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The following is a related article:
1. Simple ring and field

The definition for a simple ring:

A nonzero ring $R$ is a simple ring if the only two-sided ideals of $R$ are $R$ itself and zero.

So $\mathbb{Z}$ is not a simple ring since $2\mathbb Z$ is also an ideal.

The definition for center of a ring:

The center of $R$ is the subset $C(R) = \{x\in R \mid xr = rx , \forall r\in R\}$.

Question:
By the title, we can obtain a fact that the center of a simple ring is a ring? How to understand this? Moreover, is the center of a simple ring a simple ring itself?

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sleeve chen
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    The center of a ring $R$ equals $R$ itself if and only if $R$ is commutative. As a consequence, every noncommutative simple ring has a smaller center. – Crostul Dec 23 '16 at 08:59

1 Answers1

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the center of a simple ring is a ring

The center of any ring is a subring.

is the center of a simple ring a simple ring itself?

The center is always a commutative ring, so the center of a noncommutative ring is always strictly smaller than the ring. There are numerous examples of simple rings that are not commutative, namely square matrix rings over fields. In that case the center is the set of constant diagonal matrices.

So it is not the simple ring itself but it is itself a simple ring. For commutative rings, the simple rings are exactly the fields.

(For the title question about the center of a simple ring being a field.) It's a simple computation to show that if $x$ is in the center and $x$ is invertible, then its inverse is in the center too. By taking a nonzero element $x$ of the center of the simple ring $R$, we see that $xR$ is a nonzero ideal and hence is $R$. But this means $x$ is invertible, and its inverse is in the center. All nonzero elements of the center are then invertible, and the center is a field.

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rschwieb
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