplying $\Lambda$ by the inverse of the minor $\Lambda_{1, \ldots, m-k}$, we can assume that $\Lambda_{1, \ldots, m-k} = Id_{m-k}$, the identity. Note that each entry $b_{i,j}$, $1 \leq i \leq m-k$, $m-k+1 \leq j \leq n$, is a Plücker coordinate, because it is (up to sign) the determinant of a minor of $\Lambda$, namely the determinat of the identity $\Lambda_{1, \ldots, m-k}$ with the column $i$ replaced by the column $j$.
Now, write $A = (a_{1}, \ldots, a_{m})^{t}$, with each $a_{i}=(a_{ij})_{j} \in \mathbb{K}^{n}$. We have $\Lambda \subset \ker(A)$ (still considering $\Lambda$ in the above form) if and only if $A \cdot a_{k} = 0$, for every $k = 1, \ldots, m$. This gives us the equations that define the quasi-projective variety $\Sigma \cap U_{i}$. This proves that $\Sigma$ is a projective variety, by the fact above recalled.
To get the equations in the whole product $M_{k} \times G(m-k, n)$, homogenize the equations defining $\Sigma \cap U_{i}$. You will have lots of them. Most of time, you do not need to know them, the equations defining $\Sigma \cap U_{i}$ are sufficient.