Irreducibility of the linear equations variety $Ay=c$ Let $k$ be an algebraically closed field, $A\in k^{n^2}, y \in k^n$ and the variety $V\subset k^{n^2 + n}$ generated by equations $Ay=c$ with polynomials in $k[a_{11}, \ldots, a_{nn}, y_1, \ldots, y_n]$ and $c\in k^n$ a fixed vector. Prove that $V$ is irreducible.
My approach until now has been to consider the subset $V_1 \subset V$ where $|A| \neq 0$ which gives the rational parametrization $y = A^{-1}c$, but I'm stuck on proving that its closure $\overline{V_1}=V$. Certainly, $V_1 \neq V$ as the complement $Ay=c$ with $|A| = 0$ is not null in general.
Is there a way to prove that $\overline{V_1}=V$? Also, what other approach would lead to the irreducibility of $V$?
 A: This is true if $ c \neq 0 $ and it follows by the Jacobian criterion: I'll use variables $ x_{ij} $ for the matrix entries and then the equations are $ \sum_{j = 1}^n x_{ij} y_j = c_i $ for each $ i = 1, 2, \ldots , n $. The Jacobian matrix is in the form $ [B_1 | B_2 | \cdots | B_n | x_{ij} ] $ where $ B_k $ is the matrix which has zero entries everywhere except the $ k $-th row, which is $ (y_1, \cdots , y_n) $. Let $ A = (a_{ij}) $ and $ v = (v_k) $ be a (closed) point on this affine scheme $
V $. Since $ c \neq 0 $, there is an index $ l $ such that $ v_l \neq 0 $ so the Jacobian has full rank. So the local rings at closed points of $ V $ are regular, hence $ V $ is regular. A regular local ring is a domain, so $ V $ has all local rings integral domains. But $ V $ has an open subset $ D(Det(x_{ij})) $ mapping isomorphically to its image in $ \mathbb{A}^{n^2} $, which is irreducible, so $ V $ is irreducible as well.
This fails for $ c = 0 $ as for $ n=1$, the single equation is $ xy = 0 $, not irreducible. I don't know what happens if $ c=0 $ and $ n > 1 $.
