How do basic results in representation theory change going to positive characteristic fields and/or not algebraically closed? I'm working on a project linking graph theory, model theory and representation theory, and am interested in how some results change if we work over fields of positive characteristic or fields that are not algebraically closed.
Let us assume that the characteristic of the field does not divide the order of the group (so by Maschke's Thm the group algebra is still semi-simple).
Consider the following results:
-The sum of squares of dimensions of each irreducible component is the order of the group.
-All irreducible representations of an abelian group are 1-dimensional.
These clearly do not hold over $\mathbb{R}$ - take $\mathbb{Z}_3$ for instance: it has a 1 dimensional representation and a 2 dimensional one in this case. I'm trying to see in the proof of these results where does it use the fact that the underlying field has to be algebraically closed... any ideas?
Thanks!
 A: For the second one it's really easy to see where algebraic closure intervenes : you pick an eigenvalue for one of the elements of the group : it might not exist if the field is not algebraically closed !
For the first one you have to recall where this result on the sums of squares comes from : it comes from the fact that the group algebra $k[G]$ decomposes (as an algebra) as a product of $M_{n_i}(V_i)$ where the $V_i$'s are the irreducible representations, therefore by an analysis of dimensions we get the equality in question.
However if the field isn't algebraically closed, the Artin-Wedderburn theorem only implies that in this decomposition you get matrix algebras over larger division algebras, not over the field itself. 
To be more precise on why this first fact relies on the algebraic closedness of $k$, you need to be more specific about which proof you know in the algebraically closed case. 
Note, however, that things get much wilder if you remove the hypothesis about the characteristic.
