Invertible polynomials

Here is a classic little lemma on polynomials:

Lemma. Let $R$ be a commutative ring with nilradical $\mathfrak r$ . Then $R[t]^\times = R^\times + t\mathfrak r[t]$ .

That is, a polynomial $a_nt^n + \ldots + a_1t + a_0 \in R[t]$ is invertible if and only if $a_0$ is invertible and $a_1,\ldots,a_n$ are nilpotent. This is the formulation you find in Atiyah–MacDonald, exercise 1.2(i). Note that when $R$ is a domain (or more generally a reduced ring), this just says that $R[t]^\times = R^\times$ (via the constant polynomials).

Proof. First suppose $f = \sum_{i=0}^n a_nt^n$ with $a_0 \in R^\times$ and $a_1,\ldots,a_n \in \mathfrak r$ . Multiplying by $a_0^{-1}$ we may assume $a_0 = 1$ ; this does not change the nilpotence assumption on $a_1,\ldots,a_n$ . Then $g := 1-f = -a_1t-\ldots - a_nt^n$ is nilpotent since it is a sum of nilpotent elements $-a_it^i$ in the commutative ring $R[t]$ . Thus we get $g^{r+1} = 0$ for some $r \in \mathbf N$ , and

$(1-g)(1+g+\ldots+g^r) = 1$

shows that $f \in R[t]^\times$ . Conversely, let $f \in R[t]^\times$ , and suppose $fg = 1$ for some polynomial $g = b_mt^m + \ldots + b_1t+b_0 \in R[t]$ . Without loss of generality we may assume $a_n \neq 0 \neq b_m$ . If $R$ is a domain, then the leading term of $fg$ is $a_nb_mt^{m+n} \neq 0$ , so we must have $n=m=0$ and $f$ and $g$ are constant. In the general case, we conclude that $\phi(f) \in S^\times \subseteq S[t]$ for any ring homomorphism $\phi \colon R \to S$ to a domain $S$ . In particular, taking $S = R/\mathfrak p$ for any prime ideal $\mathfrak p \subseteq R$ shows $a_1,\ldots,a_n \in \mathfrak p$ and $a_0 \not\in\mathfrak p$ . Taking the intersection over all prime ideals gives

$\qedsymbol$ $\begin{align*}a_1,\ldots,a_n &\in \bigcap_{\mathfrak p \text{ prime}} \mathfrak p = \mathfrak r,\\ a_0 &\in \bigcap_{\mathfrak p \text{ prime}} (R \setminus \mathfrak p) = R^\times. \end{align*}$

Lovely little lemmas

Or maybe ‘amusing observations’

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