Posts Tagged ‘commutative ring’

Throughout R is a ring.

Theorem (Jacobson). If for every x \in R there exists some n > 1 such that x^n=x, then R is commutative.

The proof of Jacobson’s theorem can be found in any standard ring theory textbooks. Note that n, in Jacobson’s theorem, doesn’t have to be fixed, i.e. it could depend on x. See this post for the proof of the theorem when n is fixed. Here we only discuss a very special case of the theorem, i.e. when n=3.

Definitions. An element x \in R is called idempotent if x^2=x. The center of R is

Z(R)=\{x \in R: \ xy=yx \ \text{for all} \ y \in R \}.

It is easy to see that Z(R) is a subring of R. An element x \in R is called central if x \in Z(R). Obviously R is commutative iff Z(R)=R, i.e. every element of R is central.

Problem. Prove that if x^3=x for all x \in R, then R is commutative.

Solution.  First note that R is reduced, i.e. R has no nonzero nilpotent element. For every x \in R we have (x^2)^2=x^4 = x^2 and so x^2 is idempotent for all x \in R. Hence, by Remark 3 in this post, x^2 is central for all x \in R. Now, since

(x^2+x)^2=x^4+2x^3+x^2=2x^2+2x

we have 2x=(x^2+x)^2-2x^2 and thus 2x is central. Also, since

x^2+x=(x^2+x)^3=x^6+3x^5+3x^4+x^3=4x^2+4x,

we have 3x=-3x^2 and so 3x is central. Thus x = 3x-2x is central and so R is commutative.  \Box

A similar argument shows that if x^4=x for all x \in R, then R is commutative (see here!).

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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.

It is easy to prove that if every element of a ring is idempotent, then the ring is commutative. This fact can be generalized as follows.

Problem. 1) Let R be a ring with identity and suppose that every element of R is a product of idempotent  elements. Prove that R is commutative.
2)  Give an example of a noncommutative ring with identity R such that every element of R is a product of some elements of the set \{r \in R: \ r^n=r, \ \text{for some} \ n \geq 2 \}.

Solution. 1) Obviously we only need to prove that every idempotent is central. Suppose first that ab = 1, for some a,b \in R. We claim that a = b = 1. So suppose the claim is false. Then a = e_1e_2 \cdots e_k, where e_j are idempotents and e_1 \neq 1. Let e = e_2 \cdots e_kb. Then e_1e = 1 and hence 1 - e_1 = (1 - e_1)e_1e = 0. Thus e_1 = 1. Contradiction! Now suppose that x^2 = 0, for some x \in R. Then (1 - x)(1 + x) = 1 and therefore x = 0, by what we just proved. Finally, since (ey-eye)^2=(ye-eye)^2=0 for any idempotent e \in R and any y \in R, we have ey = ye and so e is central.
2) One example is the ring of 2 \times 2 upper triangular matrices with entries from \mathbb{Z}/2\mathbb{Z}.

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