
#1
Jan3009, 05:41 PM

P: 4

1. The problem statement, all variables and given/known data
Prove or Disprove: The set of units in a ring R with identiy is a subring of R. 2. Relevant equations 3. The attempt at a solution Let S be the the set of units in a ring R with identity. For S to be a subring of R, 0_{R} would have to be an element of S. Since S is the set of units in R, it follows that S will not a multiplicative identity, namely 0_{R}*0_{R}^{1} is not an element of S. Hence S is not a subring of R, disproving the original claim. I feel that the fact 0_{R}*0_{R}^{1} is not an element of S is the main part of the proof. I am just unsure if my argument and logic are correct. 



#2
Jan3009, 07:56 PM

P: 284

I think the identity he is referring to is the additive one (1 is trivially a unit). So your counter proof isn't really valid.
If S was to be a subgroup then it must be closed under addition and multiplication. It is easy to check that its closed under multiplication. Look at addition, when you add two units, is it always a unit? 


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