# Characterizing units in polynomial rings

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.

• If $x$ is nilpotent then $1-x$ is a unit. Commented Jan 26, 2011 at 20:47
• @Chandru1: $\ u$ unit, $z^n = 0\ \Rightarrow\ u-z\ |\ u^n - z^n = u^n,\$ so $\ u - z\$ is a unit, being a divisor of the unit $u^n\:.\$ Thus $\$ unit + nilpotent = unit. Commented Jan 30, 2011 at 19:16
• The idea in the prior comment is a special case of the method of simpler multiples Commented Feb 21, 2021 at 8:54
• Just a side remark over the integral domain we have $(R [x])^\times = R^\times$ Commented Sep 21, 2022 at 6:54

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 $$\deg(f)$$).

• Why does $a_{n-1}b_m+a_nb_{m-1}=0 \Rightarrow (a_n)^2b_{m-1}=0$ ? Commented Oct 3, 2014 at 0:45
• Wait, got it, though surely that is only true if $a_n \in (Z)R$? Commented Oct 3, 2014 at 0:46
• @Nethesis If by $Z$ you mean the center, it is probably because the ring is given commutative ;) Commented Oct 3, 2014 at 1:15
• Ah right, sorry, I was looking for an answer about a non commutative ring and din't see that in the question, my bad @N.S. Commented Oct 5, 2014 at 17:40
• @zed111 In a commutative ring you have unit-nilpotent=unit. The proof is easy: if u is unit and v is nilpotent, then $v^n=0$ which means $$u^n=u^n-v^n=(u-v)(u^{n-1}+u^{n-2}v+...+v^{n-1})$$ This shows that $$(u-v)(u^{n-1}+u^{n-2}v+...+v^{n-1})(u^{-1})^n=1$$ Commented Apr 20, 2015 at 15:47

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.

• See here for units in the Laurent polynomial ring $R[x,x^{-1}]\ \$ Commented Jun 28, 2019 at 12:43
• While this is a nice quick proof, it has the disadvantage of not working in constructive mathematics (where non-trivial commutative rings may have no prime ideals). But there is a trick to make it work at least without the axiom of choice, by reducing (with standard methods) to the case of a finitely generated and hence countable commutative non-trivial ring, which always has a maximal ideal without using the (full) axiom of choice. Commented Dec 17, 2023 at 15:34

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.