# The $2 \times 3$ matrices with rank $\leq 1$ cannot be defined by two polynomial equations

Let $X$ be the space of all ${2 \times 3}$ matrices over $\mathbb{C}$ that have rank at most 1. This is naturally a subspace of $\mathbb{C}^6.$ We can express $X$ using 3 polynomial equations, namely the three $2\times 2$ minors of the matrix. I want to prove that we can't do that with fewer than 3 equations:

There are no two polynomials $f,g$ (in 6 variables) such that their common roots are exactly $X$.

or equivalently :

There are no two polynomials $f,g$ in the variables $x,y,z,w,t,u$ such that $\{ f=g=0 \}=\{xt-wy=xu-zw=yu-zt=0\}$

Since $X$ is 4-dimensional in the space $\mathbb{C}^6$, this shows that the variety $X$ is not a set-theoretic complete intersection.

Surprisingly, I didn't find any examples of a variety which is not a set-theoretic complete intersection, not in SE nor in Shafarevich or Harris.

A starting point for manipulations is to write $f$ and $g$ as linear combinations of the $3$ minors (which generate the ideal of $X$).

• There's not an explicit question in the OP ... are you asking someone to prove the highlighted statements? Commented Jun 13, 2016 at 1:49
• Yes, That's right. Commented Jun 13, 2016 at 2:09
• The standard example of an algebraic set which is not a set-theoretic complete intersection is due to Hartshorne, the union of two 2-planes in $\mathbb{A}^4$ meeting only at the origin. Commented Jun 13, 2016 at 15:17