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I'm self teaching abstract algebra and I've come across this question...I cannot figure it out! R is not assumed to be a domain so it can't just be the constant polynomials, but whenever I try to characterise them I just run into masses of difficulty, with numerous exceptions appearing to any rule I try to set and I realise that for one thing, I don't even know what the units in $R[x]$ are if $R$ is not a domain, to be honest. Can't find an answer to this anywhere, several questions have asked similar, but none have actually been answered.

Seth
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Nethesis
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1 Answers1

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It's a standard fact, found in most commutative algebra texts (e.g. Atiyah-Macdonald), that a polynomial in $R[x]$ is a unit exactly when its constant term is a unit, and all other coefficients are nilpotents.

You can find a proof, for example, here. But here's a self-contained summary:

First, if we have such a polynomial, then we can write it as $a + n$, with $a\in R^\times$, and $n\in R[x]$ nilpotent. Then $a^N + n^N = (a+n)(a^{N-1} - \cdots) = a^N$ for large, odd $N$. Since $a^N$ is a unit, so is $a+n$.

The other direction follows from the fact that an element of $R$ is nilpotent if and only if it lies in every prime ideal. But for any prime ideal, $R/P$ is a domain. Since units map to units in the quotient, it follows that all the non-constant terms map to $0$, i.e. their coefficients lie in $P$.

The same thing holds over multiple variables, by induction—though we should be careful to check that a polynomial is nilpotent exactly when its coefficients are.

Andrew Dudzik
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