## Rings satisfying x^4 = x are commutative

Posted: April 19, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Let $R$ be a ring, which may or may not have $1.$ We proved in here that if $x^3=x$ for all $x \in R,$ then $R$ is commutative.  A similar approach shows that if $x^4=x$ for all $x \in R,$ then $R$ is commutative.

Problem. Prove that if $x^4=x$ for all $x \in R,$ then $R$ is commutative.

Solution. Clearly $R$ is reduced, i.e. $R$ has no nonzero nilpotent element. Note that $2x=0$ for all $x \in R$ because $x=x^4=(-x)^4=-x.$ Hence $x^2+x$ is an idempotent for every $x \in R$ because

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

Thus $x^2+x$ is central for all $x \in R,$ by Remark 3 in this post.  Therefore $(x^2+y)^2+x^2+y$ is central for all $x,y \in R.$ But

$(x^2+y)^2+x^2+y=x^2+x+y^2+y+ x^2y+yx^2$

and hence $x^2y+yx^2$ is central. Therefore $(x^2y+yx^2)x^2=x^2(x^2y+yx^2)$ which gives us $xy=yx. \ \Box$

## 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!).

## Zero-divisors in polynomial rings

Posted: November 1, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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

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

Then

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

Thus

$a_{\ell}b_m=0$ 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, $a_rb_s=0. \ \Box$

## von Neumann regular rings (3)

Posted: October 31, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
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We saw in part (2) that von Neumann regular rings live somewhere between semisimple and semiprimitive rings. The goal in this post is to prove a theorem of Armendariz and others which gives a necessary and sufficient condition for a ring to be both regular and reduced. This result extends Kaplansky’s result for commutative rings (see the corollary at the end of this post). We remark that a commutative von Neumann regular ring $R$ is necessarily reduced. That is because if $x^2=0$  for some $x \in R,$ then choosing $y \in R$ with $x=xyx$ we will get $x=yx^2=0.$

Definition . A von Neumann regular ring $R$ is called strongly regular if $R$ is reduced.

Theorem 1. (Armendariz, 1974) A ring  $R$ with 1 is strongly regular if and only if $R_M$ is a division ring for all maximal ideals $M$ of $Z(R).$

Proof. Suppose first that $R$ is strongly regular and let $M$ be a maximal ideal of $Z(R).$ Let $0 \neq s^{-1}x \in R_M.$ So $tx \neq 0$ for all $t \in Z(R) \setminus M.$ Since $R$ is regular, there exists some $y \in R$ such that $xyx = x.$ Then $xy=e$ is an idempotent and thus $e \in Z(R)$ because in a reduced ring every idempotent is central.  Since $(1-e)x=0$ we have $1-e \in M$ and hence $e \in Z(R) \setminus M.$ Thus $e^{-1}sy$ is a right inverse of $s^{-1}x.$ Similarly $f=yx \in Z(R) \setminus M$ and $f^{-1}sy$ is a left inverse of $s^{-1}x.$ Therefore $s^{-1}x$ is invertible and hence $R_M$ is a division ring. Conversely, suppose that $R_M$ is a division ring for all maximal ideals $M$ of $Z(R).$ If $R$ is not reduced, then there exists $0 \neq x \in R$ such that $x^2=0.$ Let $I=\{s \in Z(R): \ sx = 0 \}.$ Clearly $I$ is a proper ideal of $Z(R)$ and hence $I \subseteq M$ for some maximal ideal $M$ of $Z(R).$ But then $(1^{-1}x)^2=0$ in $R_M,$ which is a division ring. Thus $1^{-1}x=0,$ i.e. there exists some $s \in Z(R) \setminus M$ such that $sx = 0,$ which is absurd. To prove that $R$ is von Neumann regular, we will assume, to the contrary, that $R$ is not regular. So there exists $x \in R$ such that $xzx \neq x$ for all $z \in R.$ Let $J= \{s \in Z(R): \ xzx=sx \ \text{for some} \ z \in R \}.$ Clearly $J$ is a proper ideal of $Z(R)$ and so $J \subseteq M$ for some maximal ideal $M$ of $Z(R).$ It is also clear that if $sx = 0$ for some $s \in Z(R),$ then $s \in J$ because we may choose $z = 0.$ Thus $1^{-1}x \neq 0$ in $R_M$ and hence there exists some $y \in R$ and $t \in Z(R) \setminus M$ such that $1^{-1}x t^{-1}y = 1.$ Therefore $u(xy-t)=0$ for some $u \in Z(R) \setminus M.$ But then $x(uy)x=utx$ and so $ut \in J,$ which is nonsense. This contradiction proves that $R$ must be regular. $\Box$

Corollary. (Kaplansky) A commutative ring $R$ is regular if and only if $R_M$ is a field for all maximal ideals $M$ of $R. \ \Box$

At the end let me mention a nice property of strongly regular rings.

Theorem 2. (Pere Ara, 1996) If $R$ is strongly regular and $Ra+Rb=R,$ for some $a, b \in R,$ then $a+rb$ is a unit for some $r \in R.$

Definition 1. A ring $R$ is called Dedekind-finite if $\forall a,b \in R: \ ab=1 \Longrightarrow ba=1.$

Remark 1. Some trivial examples of Dedekind-finite rings: commutative rings, any direct product of Dedekind-finite rings, any subring of a Dedekind-finite ring.

Definition 2. A ring $R$ is called reversible if $\forall a,b \in R : \ ab = 0 \Longrightarrow ba = 0.$

Example 1. Every reversible ring $R$ is Dedekind-finite. In particular, reduced rings are Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Then $(ba-1)b=b(ab)-b=0$ and thus $b(ba-1)=0.$ So $b^2a=b$ and hence $ab^2a=ab=1.$ It follows that $ba=(ab^2a)ba=(ab^2)(ab)a=ab^2a=1.$ So $R$ is Dedekind-finite. Finally, note that every reduced ring is reversible because if $ab=0,$ for some $a,b \in R,$ then $(ba)^2=b(ab)a=0$ and thus $ba=0. \Box$

Example 2. Every (left or right) Noetherian ring $R$ is Dedekind-finite.

Proof. We will assume that $R$ is left Noetherian. Suppose that $ab=1$ for some $a,b \in R.$ Define the map $f: R \longrightarrow R$ by $f(r)=rb.$ Clearly $f$ is an $R$-module homomorphism and $f$ is onto because $f(ra)=(ra)b=r(ab)=r,$ for all $r \in R.$ Now we have an ascending chain of left ideals of $R$

$\ker f \subseteq \ker f^2 \subseteq \cdots.$

Since $R$ is left Noetherian, this chain stabilizes at some point, i.e. there exists some $n$ such that $\ker f^n = \ker f^{n+1}.$ Clearly $f^n$ is onto because $f$ is onto. Thus $f^n(c)=ba-1$ for some $c \in R.$ Then

$f^{n+1}(c)=f(ba-1)=(ba-1)b=b(ab)-b=0.$

Hence $c \in \ker f^{n+1}=\ker f^n$ and therefore $ba-1=f^n (c) = 0. \Box$

Example 3. Finite rings are obviously Noetherian and so Dedekind-finite by Example 2. More generally:

Example 4. If the number of nilpotent elements of a ring is finite, then the ring is Dedekind-finite. See here.

Note that Example 4 implies that every reduced ring is Dedekind-finite; a fact that we proved in Example 1.

Example 5. Let $k$ be a field and let $R$ be a finite dimensional $k$-algebra. Then $R$ is Dedekind-finite.

Proof. Every left ideal of $R$ is clearly a $k$-vector subspace of $R$ and thus, since $\dim_k R < \infty,$ any ascending chain of left ideals of $R$ will stop at some point. So $R$ is left Noetherian and thus, by Example 2, $R$ is Dedekind-finite. $\Box$

Remark 2. Two important cases of Example 5 are $M_n(R),$ the ring of $n \times n$ matrices over a field, and, in general, semisimple rings. As a trivial result, $M_n(R)$ is Dedekind-finite for any commutative domain $R$ because $M_n(R)$ is a subring of $M_n(Q(R))$, where $Q(R)$ is the quotient field of $R$.
So the ring of $n \times n$ matrices, where $n \geq 2,$ over a field is an example of a Dedekind-finite ring which is not reversible, i.e. the converse of Example 1 is not true. Now let $R_i = \mathbb{Z}, \ i \geq 1.$ Then $R= \prod_{i=1}^{\infty} R_i$ is obviously Dedekind-finite but not Noetherian. So the converse of Example 2 is not true.

Example 6 and Example 7 are two generalizations of Example 5.

Example 6. Every algebraic algebra $R$ over a field $k$ is Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Since $R$ is algebraic over $k,$ there exist integers $n \geq m \geq 0$ and some $\alpha_i \in k$ with $\alpha_n \alpha_m \neq 0$ such that $\sum_{i=m}^n \alpha_i b^i = 0.$ We will assume that $n$ is as small as possible. Suppose that $m \geq 1.$ Then, since $ab=1,$ we have

$\sum_{i=m}^n \alpha_i b^{i-1}=a \sum_{i=m}^n \alpha_i b^i = 0,$

which contradicts the minimality of $n.$ So $m = 0.$ Let $c = -\alpha_0^{-1}\sum_{i=1}^n \alpha_i b^{i-1}$ and see that $bc=cb=1.$ But then $a=a(bc)=(ab)c=c$ and therefore $ba=bc=1. \ \Box$

Remark 3. Regarding Examples 5 and 6, note that although any finite dimensional $k$-algebra $R$ is algebraic over $k,$ but $R$ being algebraic over $k$ does not necessarily imply that $R$ is finite dimensional over $k.$ For example, if $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{C},$ then it is easily seen that $\dim_{\mathbb{Q}} \overline{\mathbb{Q}}=\infty.$ Thus the matrix ring $R = M_n(\overline{\mathbb{Q}})$ is an algebraic $\mathbb{Q}$-algebra which is not finite dimensional over $\mathbb{Q}.$ So, as a $\mathbb{Q}$-algebra, $R$ is Dedekind-finite by  Example 6 not Example 5.

Example 7. Every PI-algebra $R$ is Dedekind-finite.

Proof. Let $J(R)$ be the Jacobson radical of $R.$ If $J(R)=\{0\},$ then $R$ is a subdirect product of primitive algebras $R/P_i,$ where $P_i$ are the primitive ideals of $R.$ Since $R$ is PI, each $R/P_i$ is PI too and thus, by Kaplansky’s theorem, $R/P_i$ is a matrix ring over some division algebra and thus Dedekind-finite by Example 2. Thus $\prod R/P_i$ is Dedekind-finite and so $R,$ which is a subalgebra of $\prod R/P_i,$ is also Dedekind-finite. For the general case, let $S=R/J(R).$ Now, $S$ is PI, because $R$ is PI, and $J(S)=\{0\}.$ Therefore, by what we just proved, $S$ is Dedekind-finite. Suppose that $ab = 1$ for some $a,b \in R$ and let $c,d$ be the image of $a,b$ in $S$ respectively. Clearly $cd=1_S$ and so $dc=1_S.$ Thus $1-ba \in J(R)$ and so $ba=1-(1-ba)$ is invertible. Hence there exists $e \in R$ such that $e(ba)=1.$ But then $eb=(eb)ab=e(ba)b=b$ and hence $ba=(eb)a=e(ba)=1. \Box$

## Ring of endomorphisms (3)

Posted: June 9, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
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Schur’s lemma states that if $A$ is a simple $R$ module, then $\text{End}_R(A)$ is a division ring. A similar easy argument shows that:

Example 6. For simple $R$-modules $A \ncong B$ we have $\text{Hom}_R(A,B)=\{0\}.$

Let’s generalize Schur’s lemma: let $M$ be a finite direct product of simple $R$-submodules. So $M \cong \bigoplus_{i=1}^k M_i^{n_i},$ where each $M_i$ is a simple $R$-module and $M_i \ncong M_j$ for all $i \neq j.$ Therefore, by Example 6 and Theorem 1, $\text{End}_R(M) \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i),$ where $D_i = \text{End}_R(M_i)$ is a division ring by Schur’s lemma. An important special case is when $R$ is a semisimple ring. (Note that simple submodules of a ring are exactly minimal left ideals of that ring.)

Theorem 2. (Artin-Wedderburn) Let $R$ be a semisimple ring. There exist a positive integer $k$ and division rings $D_i, \ 1 \leq i \leq ,$ such that $R \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i)$.

Proof. Obvious, by Example 1 and the above discussion. $\Box$

Some applications of Theorem 2.

1. A commutative semisimple ring is a finite direct product of fields.

2. A reduced semisimple ring is a finite direct product of division rings.

3. A finite reduced ring is a finite direct product of finite fields.

Posted: June 4, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Definition. Let $R, \ R_i, \ i \in I,$ be rings. For every $j \in I$ we let $\pi_j : \prod_{i \in I} R_i \longrightarrow R_j$ be the natural projection. Then $R$ is called a subdirect product of $R_i, \ i \in I,$ if the following conditions are satisfied:

1) There exists an injective ring homomorphism $f: R \longrightarrow \prod_{i \in I} R_i,$

2) For every $j \in I$ the map $\pi_j f: R \longrightarrow R_j$ is surjective.

Note. Suppose that $A_i, \ i \in I,$ are some ideals of $R$ and put $R_i = R/A_i.$ Then we can define $f: R \longrightarrow \prod_{i \in I} R/A_i$ by $f(r)=(r+ A_i)_{i \in I}.$ Clearly the second condition in the above definition is satisfied. Thus $R$ is a subdirect product of $R/A_i, \ i \in I,$ if and only if $f$ is injective, i.e. $\bigcap_{i \in I} A_i = \{0\}.$

Remark 6. If $P$ is a minimal prime ideal of the ring $R,$ then $S=R \setminus P$ is multiplicatively closed iff $s_1s_2 \cdots s_k \neq 0$, for all $s_i \in S, \ k \in \mathbb{N}.$

Proof. Suppose that $s_1s_2 \cdots s_k \neq 0,$ for any $s_1,s_2, \cdots, s_k \in S$ and $k \in \mathbb{N}.$ Let $T$ be the set of all elements of $R$ which are a finite product of some elements of $S.$ Clearly $T$ is multiplicatively closed, $S \subseteq T$ and $S$ is multiplicatively closed iff $S=T.$ So we’ll be done if we show that $S=T$. Let $\mathcal{C}=\{A \lhd R: \ A \cap T=\emptyset \}.$ We have $\mathcal{C} \neq \emptyset$ because $(0) \in \mathcal{C}.$ Therefore, by Zorn’s lemma, $(\mathcal{C}, \subseteq)$ has a maximal element $Q$ and $Q$ is a prime ideal of $R.$ Since $Q \cap T = \emptyset,$ we have $Q \cap S = \emptyset$ and thus $Q \subseteq P.$ Thus $Q=P$ because $P$ is a minimal prime. So $P \cap T= \emptyset$, which means $T \subseteq S.$ Hence $T=S. \ \Box$

Remark 7. If $R$ is reduced and $P \lhd R$ is a minimal prime, then $R/P$ is a domain.

Proof. Clearly $R/P$ is a domain iff $S = R \setminus P$ is multiplicatively closed. Let $T$ be as defined in Remark 6. By that remark, we only need to show that $0 \notin T.$ So suppose that $s_1s_2 \cdots s_k = 0,$ for some $s_1, s_2, \cdots , s_k \in S$, where the integer $k \geq 2$ is assumed to be minimal. Then by, Remark 1, we have $s_k R s_1s_2 \cdots s_{k-1} = \{0\}.$ Now, since $P$ is prime, $s_k R s_1$ cannot be a subset of $P$ because otherwise we’d have either $s_k \in P$ or $s_1 \in P,$ which is clearly nonsense. Thus $s_k Rs_1 \cap S \neq \emptyset.$ Let $s \in s_kRs_1 \cap S.$ Then

$ss_2 \cdots s_{k-1} \in s_kRs_1s_2 \cdots s_{k-1} = \{0\}.$

Hence $ss_2 \cdots s_{k-1}=0,$ which contradicts the minimality of $k. \ \Box$

The Structure Theorem For Reduced Rings. A ring $R$ is reduced iff $R$ is a subdirect product of domains.

Proof. If $R$ is reduced, then, by Remarks 5 and 7, $R$ is a subdirect product of the domains $R/P_i, \ i \in I,$ where $\{P_i \}_{i \in I}$ is the set of all minimal prime ideals of $R.$ Conversely, suppose that $R$ is a subdirect product of domains $R_i, \ i \in I$ and $f: R \longrightarrow \prod_{i \in I} R_i$ is an injective ring homomorphism. Suppose that $x \in R$ and $x^2=0.$ Let $f(x)=(x_i)_{\in I}.$ Then $(0_{R_i})_{i \in I} = f(x^2)=(f(x))^2=(x_i^2)_{i \in I}.$ Thus $x_i^2=0,$ for all $i \in I,$ and so $x_i = 0,$ for all $i \in I,$ because every $R_i$ is a domain. Hence $x=0$ and so $R$ is reduced. $\Box$