Polynomial Ring of Linear Algebraic Group

Question

During lectures, we defined the Linear Algebraic group as the algebraic set

$ GL(V):=k^{n^2}-V(Det) $

Where $V(Det)$ are the matrices with $0$ determinant. Then we proceed by identifying the polynomial ring of $GL(V)$ as

$ k[GL(V)]=k[T_{1,1},...,T_{n,n},1/(det(T_{i,j}))] $

What is this identification? is $T_{1,1}$ just taking the element in position 1,1 of the matrix? Why is it written in this way? I am ok that $k[V]\cong k[X_1,...,X_n]$ if $k$ is of dimension $n$ but I can't connect it with $k[GL(V)]$.

Thanks for your help!

Answer 1

$k^{n^2}$ is identified with the vector space generated by the entries of the matrices. $T_{ij}$ is the indeterminate representing the entry at position $i,j$ and this corresponds to the $i,j$th coordinate in $k^{n^2}$ . All polynomials on the general linear group can have roots except for multiples of the determinant, so the determinant is invertible.

Answer 2

$k[\mathbb{A}^{n^2}] = k[T_{1,1},\dotsc,T_{n,n}]$ should be known. If you identify $\mathbb{A}^{n^2}$ with $\mathbb{A}^{n \times n}$ (matrices), then in fact $T_{ij}$ is the regular function which evaluates the matrices at the entry $(i,j)$.

In general, if $f$ is a regular function on an affine variety $X$, then $k[X \setminus V(f)] = k[X][1/f]$. This is a basic fact which can be found in every text on algebraic geometry. Now apply this to $X=\mathbb{A}^{n \times n}$ and $f = \mathrm{det}$.

In order to illustrate this, here is an example: For $n=2$ we have $k[\mathrm{GL}_2] = k[T_{11},T_{12},T_{21},T_{22},1/(T_{11} T_{22} - T_{12} T_{21})]$. For example, $(T_{11}+T_{22})/(T_{11} T_{22} - T_{12} T_{21})$ is contained in this ring. As a regular function on $\mathrm{GL}_2$ it maps an invertible matrix to its trace divided by its determinant.