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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.

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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.

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  • $\begingroup$ Thanks again Georges! So this can really just be reduced to a correspondence theorem in algebraic geometry? Is there a source where I can read of this correspondence? $\endgroup$ Apr 30, 2012 at 6:14
  • $\begingroup$ Dear hmIII, you might have a look at Shafarevich's Basic Algebraic Geometry, vol.1, Chapter I, §2.3, page 29. Another nice reference is Fulton's Algebraic Curves, freely available online. You want Chap.2, especially sections 1 and 2. $\endgroup$ Apr 30, 2012 at 8:06
  • $\begingroup$ Thanks Georges, I'm always glad for your answers. (By the way, just skimming through some of these sources, is it also true that the kernel above is prime? If it's not a trivial deduction, I can make it a new question.) $\endgroup$ Apr 30, 2012 at 21:43
  • $\begingroup$ Yes, the ideal $ker(\phi)$ is prime because it corresponds to $S=Image(f)$, which is irreducible as an image of the irreducible variety $\mathbb C^{l+n}$ $\endgroup$ Apr 30, 2012 at 22:05

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