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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?

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  • $\begingroup$ 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. $\endgroup$ – rschwieb Mar 5 '13 at 18:43
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    $\begingroup$ 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. $\endgroup$ – Julien Mar 5 '13 at 18:44
  • $\begingroup$ Sometimes the set of non-units form not only a subring but a (maximal) ideal, in which case we have a local ring. $\endgroup$ – anon Mar 5 '13 at 18:57
  • $\begingroup$ @julien You are not alone :) $\endgroup$ – rschwieb Mar 5 '13 at 19:00
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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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