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Let $R$ be a non-commutative ring. Show that if $R$ is simple and has 1, then $Z(R) = \{a \in R | ra = ar$ for all $r \in R \}$ is a field.

I think what I need to do is to show that $Z(R)$ is simple as well. Assume $I \neq \{0\}$ is an ideal of $Z(R)$. Then, I need to show that $1 \in I$ such that $I = Z(R)$. However, I do not know how to proceed.

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Hints :

Consider $a\neq 0\in Z(R)$ then Consider the set $Ra$ which can be shown to be an ideal of $R$. Since $R$ has unity so $a\in Ra$ Thus $R$ being simple we get $Ra=R$ .

So we get $b\in R$ such that $ba=1$ Again considering $aR$ which can be shown to be an ideal again.Thus $aR=R$

Thus $\exists c\in R$ such that $ac=1$

Then $b=b.1=b.(ac)=(b.a).c=c$ thus for each non zero $a\in Z(R)$ we get $b\in R$ such that $ab=ba=1$

NOTE:use the fact $a\in Z(R)$ to show that $Ra$ is an ideal of $R$ (particularly right ideal)

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  • $\begingroup$ why is Ra and Z(R) an ideal of R. This is a noncomutative ring $\endgroup$ – user10024395 Apr 20 '15 at 3:08
  • $\begingroup$ I have given enough hints now you should complete@user2675516 $\endgroup$ – Learnmore Apr 20 '15 at 3:42

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