Characterizing units in polynomial rings

Question

I am trying to prove a result, for which I have got one part, but I am not able to get the converse part.

Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.

Proof. Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \dots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit.

I have not been able to get the converse part and would like to see the proof for the converse part.

Answer 1

Let $f=\sum_{k=0}^n a_kX^k$ and $g= \sum_{k=0}^m b_kX^k$. If $f g=1$, then clearly $a_0,b_0$ are units and:

$$a_nb_m=0 \tag1$$
$$a_{n-1}b_m+a_nb_{m-1}=0$$

(on multiplying both sides by $a_n$)

$$\Rightarrow (a_n)^2b_{m-1}=0 \tag2$$
$$a_{n-2}b_m+a_{n-1}b_{m-1}+a_nb_{m-2}=0$$ (on multiplying both sides by $(a_n)^2$) $$\Rightarrow (a_n)^3b_{m-2}=0 \tag3$$ $$.....$$ $$.....+a_{n-2}b_2+a_{n-1}b_1+a_nb_0=0$$ (on multiplying both sides by $(a_n)^m$) $$\Rightarrow (a_n)^{m+1}b_{0}=0 \tag{m+1}$$

Since $b_0$ is an unit, it follows that $(a_n)^{m+1}=0$.

Hence, we proved that $a_n$ is nilpotent, but this is enough. Indeed, since $f$ is invertible, $a_nx^n$ being nilpotent implies that $f-a_nX^n$ is unit and we can repeat (or more rigorously, perform induction on $\deg(f)$).

Answer 2

If $R$ is a domain then easily $f(X)$ a unit implies that $a_i = 0$ for $i>0$. Now $R\to R/\mathfrak p$, for $\mathfrak p$ prime, reduces to the domain case, yielding that the $a_i$, $i>0$ are in every prime ideal. But the intersection of all prime ideals is the nilradical, the set of all nilpotent elements - as you proved a few days ago.

Remark $ $ This is a prototypical example of reduction to domains by factoring out prime ideals.

Answer 3

Here is a different proof.

If $f = \sum_n a_n X^n \in R[X]$ is a unit, then also $f(0)=a_0 \in R$ is a unit. By considering $a_0^{-1} f$, we may assume $a_0 = 1$ and write $f = 1 - g$ for some $g \in \langle X \rangle$. In the ring of formal power series $R[[X]]$ every such element is invertible, namely with $(1-g)^{-1} = 1 + g + g^2 + \cdots$. In fact, this series converges since the coefficient of $X^n$ remains stable after $g^{n+1},\dotsc$, or alternatively use that $R[[X]]$ is $X$-adically complete. So now the only assumption we have is that $(1-g)^{-1} \in R[[X]]$ is actually contained in $R[X]$. This clearly implies $g^n=0$ for some $n$. Then the constant term of $g$ is nilpotent, and it follows inductively that all coefficients of $g$ are nilpotent.