Question about Segre embedding proof in Karen's Smith book.

Question

The main problems that I am having with the proof is linear algebra. First, why is the Segre mapping contained in the set defined by 2x2 minors of the matrix $(z_{ij})$ why does it follow from the fact that 2x2 sub- determinants vanish ? In the converse why is it the case that when all 2x2 subdeterminants vanish, then rank of the matrix at most 1 ? Finally, why (m + 1)(n + 1) of rank k factors according to what the book mentioned ?

Answer 1

1. To answer your first question, assume for contradiction that some $2\times2$ minor does not vanish. After reordering the rows and columns of your matrix (which does not change its rank), you may assume that it is the top left $2\times 2$ submatrix. Now note that the second column can not be a scalar multiple of the first, because otherwise the top left $2\times 2$ block would have the form $$ \begin{pmatrix} a & \lambda a \\ b & \lambda b \end{pmatrix} $$ for some $\lambda\in \Bbbk$; and the determinant of this matrix vanishes.

   Remark. Of course, this works more generally. If you have a non-vanishing $k\times k$ minor, then you can conclude that there are at least $k$ linearly independent rows, which means that your matrix has rank at least $k$.

2. To answer your second question, assume that all $2\times 2$ minors of the matrix $z=(z_{ij})$ vanish. We may assume that the matrix is nonzero, otherwise we are done (it has rank less than one). Without loss of generality, assume that the entry $z_{11}$ is nonzero. Now, for any $i$ and $j$, you have that the matrix $$ \begin{pmatrix} z_{11} & z_{1j} \\ z_{i1} & z_{ij} \end{pmatrix} $$ has vanishing determinant, so the second column must be a multiple of the first, i.e. there is some $\lambda\in\Bbbk$ with $z_{1j}=\lambda z_{11}$ and $z_{ij}=\lambda z_{i1}$. Since this holds for all $i$, it follows that the $j$-th column of $z$ is a scalar multiple of the first column. Since that holds for all $j$, all columns are scalar multiples of the first column. Hence, the matrix has rank one.
3. To see that any rank $k$ matrix factors like this, recall that the rank of a linear map is the dimension of its image. Hence, your $(m+1)\times(n+1)$ matrix describes a linear map $f:\Bbbk^{n+1}\to\Bbbk^{m+1}$ such that $V:=f(\Bbbk^{n+1})\cong\Bbbk^k$. In other words, the map factors as $\Bbbk^{n+1}\to V\to\Bbbk^{m+1}$, where the first map is the co-restriction of $f$ and the second is just the inclusion. Choosing appropriate bases, you may describe this as $\Bbbk^{n+1}\to\Bbbk^k\to\Bbbk^{m+1}$, i.e. the first map is given by a $k\times(n+1)$ matrix and the second one by an $(m+1)\times k$ matrix.