Why is the kernel of this strange polynomial homomorphism what it is? I've been trying to delve a little further into linear algebra, but I'm not following something I think is supposed to be obvious. 
Suppose $M_{m,n}(\mathbb{C})$ is the set of rectangular $m\times n$ matrices over $\mathbb{C}$, and let $S$ be the set of rank $1$ matrices. Furthermore, let $K\subset\mathbb{C}[S_{11},\dots,S_{mn}]$ be the ideal associated to $S$. Then the homomorphism $\mathbb{C}[S_{11},\dots,S_{mn}]\to\mathbb{C}[X_1,\dots,X_m,Y_1,\dots,Y_n]$ such that $S_{ij}\mapsto X_iY_j$ has kernel $K$.
I don't follow the last claim. I'm used to the associated ideal of $S$ to be the polynomials in $\mathbb{C}[S_{11},\dots,S_{mn}]$ to be the ideal of polynomials which vanish on all points of $S$ for an algebraic set of zeroes, but that doesn't quite make sense with a set of rank $1$ matrices. Would someone be nice enough to explain why the kernel above is what it is? Thank you.
 A: I think part of the problem stems from your notations, which I  will take the liberty to change.   
Recall that a matrix $M=(a_{ij})\in M_{l,n}(\mathbb  C)$  corresponds to a linear map $\mu:\mathbb C^n\to \mathbb  C^l$ and that $\mu$ will be of rank $\leq 1$ iff  $M=a\cdot b^T$ for some column vectors $a\in \mathbb  C^l$, $b\in \mathbb C^n $.
Hence the map $f:\mathbb  C^l\times \mathbb C^n \to M_{l,n}(\mathbb  C): (a,b)\mapsto a\cdot b^T=(a_ib_j)$ has as image exactly the set $S\subset M_{l,n}(\mathbb  C)$ of matrices of rank $\leq 1$.
We may see $f$ as the morphism of affine spaces  corresponding to the $\mathbb C$-algebra morphism of polynomial rings $\phi: \mathbb C[z_{ij}] \to \mathbb C[x_i;y_j]\ :z_{ij} \mapsto x_i y_j$.      
The claim you want is then that the ideal of polynomials in  $I(S)\subset \mathbb C[z_{ij}]$ consisting of the polynomials vanishing on $S=\operatorname{Im}(f)$ is $\ker(\phi)$, in conformity with the basic correspondence in algebraic geometry between morphism of rings and associated morphisms of varieties. 
