Simplicity of (complex) orthogonal groups I need a reference for the proof that the complex orthogonal group
$SO_{2n+1}($ℂ$) = \{A\in SL_{2n+1}($ℂ$): A^TA = Id\}$ is simple in a group theoretical sense (if it is true). 
How about the simplicity of $SO_{2n+1}(K)$ in general (i.e. $K$ an arbitrary infinite field)? 
It there any criterion? It seems that if $K$ is "big", then $SO_{2n+1}(K)$ is not simple, e.g. if $K$ has a valuation, then (I think) one can find somehow a proper normal subgroup.
 A: You have to be kind of careful.  For the complex numbers, there is no problem as BCnrd wrote above.  For a general field $K$, you are asking a basic question from the field of geometric algebra (there's an AMS GTM volume by Grove on the topic, and Dieudonne's books on classical groups).  
If you take a nondegenerate quadratic form $q$, then we know:


*

*If $q$ is isotropic over $K$, then $Spin(q)(K)$ is "projectively simple" (i.e., the quotient by its finite center is simple)

*If $q$ is anisotropic, then $Spin(q)(K)$ can be far from simple.  You can construct examples using valuations, as you suggest.


But you asked about $SO(q)(K)$.  Then you use Galois cohomology (where I assume that $K$ has characteristic different from 2 for simplicity):

$1 \to \mu_2(K) \to Spin(q)(K) \to SO(q)(K) \to K^{\times}/K^{\times 2}$

where the last map is the spinor norm.  You should think of the spinor norm as usually having a big image, so $SO(q)(K)$ will be pretty far from simple.

Your specific group
You asked specifically about $SO(q)$ where $q$ is a sum of squares.  Then the spinor norm map has image products of sums of squares (typically not actually squares themselves), so you should expect the image of $Spin(q)(K)$ to be a normal subgroup in $SO(q)(K)$ of large index.  (That is all very imprecise, but a precise answer depends on the arithmetic of $K$ and the dimension of $q$.)  But now you can ask: Is $Spin(q)(K)$ modulo its center a simple group?
Here you have a strong advantage.  If $q$ is isotropic over $K$ (i.e., you can write $0$ as a sum of a small enough number of nonzero squares), then you know from classical results that the answer is "yes".
If $q$ is not isotropic over $K$, then $Spin(q)$ is anisotropic (as an algebraic group) but it is split by the quadratic extension $K(\sqrt{-1})$.  For such groups, under some hypotheses on $K$ (maybe as weak as characteristic zero), you know that the answer is again "yes" by Chernousov.  See Theorem 9.7 on page 514 of the book "Algebra & Number Theory" by Platonov and Rapinchuk.  You will have to inspect the proof (which starts on p.546) to see the precise hypotheses they need on $K$, or you can consult some of the original papers, where the proof is slightly different.
A: The structure of classical groups goes back a long way and has been treated in a number of books, but in varying generality (arbitrary fields, various commutative rings, etc.).   One older source in French is J.A. Dieudonne's concise Springer Ergebnisse volume La geometrie des groupes classiques (1963).   A probably more readable modern textbook with limited aims is Larry Grove's Classical Groups and Geometric Algebra (AMS, 2002), just cited by Skip.   There is also Emil Artin's old book Geometric Algebra and a much larger book by Hahn-O'Meara oriented more to algebraic K-theory.   Anyway your group is simple both as an algebraic and as an abstract group (special orthogonal groups in odd dimension are also adjoint groups).   There's no real need to get into algebraic groups, BN-pairs, or the like, though this is the "correct" general setting as Tits showed.
Actually, simplicity of various classical groups is proved sometimes in graduate algebra textbooks (which I don't have at hand).    It depends how far you want to go.  Over more general fields, especially of characteristic 2, a little more care is needed but these groups are still simple or very close to it even over most finite fields.  
A: A reference with a small amount of patching to do is Bourbakie: Groupes et algèbres de Lie, Chap IV, 2.7, it uses the theory of BN-pairs (which there are called Tits systems). It is shows that the only non-trivial normal subgroup is that consisting of the scalar matrices (which is non-trivial exactly when $n$ is even). As I said above this is probably overkill for this classical situation.
A: You need to be more careful, or else there may be a non-trivial central element. Note that for $n$ even you have $-I$ in the special orthogonal group.
