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- April 21st 2009, 11:03 AMmathman88Units
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- April 21st 2009, 04:20 PMchiph588@
Suppose is a unit.

Then there exists s.t. . It is evident .

Therefore and .

Since is an integral domain and , .

The only way for is if , where . So or

Hence or . Which means . - April 21st 2009, 09:42 PMNonCommAlg
i don't think you can make this conclusion!

well, if and n > 0 or m > 0, then we must have which is not possible in an integral domain. so n = m = 0, and the rest is

obvious. this also shows that, in general, for any integral domain the set of units of is exactly equal to the set of units of

Remark: there's a well-known result that in any commutative ring with unity a polynomial is a unit iff is a unit of and all are nilpotent.