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In the context of linar algebraic groups, I read in my notes from the lecture that's already some while ago that $GL_n(k)$ is an algebraic variety because $GL_n=D(\det)$, $ \det \in k [ (X_{ij})_{i,j} ]$. Now, $k$ is an algebraically closed field, $\det$ is the determinant and $k [ (X_{ij})_{i,j} ]$ are the polynomials in unkwnowns $X_{ij}$. But I cannot find what $D$ meant, maybe it's also a typo or uncommon notation -.-

How can I interpret $GL_n (k)$ as a variety? (Or what does this $D$ stand for?)

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up vote 8 down vote accepted

I think $D$ becomes a more common notation when working with schemes, but here $D(f)$ for $f \in k[X_1, \ldots, X_n]$ should mean $\{a \in k^n : f(a) \neq 0\}$. So it's the open set where $f$ does not vanish. The important thing is that $D(f)$ is isomorphic to the algebraic set defined by $fX_{n + 1} - 1$ in $k^{n + 1}$ via the map $(a_1, \ldots, a_n) \mapsto (a_1, \ldots, a_n, 1/f(a_1, \ldots, a_n))$.

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So, with the transformation $f\mapsto fX_{n+1}-1$ in one dimension more I have $D(f) = V(fX_{n+1}-1)$ where $V(P_i) = \{ x | P_i(x) =0 \}$ (the set where f vanishes), and one can "in a certain sense" identify the sets where f does not vanish with this choice of vanishing sets in one dimension more. And I suppose your $A$ is a ring (maybe additonally with 1 and so on), so I can use this for $A=k$ – Suedklee Jul 12 '12 at 14:43
@user11163 Oh, sorry, $\mathbf A^n$ is just common notation for $k^n$. There's some fuss about how this means that we aren't giving the origin any special privilege. I'll remove it. I also made a badd typo, so definitely check the answer again. – Dylan Moreland Jul 12 '12 at 14:52

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