## Rings satisfying x^3 = x are commutative

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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 standard ring theory textbooks. Here we only discuss a very special case of the theorem, i.e. when $x^3=x$ for all $x \in R.$

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.  Clearly $R$ 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 hence $3x$ is central. Therefore $x = 3x-2x$ is central. $\Box$

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

## Units in polynomial rings

Posted: September 2, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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. $\Box$

## Representation of polynomials

Posted: May 23, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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.

## “Almost Boolean” rings are commutative

Posted: March 12, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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 $\mathbb{Z}/2\mathbb{Z}.$

## Module-finiteness of Laurent polynomial rings

Posted: March 11, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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 $S=xR[f] + R[f]. \ \Box$