Centre of algebra under field extension

Question

Suppose $F$ is a field extension of a field $k$ and $R$ is an algebra over $k$. Is it true that $Z(R \otimes_kF)=Z(R)\otimes_k F$, where $Z(R)$ is the centre of $R$?

Answer 1

This is correct.

Let $ a = \sum r_i \otimes x_i $ be an element in the center, $ r_i \in R $ and $ x_i \in F $. We may assume that the $ x_i $ are linearly independent over $ k $ and hence, are a part of a $ k $-basis of $ F $. Then the elements $ 1 \otimes x_i $ are part of a basis of $ R \otimes F $ as a free $ R $-module. We have $ a(r \otimes 1) = (r \otimes 1)a $ for all $ r \in R $. This gives $$ \sum (rr_i-r_ir) \otimes x_i = 0 = \sum (rr_i-r_ir)(1 \otimes x_i) $$ where in the last sum, we are viewing $ R \otimes F $ as a free $ R $-module. This means $ rr_i=r_ir $ for all $ i $ and $ r \in R $ hence $ r_i \in Z(R) $ giving the conclusion.