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*}

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