Is the Set of Units in a Ring with Identity a Subring?

  • Thread starter Thread starter capsfan828
  • Start date Start date
capsfan828
Messages
3
Reaction score
0

Homework Statement



Prove or Disprove: The set of units in a ring R with identiy is a subring of R.


Homework Equations





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, 0R 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 0R*0R-1 is not an element of S. Hence S is not a subring of R, disproving the original claim.


I feel that the fact 0R*0R-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.
 
Physics news on Phys.org
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top