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If $u$ is a unit of $R$, then it means that an inverse belongs to $R$. Thus it belongs to any ring containing $R$. Is there a better answer to this question?

J. Salieri
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LM10
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    This is totally fine. Since R is in R[x] u^{-1} is too. However, you might proof that using the definition of a polynomial ring. – Justin P. Aug 28 '18 at 12:26
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    What is $R$? A field? – dmtri Aug 28 '18 at 12:34
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    @dmtri The context makes it clear that $R$ is a ring, not to mention the letter $R$ which rather suggests that it is indeed a ring. – J.-E. Pin Aug 28 '18 at 12:38
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    Why did you ask the same question twice making this question's body virtually a duplicate of a comment on that question? There is no good reason to do that. Please don't do it again. You might also benefit from reviewing the FAQ about posting on our site. – rschwieb Aug 28 '18 at 13:29
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    @dmtri ....R is a ring. – LM10 Aug 29 '18 at 11:04

1 Answers1

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Your argument is correct, but beware that there might be other units in $R[X]$. For instance, if $R = \mathbb{Z}/4\mathbb{Z}$, then $1 + 2X$ is a unit since $(1 + 2X)(1+ 2X) = 1$.

J.-E. Pin
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