Let $k$ be an infinite field (possibly algebraically closed).
I'm trying to prove that the set of $n\times n$ matrices with entries on $k$ and determinant $1$, $\operatorname{SL}_n(k)$, is an affine variety (by affine variety I will refer to an irreducible algebraic set of $\mathbb{A}_k^{n^2}$). I've been instructed to prove it using these facts:
- Morphisms between quasi-affine varieties are continuous (which respect to the Zariski topologies).
- Continuity preserves irreducibility (that is, a continuous function between irreducible topological spaces has irreducible image).
- Open sets of an irreducible topological space are also irreducible.
I've been given also the hint to look for a morphism between $\operatorname{GL}_n(k)$ (the general linear group of degree $n$ over $k$) and $\operatorname{SL}_n(k)$.
I already know that $\operatorname{GL}_n(k)$ is a quasi-affine variety (i.e., an open set of an affine variety) and that $\operatorname{SL}_n(k)$ is an algebraic set. It remains for me to see that $\operatorname{SL}_n(k)$ is irreducible. By the third previous fact, $\operatorname{GL}_n(k)=\operatorname{M}_n(k)\setminus\det^{-1}(0)$ is an irreducible topological space (for $\operatorname{M}_n(k)\cong\mathbb{A}_k^{n^2}$, the set of $n\times n$ matrices with entries in $k$, is irreducible since $k$ is infinite). Thus, by the second previous fact, to give a morphism $\phi:\operatorname{GL}_n(k)\to\operatorname{GL}_n(k)$ whose image coincides with $\operatorname{SL}_n(k)$ suffices for the proof. However, I can only come up with this definition for the morphism: $\phi(A)=\frac{A^n}{\det A}$ (recall that $\det$ is homogeneous of degree $n$). $\phi$ is indeed a quasi-affine varieties morphism and by construction $\operatorname{Im}\phi\subset\operatorname{SL}_n(k)$. Nevertheless, proving the reverse inclusion would require to prove that the $n$-th root for a matrix of $\operatorname{SL}_n(k)$ always exists. For what I've been looking up on the internet and MSE, this may be non-trivial or even not true in general. From here, I don't see how to continue.
Edit: Thanks to KReiser's comments I've found a proof for the irreducibility of $\operatorname{SL}_n(k)$ which passes through proving that $\det X-1$ is irreducible (where $X=(x_{ij})$ is an $n\times n$ matrix on the indeterminates $x_{ij}$). Problem is this proof is purely algebraic, and I'm still interested on finding a proof which uses the technique stated above: to look for a quasi-affine varieties morphism $\operatorname{GL}_n(k)\to \operatorname{GL}_n(k)$ with image equal to $\operatorname{SL}_n(k)$. Still, I have no clue about how to come up with the morphism.
Any help on the topic will be appreciated.
Best Answer
Kreiser comments give the answer: one could just simply prove that $\det -1$ is irreducible (proof here and here). The other way, following the original path, is to consider the morphism $\operatorname{GL}_n(k)\to \operatorname{GL}_n(k)$ which takes each matrix $A\in \operatorname{GL}_n(k)$ and maps it to a matrix equal to $A$ except for the first row, which is equal to that of $A$ divided by the determinant of $A$. It's straighforward to check that the image of this morphism is indeed $\operatorname{SL}_n(k)$.