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Background:

Exercise 21: If every nonzero element of $R$ (integral domain) is either irreducible or a unit, prove that $R$ is a field.

Questions:

In the above exercise, the hint suggested that the reader should consider for every nonzero $a\in R,$ to consider $a^2\in R.$ I know that $a^2\neq 0.$ But why is it i can assume that $a^2\cdot u=1$ for some unit $u\in R$?

Thank you in advance

Arturo Magidin
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Seth
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1 Answers1

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If $a$ is a unit, then obviously $a^2$ is also a unit.

If $a$ is not a unit, then $a^2$ is clearly reducible, and therefore by the definition of our ring, a unit.

Arthur
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  • if $a$ is not a unit, why does it automatically make it irreducible? – Seth Jan 29 '24 at 14:09
  • @Seth The problem statement says "every nonzero element of $R$ (integral domain) is either irreducible or a unit". But I deleted that part, as the irreducibility of $a$ is actually irrelevant. – Arthur Jan 29 '24 at 14:10
  • oh kk. Sorry my bad. Sometimes these little details by passes my brain from some reasons. – Seth Jan 29 '24 at 14:11
  • @Seth It happens to everyone from time to time. Case in point: I made a comment on this site just yesterday asking about a detail in an answer I couldn't figure out, because I had forgotten the main premise of the whole solution, in much the same way. – Arthur Jan 29 '24 at 14:12
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jan 29 '24 at 15:50
  • @Arthur thank you for that. – Seth Jan 29 '24 at 15:54