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I have a statement that says that if $R$ is a simple ring, then $Z(R)$ must be a field.

Since $R$ is a ring we get most of the field axioms (associativity and commutativity of addition, multiplication$\text{--}$since it's the center$\text{--}$, presence of the identities) immediately without need for explanation.

But I don't see how we can get the invertibility of multiplication.

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Every central element is invertible. Suppose $a \in Z(R)$ is not. Then image of multiplication by $a$ does not contain $1$; but then it's a nontrivial two-sided ideal, because for all $r, r'$ $r'ar = ar'r$ and $arr'$ bot lie in $aR$.

If central element is invertible, then it's inverse also central. Take central $a$ and it's inverse $a^{-1}$. Then $a^{-1}r = a^{-1}(ra)a^{-1} = a^{-1}(ar)a^{-1} = ra^{-1}$.

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  • $\begingroup$ Wait, so if an ideal contains 1 it must be a trivial ideal? $\endgroup$ – Makogan Apr 1 '18 at 20:16
  • $\begingroup$ Of course, because if $1$ is an element of ideal, then for any $r \in R$ $r \cdot 1 = r$ is also in that ideal. $\endgroup$ – xsnl Apr 1 '18 at 20:17
  • $\begingroup$ Because ideals must be closed under product? $\endgroup$ – Makogan Apr 1 '18 at 20:18
  • $\begingroup$ Yes. en.wikipedia.org/wiki/Ideal_(ring_theory) $\endgroup$ – trollkotze Apr 2 '18 at 3:54
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Let $z\ne 0$ be an element of $Z(R)$. We know such a $z$ exists because $1\in Z(R)$.

Consider the ideal $I$ generated by $z$. Since $z\ne 0$ and $R$ is simple $I = R$. Therefore, there exists $s\in R$ such that $zs = 1$. Since $z\in Z(R)$ $zs = sz$ so $s=z^{-1}$.

It remains to show that $z^{-1}\in Z(R)$.

For any $r\in R$, $z^{-1}r = z^{-1}r(zz^{-1})=(z^{-1}z)rz^{-1} = rz^{-1}$.

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