Center of the algebraic group G_{\mathbb{R}} for a centerless G This must be an easy question but I don't have a good argument for it and have not found a counterexample: Let $G$ be a connected semisimple algebraic group over $\mathbb{Q}$ such that the center of $G$, $Z(G)$ is trivial, i.e. $Z(G)=\{1\}$ is it true that the center of $G_{\mathbb{R}}$, denoted $Z(G_{\mathbb{R}})$ is also trivial?
 A: The short answer is that the center Z(G) of a semisimple algebraic group is a well-defined (finite) algebraic subgroup which commutes with extension of the base field, so if it's trivial over $\mathbb{Q}$, then its trivial over every field. But presumably, that is not what you meant to ask. Perhaps you mean: does $Z(\mathbb{Q})$ trivial imply $Z(\mathbb{R})$ trivial? The answer is no. You have a finite group $Z(\bar{\mathbb{Q}})$ with an action of the absolute Galois group of $\mathbb{Q}$ and you are asking: if only 1 is fixed by the full Galois group is only 1 fixed by complex conjugation? Obviously, not necessarily. It's easy to makes lots of different Galois modules as centers of semisimple groups over $\mathbb{Q}$. Or perhaps you mean: if $Z(G)$ is trivial (as an algebraic subgroup), is the center of $G(\mathbb{R})$ trivial? The answer is yes, because $G(\mathbb{R})$ is dense in $G$ for the Zariski topology. 
A: Since the question and its answer have both gone through a number of edits, and are accompanied by numerous comments, it may be helpful (or not) to sort the question out further.  
1) The interface between semisimple Lie groups (real or complex) and semisimple algebraic groups over $\mathbb{C}$ is somewhat tricky and has been approached historically from different directions.  (Borel for instance got heavily involved in correlating these theories.)   There is no single reference which takes all viewpoints into account, but it may be helpful to look at Chapter 5 of Onishchik-Vinberg Lie Groups and Algebraic Groups.    See for example Section 1 and the first few problems there, with hints and cross-references given later.
How you think about the question here is partly determined by your upbringing.   
2) It's worth noting that "defined over $\mathbb{Q}$" in this context is not really needed, since a connected semisimple algebraic group in characteristic 0 always arises (by Chevalley's work) from a $\mathbb{Z}$-form and is automatically defined over $\mathbb{Q}$.   Aside from this, the essential problem is to relate the complex group $G$ to its subgroup of real points.   Here is where the interaction with Lie groups comes in, adding a layer of complication in terminology and notation.
3) To amplify Emerton's comment in matrix terms, think of each center as the kernel of an adjoint representation.  Say $G$ is a complex algebraic matrix group, so its Lie algebra (in the sense of either algebraic groups or Lie groups) contains  the Lie algebra of the Lie group $G(\mathbb{R})$.  Under the adjoint (conjugation) action any real matrix in the center of $G(\mathbb{R})$ then centralizes the complex Lie algebra of $G$ and thus lies in the (trivial) center of $G$.  
4) This kind of relationship between real and complex groups goes back to E. Cartan and Chevalley, but the interaction with algebraic group language owes most to Borel.    In any case, the theory involved was in place over half a century ago.   It doesn't require the explicit classificaiton of semsiimple groups or Lie algebras, nor does it require any scheme language.   (In the characteristic 0 Borel-Chevalley theory, it's usually safe to identify an algebraic group with its points over $\mathbb{C}$.)    
