Characterization of all those integral domain for which $1$ is the only element that is its own multiplicative inverse?

Question

For any element $x$ which is self invertible, we have $x^{2}=1$. And $(x-1)(x+1)=0$ which implies $x$ must be either $1$ or $-1$ as we have considered an integral domain.

I was wondering whether there are more such elements or these are the only ones? With the help of the above, can we characterize all those integral domain for which $1$ is the only element that is its own multiplicative inverse?

Does this answer your question? Proving that an integral domain has at most two elements that satisfy the equation $x^2 = 1$. — player3236, Sep 04 '20 at 05:53
Actually I was trying to characterize all those integral domains for which $1$ is the only element that is its own multiplicative inverse. — Learner_Shas, Sep 04 '20 at 06:01
@curious_soul Trivially, yes. It would necessitate that $1=-1$, that is, the domain has characteristic $2$. Then you have your characterization: domains with characteristic 2. — rschwieb, Sep 04 '20 at 13:34

score 1 · Answer 1 · answered Sep 04 '20 at 05:51

1

There are no more such elements and your proof is sufficient.

answered Sep 04 '20 at 05:51

player3236

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Characterization of all those integral domain for which $1$ is the only element that is its own multiplicative inverse?

1 Answers1