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Let $F$ be an infinite field (that is not necessarily algebraically closed) and consider the algebraic variety $\mathrm{SL}(n,F)=\mathcal{V}(\det-1)$ of $F^{n^2}$, where $$\mathcal{V}(S)=\{\alpha\in F^k\,|\,f(\alpha)=0\ \forall f\in S\}.$$ We say that a Zariski-closed subset $X$ of $F^{n^2}$ is irreducible, i.e., that if there are two non-empty, proper, Zariski-closed subsets $X_1,X_2$ of $F^{n^2}$ such that $X=X_1\cup X_2$, then $X\subseteq X_i$ for some $i=1,2$.

Is there an easy proof that $\mathrm{SL}(n,F)$ is irreducible?

I tried following these notes (see Example 2.25), but I can't quite understand their method. Basically, the proof there says that $\mathrm{SL}(n,F)$ is irreducible if and only if $\det-1$ is an irreducible polynomial in $n^2$ variables.

As mentioned in Theorem 2.27 of the notes, an algebraic variety $X\subseteq F^k$ is irreducible if and only if its ideal $$\mathcal{I}(X)=\{f\in F[x_1,\ldots,x_k]\,|\,f(\alpha)=0\ \forall \alpha\in X\}$$ is prime, so what I really need is a proof that the ideal of $\mathcal{V}(\det-1)$ is prime. Hence, what I'm really asking is this:

Why does irreducibility of $\det-1$ imply that the ideal $\mathcal{I}(\mathcal{V}(\det-1))$ is prime?

EDIT: This also answers the question: Over $\mathbb{R}$, if $Z(p') \subset Z(p)$ when does $p' \vert p$?

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    $\begingroup$ In a unique factorization domain, a principal ideal generated by an irreducible element is prime. The ideal $\mathcal{I}(\mathcal{V}(\det-1))$ is the radical of the ideal $(\det-1)$, which is $(\det-1)$ itself whenever $(\det-1)$ is prime. $\endgroup$ – Servaes Apr 21 '14 at 16:26
  • $\begingroup$ But aren't we only allowed to say that $\mathcal{I}(\mathcal{V}(\det-1))$ is the radical of $(\det-1)$ if we're working within an algebraically closed field? $\endgroup$ – Bryder Apr 21 '14 at 16:30
  • $\begingroup$ @Servaes: I need the result for $F=\mathbb{R}$ at the very least, so Hilbert's Nullstellensatz doesn't apply (as far as I can see). $\endgroup$ – Bryder Apr 21 '14 at 16:37
  • $\begingroup$ If your field isn't algebraically closed, then you should be careful about what you mean by $\mathcal{I}(\mathcal{V}(\det-1))$. What are your definitions of $\mathcal{V}(I)$ and $\mathcal{I}(V)$ for ideals $I$ and algebraic sets $V$? $\endgroup$ – Servaes Apr 21 '14 at 16:40
  • $\begingroup$ Ah, I see. Sorry about that! I'll edit my original question. $\endgroup$ – Bryder Apr 21 '14 at 16:43
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You are right that some special properties of SL$_n$ are needed.

First some counterexamples in the general case.

  1. If $F=\mathbb Q$. Consider $x^3+y^3-1\in \mathbb Q[x,y]$. It is an irreducible polynomial, but its zero set is finite and reducible.

  2. Over $\mathbb R$, consider $x^2(x-1)^2+y^2$. It is irreducible, but its zero set is reducible.

In the first case, the field is too small, and in the second case the variety is singular at its real points.

Now in your case, (you are supposed to know that) $f:=\det -1$ is irreducible (over $\mathbb C$), and it defines a smooth variety (it is a Lie group). I claim that under these assumption, the ideal of $V(f)$ is irreducible. In what follows, by algebraic varieties over $\mathbb R$, I mean schemes of finite type over $\mathbb R$, not only real points.

Lemma. Let $X$ be a smooth geometrically irreducible variety over $\mathbb R$ such that $X(\mathbb R)\ne \emptyset$, then $X(\mathbb R)$ is Zariski dense in $X(\mathbb C)$.

Proof: Let $g$ be a regular function on $X$ vanishing at $X(\mathbb R)$. Take any $x_0\in X(\mathbb R)$. By the implicit function theorem, in a small analytic neighborhood of $x_0$, $X(\mathbb C)$ is isomorphic (as complex analytic manifold) to an open disc $D$ in $\mathbb C^d$ with $d=\dim X$. Moreover, this isomorphism is defined using real coefficients, so it induces an isomorphism on the real points. Therefore, $g$ can be viewed as a holomorphic function on $D$ which vanishes at the real points of $D$. This implies that $g=0$ on $D$. But $D$ is Zariski open in $X(\mathbb C)$, so $g=0$.

Now we can prove the claim. Let $g$ be a real polynomial such that $g(V(f))=0$. Let $X$ be the algebraic variety over $\mathbb R$ defined by $f$. Then $g$ is regular on $X$ and vanishes on the real points of $X$. So $g=0$ on $X$. By the usual Nullstellensatz, this implies that $$g\in (f\mathbb C[x_1, \dots, x_n])\cap \mathbb R[x_1, \dots, x_n]=f\mathbb R[x_1, \dots, x_n].$$

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  • $\begingroup$ Do you have a reference for this? Also: 1) I'm not sure what you mean by $X(\mathbb{R})$ or $X(\mathbb{C})$? 2) What makes the algebraic variety defined by $f$ a smooth geometrically irreducible variety, if the second counterexample isn't? $\endgroup$ – Bryder Apr 22 '14 at 8:05
  • $\begingroup$ When $X$ is defined by a real polynomial $f$, then $X(\mathbb R)$ is the real zeros of $f$ and $X(\mathbb C)$ the complex ones. For your $f$, $X$ is smooth because it is a group variety (this could also be check by Jacobian criterion with the Jacobian matrix). In the second counterexample, the Jacobian matrix is zero at the only real points $(0,0)$ and $(1,0)$. $\endgroup$ – user143488 Apr 22 '14 at 8:08
  • $\begingroup$ I think I understand. Could you perhaps point me to something that contains this proof? It's for a thesis. $\endgroup$ – Bryder Apr 22 '14 at 8:19
  • $\begingroup$ These are somehow basic algebraic geometry results, I don't know whether there is a place where this is written. Maybe you can just quote this answer ? $\endgroup$ – user143488 Apr 22 '14 at 8:26
  • $\begingroup$ I could! Let me just make sure: you prove that $\mathcal{I}(\mathcal{V}(f))$ is the principal ideal of $f$, right? And this entails that $\mathcal{V}(f)$ is irreducible. At least for the case $F=\mathbb{R}$, which is what I need. (Would you mind explaining what regularity and geometric irreducibility means?) $\endgroup$ – Bryder Apr 22 '14 at 8:45

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