## Posts Tagged ‘division ring’

Throughout this post, $R$ is a ring with $1.$

Theorem (Jacobson). If $x^n=x$ for some integer $n > 1$ and all $x \in R,$ then $R$ is commutative.

In fact $n,$ in Jacobson’s theorem, doesn’t have to be fixed and could depend on $x,$ i.e. Jacobson’s theorem states that if for every $x \in R$ there exists an integer $n > 1$ such that $x^n=x,$ then $R$ is commutative. But we are not going to discuss that here.
In this post, we’re going to prove Jacobson’s theorem. Note that we have already proved the theorem for $n=3, 4$ (see here and here) and we didn’t need $R$ to have $1,$ we didn’t need that much ring theory either. But to prove the theorem for any $n > 1,$ we need a little bit more ring theory.

Lemma. If Jacobson’s theorem holds for division rings, then it holds for all rings with $1.$

Proof. Let $R$ be a ring with $1$ such that $x^n=x$ for some integer $n > 1$ and all $x \in R.$ Then clearly $R$ is reduced, i.e. $R$ has no non-zero nilpotent element. Let $\{P_i: \ i \in I\}$ be the set of minimal prime ideals of $R.$
By the structure theorem for reduced rings, $R$ is a subring of the ring $\prod_{i\in I}D_i,$ where $D_i=R/P_i$ is a domain. Clearly $x^n=x$ for all $x \in D_i$ and all $i \in I.$ But then, since each $D_i$ is a domain, we get $x=0$ or $x^{n-1}=1,$ i.e. each $D_i$ is a division ring. Therefore, by our hypothesis, each $D_i$ is commutative and hence $R,$ which is a subring of $\prod_{i\in I}D_i,$ is commutative too. $\Box$

Example. Show that if $x^5=x$ for all $x \in R,$ then $R$ is commutative.

Solution. By the lemma, we may assume that $R$ is a division ring.
Then $0=x^5-x=x(x-1)(x+1)(x^2+1)$ gives $x=0,1,-1$ or $x^2=-1.$ Suppose that $R$ is not commutative and choose a non-central element $x \in R.$ Then $x+1,x-1$ are also non-central and so $x^2=(x+1)^2=(x-1)^2=-1$ which gives $1=0,$ contradiction! $\Box$

Remark 1. Let $D$ be a division ring with the center $F.$ If there exist an integer $n \ge 1$ and $a_i \in F$ such that $x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0=0$ for all $x \in D,$ then $F$ is a finite field. This is obvious because the polynomial $x^n+a_{n-1}x^{n-1}+ \cdots + a_1x+a_0 \in F[x]$ has only a finite number of roots in $F$ and we have assumed that every element of $F$ is a root of that polynomial.

Remark 2. Let $D$ be a domain and suppose that $D$ is algebraic over some central subfield $F.$ Then $D$ is a division ring and if $0 \ne d \in D,$ then $F[d]$ is a finite dimensional division $F$-algebra.

Proof. Let $0 \ne d \in D.$ So $d^m +a_{m-1}d^{m-1}+ \cdots + a_1d+a_0=0$ for some integer $m \ge 1$ and $a_i \in F.$ We may assume that $a_0 \ne 0.$ Then $d(d^{m-1} + a_{m-1}d^{m-2}+ \cdots + a_1)(-a_0^{-1})=1$ and so $d$ is invertible, i.e. $D$ is a division ring.
Since $F[d]$ is a subring of $D,$ it is a domain and algebraic over $F$ and so it is a division ring by what we just proved. Also, since $d^m \in \sum_{i=0}^{m-1} Fd^i$ for some integer $m \ge 1,$ we have $F[d]=\sum_{i=0}^{m-1} Fd^i$ and so $\dim_F F[d] \le m. \ \Box$

Proof of the Theorem. By the above lemma, we may assume that $R$ is a division ring.
Let $F$ be the center of $R.$ By Remark 1, $F$ is finite. Since $R$ is a division ring, it is left primitive. Since every element of $R$ is a root of the non-zero polynomial $x^n-x \in F[x], \ R$ is a polynomial identity ring.
Hence, by the Kaplansky-Amtsur theorem, $\dim_F R < \infty$ and so $R$ is finite because $F$ is finite. Thus, by the Wedderburn’s little theorem, $R$ is a field. $\Box$

## 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 $r \in R.$

## Rings with a finite number of non-units

Posted: September 17, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem. Let $R$ be an infinite ring with 1 and let $U(R)$ be the set of units of $R.$ Prove that if $R \setminus U(R)$ is finite, then $R$ is a division ring.

Solution. Suppose, to the contrary, that there exists some $0 \neq x \in R \setminus U(R).$ First note that if $I \neq R$ is a left or right ideal of $R,$ then $I$ is finite because otherwise $I \cap U(R) \neq \emptyset$ and so $I=R.$ Therefore $Rx$ and $xR$ cannot both be infinite. Suppose that $Rx$ is finite and let $I=\{r \in R: \ rx = 0 \}.$ Then $I$ is a left ideal of $R$ and $I \neq R$ because $x \neq 0.$ Hence $I$ is finite. Now the $R$-module isomorphism $R/I \cong Rx$ implies that $R$ is finite. Contradiction!

## Primitive rings; definition & examples

Posted: December 17, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Let $R$ be a ring and let $M$ be a left $R$-module. Recall the following definitions:

1) $M$ is called faithful if $rM=(0)$ implies $r=0$ for any $r \in R.$ In other words, $M$ is called faithful if $\text{ann}_RM = \{r \in R : \ rM=(0) \}=(0).$
2) $M$ is called simple if $(0)$ and $M$ are the only left $R$ submodules of $M.$

Faithful and simple right $R$-modules are defined analogously.

Definition. A ring $R$ is called left (resp., right) primitive if there exists a left (resp., right) $R$-module $M$ which is both faithful and simple.

Remark. We will show later that left and right primitivity are not equivalent.  From now on, I will only consider left primitive rings. If a statement is true for left but not for right, I will mention that.

Example 1. If $R$ is a ring and $M$ is a left simple $R$-module, then $R_1=R/\text{ann}_RM$ is a left primitive ring.  This is clear because $M$ would be a faithful simple left $R_1$-module.

Example 2. Every simple ring is left primitive. That’s because we can choose a maximal left ideal $\mathbf{m}$ of $R$ and then $M=R/\mathbf{m}$ would be a faithful simple left $R$-module. The reason that $M$ is faithful is that $\text{ann}_R M$ is a two-sided ideal contained in $\mathbf{m}$ and therefore $\text{ann}_R M = (0),$ because $R$ is simple. One special case of this example is $M_n(D),$ the ring of $n \times n$ matrices with entries from a division ring $D.$ If $V$ is an infinite dimensional vector space over a field $F$, then $\text{End}_F V$ is an example of a left primitive ring which is not simple [see Example 4 and this post].

Example 3. If $R$ is a left primitive ring and $0 \neq e \in R$ an idempotent, then $R_1=eRe$ is left primitive: let $M$ be a faithful simple $R$-module. The claim is that $M_1=eM$ is a faithful simple left $R_1$-module. Note that $M_1 \neq (0)$ because $e \neq 0$ and $M$ is faithful. Clearly $M_1$ is a left $R_1$-module.  To see why it’s faithful, let $r_1 = ere \in R_1$ with $r_1M_1=(0).$ Then $(0)=ere^2M=ereM=r_1M.$ So $r_1=0,$ because $M$ is faithful. To prove that $M_1$ is a simple $R_1$-module let $0 \neq x_1 \in M_1.$ We need to show that $R_1x_1=M_1.$ Well, since $x_1 = ex,$ for some $x \in M,$ we have $ex_1=ex=x_1.$ Thus $R_1x_1=eRex_1=eRx_1=eM=M_1.$

Example 4. Let $D$ be a division ring and let $M$ be a right vector space over $D.$ Then $R=\text{End}_D M$ is a left primitive ring. Here is why: $M$ is a left $R$-module if we define $fx = f(x),$ for all $f \in R$ and $x \in M.$ It is clear that $M$ is faithful as a left $R$-module. To see why it is simple, let $x,y \in M$ with $x \neq 0.$ Let $B=\{x_i: \ i \in I \}$ be a basis for $M$ over $D$ such that $x=x_k \in B,$ for some $k \in I.$ Define $f \in R$ by $f(\sum_{i \in I}x_id_i) = yd_k.$ Then $f(x)=f(x_k)=y.$ So, we’ve proved that $Rx = M,$ which shows that $M$ is a simple $R$-module. If $\dim_D M = n < \infty,$ then $R \cong M_n(D),$ which we already showed its primitivity in Example 2. Note that if $M$ was a “left” vector space over $D$ with $\dim_D M = n < \infty,$ then $R$ would be isomorphic to the ring $M_n(D^{op})$ rather than the ring $M_n(D).$