# Proof Critique - Atiyah-MacDonald Question 1.2

Here is my proposed proof for part of Exercise 1.2 of Atiyah-MacDonald's Commutative Algebra:

Let $R$ be a ring and let $R[x]$ be the ring of polynomials in an indeterminate $x$, with coefficients in $R$. Let $f = a_nx^n + \cdots +a_1x+a_0 \in R[x]$. Prove that

(i) $f$ is a unit in $R[x]$ if and only if $a_0$ is a unit in $R$ and $a_1, ..., a_n$ are nilpotent.

Suppose that $f \in R[x]$ is a unit. Then, by considering the reduction map $R[x] \to (R/\mathfrak{p})[x]$, we have that $f$ is a unit in $(R/\mathfrak{p})[x]$, since any ring morphism takes units to units. We observe that since $\mathfrak{p}$ is a prime ideal, $(R/\mathfrak{p})[x]$ is an integral domain, which by definition, contains no zero divisors. Now suppose that $g \in (R/\mathfrak{p})[x]$ is the inverse of $f$. Since we are in an integral domain, we can guarentee that $\deg(fg) = \deg(f) +\deg(g)$. Therefore, $fg = 1$ implies that $f$ and $g$ are constant. Thus $a_1, ..., a_n$ are nilpotent, given that $a_1, ..., a_n =0 \in (R/\mathfrak{p})[x]$. The fact that $fg =1$ also allows us to conclude that $a_0b_0 =1$ if we let $b_0$ denote the constant term of $g$. Hence $a_0$ is a unit.

Conversely, suppose that $a_1, ..., a_n$ are nilpotent and that $a_0$ is a unit in $R$. Then $a_nx^n, ..., a_1x$ are nilpotent and $a_0$ is a unit in $R[x]$. With the sum of nilpotent elements being nilpotent, we have that $a_nx^n + \cdots + a_1x$ is nilpotent in $R[x]$. Therefore, Exercise 1.1 allows us to conclude that $f(x) = a_nx^n + \cdots + a_1x+a_0$ is a unit in $R[x]$, since it is the sum of a nilpotent element and a unit.

(ii) $f$ is nilpotent if and only if $a_0, a_1, ..., a_n$ are nilpotent.

Suppose that $f \in R[x]$ is nilpotent. Then, by considering the reduction map $R[x] \to (R/\mathfrak{p})[x]$, we have that $f =0 \in (R/\mathfrak{p})[x]$, since $(R/\mathfrak{p})[x]$ is an integral domain. This implies that $f(x) = a_nx^n + \cdots +a_1x + a_0 \in \mathfrak{p}[x]$. Hence, $a_n, ..., a_0 \in \mathfrak{p}$ and therefore $a_n, ..., a_0$ are nilpotent.

Conversely, suppose that $a_0, a_1, ..., a_n$ are nilpotent in $R[x]$. Then $a_0, a_1x, ..., a_nx^n$ are nilpotent. By taking the sum, which remains nilpotent, we see that $f(x) = a_nx^n + \cdots + a_1x +a_0$ is nilpotent.

The proof is good, but perhaps the confusion between $f$ and its image modulo $\mathfrak{p}$ needs to be sorted out.
If $\mathfrak{p}$ is a prime ideal, denote by $f_{\mathfrak{p}}$ the reduction of $f$ modulo $\mathfrak{p}$.
• Assuming $f$ is a unit, we have that $f_{\mathfrak{p}}$ is a unit in a polynomial ring over a domain, so it has degree $0$. Hence $a_1,\dots,a_n$ are zero modulo $\mathfrak{p}$, so they belong to it. Since $\mathfrak{p}$ is arbitrary, $a_1,\dots,a_n$ are nilpotent.
• Assuming $f$ is nilpotent, so is its reduction $f_{\mathfrak{p}}$, for every prime ideal $\mathfrak{p}$, which implies $f_{\mathfrak{p}}$ is zero. Therefore, as before, all coefficients of $f$ are nilpotent.