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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 ?

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By $Z({\rm whatever})$, you mean the center? – Gerry Myerson Mar 5 '12 at 6:15
@GerryMyerson yes,$Z(G)$={$g\in G$| ga=ag for any $a\in G$} – Babak Miraftab Mar 5 '12 at 8:58
Just a thought (might be wrong): $Z(R) = Z_R(R) = Z_R(D\otimes D') = Z_D(D)\otimes Z_{D'}(D') = F \otimes_F Z_{D'}(D') = Z_{D'}(D')$. Some equalities are isomophisms... – Thomas Apr 22 '12 at 1:21
up vote 1 down vote accepted

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.

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