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 .
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.