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Let $R$ be an integral domain. Show that $U(R)$ has at most $n$ elements of order $n$, for every positive integer $n$. Also give an example of a commutative ring $R$ with identity which is not an integral domain for which this is not true.

My attempt: If $a$ is an element in $ U(R)$ and has order $n$, then $a^n=1$. I think that a possible way to do this in the case that $R$ is finite is by considering the polynomial $f(x)=x^n-1$ in $R[x]$, since in this case $R$ is a field, and thus we can ensure that $f(x)$ has at most $n$ roots. Since $a$ is a root of $f(x)$, the result hold. But, what about the case when $R$ is infinite?

On the other hand, in this question appears what would be a solution for the second part. It should be said that this does not completely answer my question, since here $R$ was considered directly as a field. Or at least I think so... Thanks in advance

Hopmaths
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    A polynomial of degree $n$ over a domain has at most $n$ roots. Whether $R$ is finite or infinite, you can always embed a domain in its field of fractions, and it seems you know this result over a field. – Viktor Vaughn Sep 16 '21 at 19:29
  • Two posts on this result: 1, 2 – Viktor Vaughn Sep 16 '21 at 19:32
  • @ViktorVaughn So solving this exercise is essentially showing the result of the roots for the integral domains? – Hopmaths Sep 16 '21 at 19:37
  • Yes, I think you've solved the first part. – Viktor Vaughn Sep 16 '21 at 20:01
  • @ViktorVaughn I have a question. I understand that $R$ is contained in its field of fractions, namely $F$. But at the time of establishing the polynomial $f(x)$ is it where it lives in $R[x]$ or in $F[x]$? – Hopmaths Sep 16 '21 at 22:31
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    I'm not sure I understand your question. Every root of $f$ in $R$ is also in $F$. So if you had $n+1$ roots of $f$ in $R$, this would mean there are $n+1$ roots of $f$ in $F$, contradicting the result you referenced in your post. – Viktor Vaughn Sep 16 '21 at 22:43
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    @ViktorVaughn Oh, really thank you. I was getting a bit confused. Thanks again for your help. – Hopmaths Sep 16 '21 at 22:47

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