About the center of an algebra over different fields

Question

Is it true that the center of an algebra is bigger when the ground field has positive characteristic than the center of the same algebra over a field of characteristic zero? And why? Can someone explain it to me? (In both cases YES or NO)

Thanks!

Answer 1

An algebra $A$ over a field $F$ of positive characteristic cannot be an algebra over a field $K$ of characteristic $0$.

Being an algebra over $F$ means that there exists an injective ring homomorphism $F\to Z(A)$ (the center of $A$). In particular, if $p>0$ is the characteristic of $F$, then the $p$-element field $\mathbb{F}_p$ is a subring of $A$.

An algebra over a field $K$ with characteristic $0$ cannot have a subring with $p$ elements, because the minimal subring of $A$ is $\mathbb{Z}$.

A ring $A$ can be an algebra over different fields; for instance $F(x)$ (the ring of rational functions over the field $F$) is both an algebra over itself and an algebra over $F$. The center of $A$ doesn't depend on the field.

Answer 2

In my opinion, your question is not well posed.

First: What do you mean by "same" algebra? Ok, you could say that you are considering a quotient of a free algebra in a set of generators $X$ and ideal (of relations) $I$, i.e. $A=F\langle X \rangle / I$, but then you have to tell me how to carry them over to a field of positive characteristic. One way would be to consider the subring $R$ of $F$ generated by all coefficients of generators of $I$ over $\mathbb{Z}$ (assuming $char(F)=0$). So you might consider the ideal $I'$ in $R\langle X\rangle$, generated by the generators of $I$ and $A'=R\langle X\rangle/\langle I'\rangle$. Then cut down to characteristic $p$ by looking at $A''=A'/pA'$ and tensor it with your favourite field $K$ of characteristic $p$, i.e. $A'''=K\otimes_{\mathbb{Z}_p} A''$. Is this what you want?

Second: Comparing the cardinality is not a good choice. Simply consider the $2\times 2$-matrix ring $A=M_2(\mathbb{Q})$ over the rationals. Then $|Z(A)|=|\mathbb{Q}|=\aleph_0$ is countably infinite. However, if you look at $A'=M_2(K)$ for field of characteristic $p$ and $|K|=\aleph_1$, then certainly $\aleph_1=|Z(A')|>|Z(A)|=\aleph_0.$ Maybe you want to look at the dimension of the center over its base field?