Posts Tagged ‘polynomial ring’

Problem. (McCoy) Let R be a commutative ring with identity and let f(x)= \sum_{i=0}^n a_ix^i \in R[x].

1) Prove that f(x) is a zero-divisor if and only if there exists some 0 \neq c \in R such that cf(x) = 0.

2) Prove that if R is reduced and f(x)g(x)=0 for some g(x)=\sum_{i=0}^m b_ix^i \in R[x], then a_ib_j=0 for all 0 \leq i \leq n and 0 \leq j \leq m.

Solution. 1) If there exists 0 \neq c \in R such that cf(x)=0, then clearly f(x) is a zero-divisor of R[x]. For the converse, let g(x)=\sum_{i=0}^m b_ix^i, \ b_m \neq 0, be a polynomial with minimum degree such that f(x)g(x)=0. I will show that m = 0. So, suppose to the contrary, that m \geq 1. If a_jg(x)=0, for all j, then a_jb_m=0, for all j, and so b_mf(x)=0 contradicting the minimality of m because \deg b_m = 0 < m. So we may assume that the set \{j: \ a_jg(x) \neq 0 \} is non-empty and so we can let

\ell=\max \{j : \ a_jg(x) \neq 0 \}.


0=f(x)g(x)=(a_{\ell}x^{\ell} + \cdots + a_0)(b_mx^m + \cdots + b_0).


a_{\ell}b_m=0 and so a_{\ell}g(x)=a_{\ell}b_{m-1}x^{m-1} + \cdots + a_{\ell}b_0.

Hence \deg a_{\ell}g(x) < m=\deg g. But we have f(x)(a_{\ell}g(x))=a_{\ell}f(x)g(x)=0, which is impossible because g(x) was supposed to be a polynomial with minimum degree satisfying f(x)g(x)=0.

2) The proof of this part is by induction over i+j. It is obvious from f(x)g(x)=0 that a_0b_0=0. Now let 0 < \ell \leq m+n and suppose that a_rb_s=0 whenever 0 \leq r+s < \ell. We need to show that a_rb_s=0 whenever r+s=\ell. So suppose that r+s=\ell. The coefficient of x^{\ell} in f(x)g(x) is clearly

0=\sum_{i < r, \ i+j=\ell}a_ib_j+ a_rb_s + \sum_{i > r, \ i+j=\ell} a_ib_j,

which after multiplying both sides by a_rb_s gives us

\sum_{i < r, \ i+j=\ell} a_rb_sa_ib_j+ (a_rb_s)^2 + \sum_{i > r, \ i+j=\ell} a_rb_sa_ib_j=0.

Call this (1). Now in the first sum in (1), since i < r, we have i+s < r+s=\ell and hence by the induction hypothesis a_ib_s=0. Thus a_rb_sa_ib_j=0. So the first sum in (1) is 0. In the second sum in (1), since i > r and i+j=r+s=\ell, we have j < s. Therefore by the induction hypothesis a_rb_j=0 and hence a_rb_sa_ib_j=0. So the second sum in (1) is also 0. Thus (1) becomes (a_rb_s)^2=0 and so, since R is reduced, a_rb_s=0. \ \Box

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

This is a generalization of the ordinary representation of polynomials:

 Problem. Let R be a commutative ring with 1 and A \in R[x] have degree n \geq 0 and let B \in R[x] have degree at least 1. Prove that if the leading coefficient of B is a unit of R, then there exist unique polynomials Q_0,Q_1,...,Q_n \in R[x] such that \deg Q_i < \deg B, for all i, and A = Q_0+Q_1B+...+Q_nB^n

SolutionUniqueness of the representation : Since the leading coefficient of B is a unit, for any C \in R[x] we have \deg (BC)=\deg B + \deg C. Now suppose that Q_0 + Q_1B + \cdots + Q_nB^n = 0, with Q_n \neq 0. Let \alpha, \ \beta be the leading coefficients of Q_n and B repectively. Then the leading coefficient of Q_0 + Q_1B + \cdots +Q_nB^n is \alpha \beta^n. Thus \alpha \beta^n = 0. Since \beta is a unit, we’ll get \alpha = 0, which contradicts Q_n \neq 0. Therefore Q_0 = Q_1= \cdots = Q_n=0. 

Existence of the representation : We only need to prove the claim for A=x^n. The proof is by induction over n. It is clear for n = 0, Suppose that the claim is true for any k < n. If n < \deg B, then choose A=Q_0 and Q_1 = \cdots = Q_n=0. So we may assume that n \geq \deg B. Let B=b_mx^m + b_{m-1}x^{m-1}+ \cdots + b_0. Therefore, since b_m is a unit, we will have x^m=b_m^{-1}B-b_m^{-1}b_{m-1}x^{m-1} - \cdots - b_m^{-1}b_0, which will give us x^n = b_m^{-1}x^{n-m}B - b_m^{-1} b_{m-1}x^{n-1} - \cdots - b_m^{-1}b_0 x^{n-m}. Now apply the induction hypothesis to each term x^{n-k}, \ 1 \leq k \leq m, to finish the proof.

Let R be a commutative ring with identity and S=R[x,x^{-1}], the ring of Laurent polynomials with coefficients in R. Obviously S is not a finitely generated R-module but we can prove this:

Problem. There exists f \in S such that S is a finitely generated R[f]-module.

Solution. Let f=x+x^{-1}. Then x=f - x^{-1} and x^{-1}=f-x. Now an easy induction shows that x^n \in xR[f]+R[f] for all n \in \mathbb{Z}. Hence S=xR[f] + R[f]. \ \Box

There are many commutative rings R satisfying this property that the indeterminate x is reducible in the polynomial ring R[x]. Here are two examples: in (\mathbb{Z}/6\mathbb{Z})[x] we have x=(3x+4)(4x+3) and in (\mathbb{Z}/10\mathbb{Z})[x] we have x=(5x+4)(4x+5). In general, if a commutative ring R  has an idempotent e \neq 0,1, then

x=(ex + 1-e)((1-e)x + e).

So if an integer n > 1 has at least two distinct prime divisors, then x will be reducible in (\mathbb{Z}/n\mathbb{Z})[x].