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Nilpotent Elements and Units

  • Thread starter Bashyboy
  • Start date
  • #1
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5

Homework Statement


First Claim: If ##u \in R## is a unit, then it cannot be nilpotent

Second Claim: If ##u \in R## is nilpotent, then it cannot be a unit

Homework Equations




The Attempt at a Solution



I realize these are simple problems, but I have no one to verify my work and I want to be certain I am doing things correctly. Here is a proof of the first claim:

Suppose the contrary, that ##u## is a unit and also nilpotent. This implies there exists an element ##v## that acts as a multiplicative inverse and a natural number ##n## such that ##u^n = 0##. By the well ordering property, we can take ##n## to be the smallest natural number for which ##u^n=0##. Then

##u^n = 0##

##u u^{n-1} = 0##

##vu u^{n-1} = v0##

##u^{n-1} = 0##,

contradicting the minimality of ##n##.

Does this seem right? If I am not mistaken, then proof of the second claim is identical.
 

Answers and Replies

  • #2
13,084
9,854
That's correct, although I wouldn't have used ordering and minimality, which looks kind of artificial to me, like a little stone in the shoe.
You could have simply multiplied by ##v^n## instead.
 

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