let $D$ and D' be $F$-algebras,where as $F=Z(D)$ is field,and R:=D\otimes D' and D':=1\otimes D' and $D:=D\otimes 1$
show that: Z(R)=Z(D')
I can show that Z_R(D)=D' but I can't show that. I have not any idea how to deal with it. any suggestions ?
let $D$ and D' be $F$-algebras,where as $F=Z(D)$ is field,and R:=D\otimes D' and D':=1\otimes D' and $D:=D\otimes 1$
show that: Z(R)=Z(D')
I can show that Z_R(D)=D' but I can't show that. I have not any idea how to deal with it. any suggestions ?
Assume that x=\sum_{i=1}^nd_i\otimes d_i' is in the center of D\otimes D', and assume that this presentation of $x$ as a sum of elementary tensors has a minimal number $n$ of terms. It follows that the elements d_i', i=1,2,\ldots,n, are linearly independent over $F$, for otherwise we could reduce the number of terms in the obvious way. Let $r\in D$ be arbitrary. Using the fact that the sum of the subspaces D\otimes d_i', i=1,2,\ldots,n is direct and comparing the products $(r\otimes 1)x$ and $x(r\otimes 1)$ shows that all the elements $d_i$ must commute with $r$. Therefore $d_i\in Z(D)=F$ for all $i=1,2,\ldots,n$. Therefore $n=1$. The rest is easy.