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

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Comments
  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.

  4. Sugata Adhya says:

    Using the concept of matrix in the converse of the first one really made it simpler. I was trying to solve the problem for last few days. Thanks for the post.

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