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Let $n$ be a positive integer and let $k$ be an algebraically closed field. What is the coordinate ring of $GL(n,k)$ (the set of all $n \times n$ matrices with entries in $k$)? Here we identify this set as a subset of $k^{n^{2}}$.

Would it suffice to say that the coordinate ring is the localization of $k[x_{11},x_{12},..,x_{nn}]$ at the determinant function? Is there a way to "simplify" this?

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Yes, that is precisely what the coordinate ring is. I think one can argue that this is the most simple presentation as follows: the dimension of $\textrm{GL}(n, k)$ is $n^2$, but it is not affine space, so it must have at least $n^2 + 1$ generators and at least one relation... – Zhen Lin Jun 19 '12 at 10:35
up vote 6 down vote accepted

Maybe you already know this, but I think the most natural way to simplify it (if you think taking quotient is simpler than taking localization) is to view $GL(n,k)$ as a subvariety in $\mathbb{A}^{n^2+1}$: Now view $det$ as a polynomial $D(t_1,...,t_{n^2})$ of $n^2$ variables, then the coordinate ring is $k[t_1,...,t_{n^2},y]/(yD-1)$.

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Why did you write $yD-1$ rather than $D-y$? – math-visitor Jun 19 '12 at 18:32
$D-y$ is wrong. $y$ should be a multiplicative inverse of the determinant, not equal to it. – Thomas Apr 14 '15 at 13:44

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