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TRUE or FALSE: If R is a commutative ring with unity, then the set of units in R forms a subring

As I though if R is a ring, then the set of all units of R is not a subring of R because the zero element is not a unit. How about R is a commutative with unity?

Qiaochu Yuan
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  • You appear to be overlooking a basic fact of logic here. If you show that it is *not true for all rings with identity* then asking the same question about special cases of rings (like commutative rings) is a waste of time. – rschwieb Mar 05 '13 at 18:43
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    If $R$ has a $1$, the units form a group with respect to the multiplication. That's all. And you do not need $R$ to be commutative. – Julien Mar 05 '13 at 18:44
  • Sometimes the set of *non*-units form not only a subring but a (maximal) ideal, in which case we have a local ring. – anon Mar 05 '13 at 18:57
  • @julien You are not alone :) – rschwieb Mar 05 '13 at 19:00

2 Answers2

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You are correct that zero poses a problem. Note also that the sum of two units need not be a unit.

Example:

$R=\mathbb{Z}$. Then $1$ is a unit, but $1+1=2$ is not.

However, as others have mentioned, the set of nonzero units form a group under multiplication.

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As said, there are easy counterexamples. But there are also interesting cases where it's true, e.g this question has proofs that a finite ring is a field if its units $\cup \{0\}\,$ form a field of characteristic $\ne 2.$

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