## Units in polynomial rings

Posted: September 2, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

Lemma. Let $R$ be a commutative ring with 1. If $a \in R$ is nilpotent and $b \in R$ is a unit, then $a+b$ is a unit.

Proof. So $a^n = 0$ for some integer $n \geq 1$ and $bc = 1$ for some $c \in R.$ Let

$u = (b^{n-1}- ab^{n-2} + \ldots + (-1)^{n-2}a^{n-2}b + (-1)^{n-1}a^{n-1})c^n$

and see that $(a+b)u=1. \ \Box$

Problem. Let $R$ be a commutative ring with 1. Let $p(x) = \sum_{j=0}^n a_j x^j, \ a_j \in R,$ be an element of the polynomial ring $R[x].$ Prove that $p(x)$ is a unit if and only if $a_0$ is a unit and all $a_j, \ j \geq 1,$ are nilpotent.

First Solution. ($\Longrightarrow$) Suppose that $a_1, \cdots , a_n$ are nilpotent and $a_0$ is a unit. Then clearly $p(x)-a_0$ is nilpotent and thus $p(x)=p(x)-a_0 + a_0$ is a unit, by the lemma.

($\Longleftarrow$) We’ll use induction on $n,$ the degree of $p(x).$ It’s clear for $n = 0.$ So suppose that the claim is true for any polynomial which is a unit and has degree less than $n.$ Let $p(x) = \sum_{j=0}^n a_jx^j, \ n \geq 1,$ be a unit. So there exists some $q(x)=\sum_{j=0}^m b_jx^j \in R[x]$ such that $p(x)q(x)=1.$ Then $a_0b_0=1$ and so $b_0$ is a unit. We also have

$a_nb_m = 0, \ a_nb_{m-1}+ a_{n-1}b_m = 0, \ \cdots , a_nb_0 +a_{n-1}b_1 + \cdots = 0.$

So $AX=0,$ where

$A=\begin{pmatrix}a_n & 0 & 0 & . & . & . & 0 \\ a_{n-1} & a_n & 0 & . & . & . & 0 \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ * & * & * & . & . & . & a_n \end{pmatrix}, \ \ X=\begin{pmatrix}b_m \\ b_{m-1} \\ . \\ . \\ . \\ b_0 \end{pmatrix}.$

Thus $a_n^{m+1}X =(\det A)X = \text{adj}(A)A X = 0.$ Therefore $a_n^{m+1}b_0=0$ and hence $a_n^{m+1} = 0$ because $b_0$ is a unit. Thus $a_n,$ and so $-a_nx^n,$ is nilpotent. So $p_1(x)=p(x) -a_nx^n$ is a unit, by the lemma. Finally, since $\deg p_1(x) < n,$ we can apply the induction hypothesis to finish the proof. $\Box$

Second Solution. ($\Longrightarrow$) This part is the same as the first solution.

($\Longleftarrow$) Let $p(x) = \sum_{j=0}^n a_jx^j, \ a_n \neq 0,$ be a unit of $R[x]$ and let $q(x)=\sum_{i=0}^m b_i x^i \in R[x], \ b_m \neq 0,$ be such that $p(x)q(x)=1.$ Then $a_0b_0=1$ and so $a_0$ is a unit. To prove that $a_j$ is nilpotent for all $j \geq 1,$ we consider two cases:

Case 1 . $R$ is an integral domain. Suppose that $n > 0.$ Then from $p(x) q(x)=1$ we get $a_n b_m = 0,$ which is impossible because both $a_n$ and $b_m$ are non-zero and $R$ is an integral domain. So $n=0$ and we are done.

Case 2 . $R$ is arbitrary. Let $P$ be any prime ideal of $R$ and let $\overline{R}=R/P.$ For every $r \in R$ let $\overline{r}=r+P.$ Let

$\overline{p(x)}=\sum_{j=0}^n \overline{a_j}x^j, \ \ \overline{q(x)}=\sum_{i=0}^m \overline{b_i}x^i.$

Then clearly $\overline{p(x)} \cdot \overline{q(x)}=\overline{1}$ in $\overline{R}[x]$ and thus, since $\overline{R}$ is an integral domain, $\overline{a_j}=\overline{0}$ for all $j \geq 1,$ by case 1. Hence $a_j \in P$ for all $j \geq 1.$ So $a_j, \ j \geq 1,$ is in every prime ideal of $R$ and thus $a_j$ is nilpotent. $\Box$

1. Anand says:

For the lemma : given $b.c=1$ for some $c$ then , if suppose there is no element such that $(a+b)u \not =1$ then by multiplying by $a^{n-1} \not =$ we get a contradiction that $a^{n-1} \not = a^{n-1}$ . does that sound ok ?

2. Todd Trimble says:

Hi, I just saw this. Do you know what can be said in the case where $R$ is non-commutative?

3. Georgios Anthitsis says:

Thank you for posting this proof. There’s an insignificant mistake: $\det(A)=a_n^{m+1}$ since $A$ is a $(m+1) \times (m+1)$ matrix.

• Yaghoub says:

You’re right, thank you. I fixed it.