Given $A^6_{C}$ with coordinates $(u,v,w,x,y,z)$. Let $D ⊆ A^6$ be the Zariski closed subset given by the vanishing of the 2 × 2 minors of the matrix \begin{bmatrix} u & v & w \\ x & y & z \\ \end{bmatrix} (so $D = V (uy − vx, uz − wx, vz − wy)$). How can it be shown that D is irreducible? Any advice?
How can it be shown that D (Zariski closed subset) is irreducible?
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One can show that if an $m\times n$ matrix $(a_{ij}) \in M_{m\times n}(F)$ satisfies $$a_{ij} \cdot a_{kl} = a_{il} \cdot a_{kj}$$ for all $1\le i,k \le m$, $1\le j,l \le n$ there exist $b_1, \ldots, b_m$, $c_1, \ldots c_n$ in $F$ such that $$a_{ij} = b_i \cdot c_j$$
(equivalently: if a linear map $T\colon F^n \to F_m$ is of rank at most $1$, there exists $U\colon F^n \to F$, $V\colon F\to F^m$ such that $T = V\circ U$). Note that the statement can be proved by hand, works for unique factorization domains too.
So now we see that there exists a (bilinear, so polynomial) map from $F^m\times F^n$ with image the matrices of rank $\le 1$. Since $F^m\times F^n$ is irreducible, so is the image.
Note: the generalization to matrices of rank $\le k$ is immediate. The set will be the image of $M_{m\times k}\times M_{k\times n}$ under composition.
It's a standard result from linear algebra that your set $D$ is the set of all rank zero or one matrices. Multiplying on the left by elements of $\operatorname{GL}_2(\mathbb C)$, and on the right by elements of $\operatorname{GL}_3(\mathbb C)$, every matrix in $D$ is equivalent to either
$$z_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ or $$z_2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$
The set $D' = D - \{z_1\}$ of rank one matrices is open and dense in $D$. It's not difficult to check that a subspace is irreducible if and only if its closure is. Thus it suffices to show that $D'$ is irreducible. This is because $$D' = \{ gz_2h : g \in \operatorname{GL}_2(\mathbb C), h \in \operatorname{GL}_3(\mathbb C)\} \tag{1}$$
Can you argue that $D'$ is irreducible from its description (1)?