Does the choice of the algebraically closed field of characteristic $p$ have influence on the module category? Let $G$ be a finite group and $p$ be a prime number dividing $|G|$.
Let $k$ be the algebraic closure of $\mathbb{F}_p$.
Let $K$ be another algebraically closed field of characteristic $p$ which is not isomorphic to $k$.
Is there a difference between the module categories $kG$-mod and $KG$-mod ?
 A: The question is not completely clear about what is meant by a difference.  Let me give one interpretation. Every representation of a finite group $G$ over an algebraically closed field is defined over the algebraic closure of $\mathbb Q$ in that field, and so in a sense there is no difference between working over different algebraically closed fields of characteristic $0$.  Of course, the module categories are not equivalent for the reasons in the comments but the basic structure doesn't change.
If $k$ is an algebraically closed field whose characteristic divides the order of $G$, then it's true that every simple and projective indecomposable $kG$-module is defined over the algebraic closure of the prime field, but this seems not to be the case for arbitrary representations (and hence for arbitrary indecomposable representations).  This seems to have been proved as part of the theory of essential dimension, of which I know little about, and so there is a non-zero probability that I have misunderstood this.  But from what I have gathered, Proposition 14.1 of A NUMERICAL INVARIANT
FOR LINEAR REPRESENTATIONS
OF FINITE GROUPS by NIKITA A. KARPENKO AND ZINOVY REICHSTEIN implies that if $G$ is a finite group containing $\mathbb Z/p\mathbb Z\times \mathbb Z/p\mathbb Z$ and if $k$ is a field of characteristic $p$, then for any $n\geq 0$, there is an extension field $K/k$ of transcendence degree at least $n$ such that there is a $KG$-module which is not defined over any intermediate subfield $k\subseteq L\subseteq K$ with $L$ having transcendence degree less than $n$ over $k$.  In particular, there are indecomposable modular representations of $G$ not defined over the algebraic closure of the prime field of $G$ if I understood this correctly.
