## Archive for the ‘Primitive Rings’ Category

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 I’m not going to prove 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. $\Box$

## Kaplansky-Amitsur theorem (2)

Posted: December 24, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Theorem. (Kaplansky-Amitsur) Let $A$ be a left primitive algebra, $M$ a faithful simple $A$ module and $D=\text{End}_A(M).$ If  $A$ satisfies a polynomial of degree $d,$ then

i) $A \cong M_n(D),$ where $n = \dim_D M \leq [d/2].$ So $Z:=Z(A) \cong Z(D)$ is a field.

ii) $\dim_Z A \leq [d/2]^2.$

We proved i) in the previous section. So we just need to prove ii). First two easy remarks.

Remark 1. Let $F$ be a field. Then $M_m(A) \otimes_F M_n(B) \cong M_{mn}(A \otimes_F B),$ for any $F$ algebras $A,B.$

Remark 2. Let $C$ be a commutative ring, $A$ a PI $C$-algebra and $K$ a commutative $C$-algebra. If $A$ satisfies a multi-linear polynomial $f,$ then $A \otimes_C K$ will also satisfy $f.$

Proof of ii). So $A$ satisfies some multi-linear polynomial $f$ of degree at most $d.$ Clearly $D$ satisfies $f$ too because it’s a subring of $M_n(D) \cong A.$ Let $K$ be a maximal subfield of $D$ and $R=D \otimes_Z K.$ By Remark 2, $R$ also satisfies $f.$ But, by Azumaya theorem, $R$ is left primitive, $D$ is a faithful simple left $R$-module and $\text{End}_R(D) \cong K.$ Thus, by part i) of the theorem, $R \cong M_m(K),$ for some positive integer $m.$  Therefore by Remark 1

$A \otimes_Z K \cong M_n(D) \otimes_Z K \cong M_n(R) \cong M_{mn}(K).$

Hence $\dim_Z A = \dim_K A \otimes_Z K = (mn)^2.$ On the other hand, by Remark 2, $A \otimes_Z K$ satisfies $f$ and so $d \geq \deg f \geq 2mn.$ Therefore $mn \leq d/2$ and so $mn \leq [d/2].$ Finally we have

$\dim_Z A =(mn)^2 \leq [d/2]^2. \ \Box$

## Kaplansky-Amitsur theorem (1)

Posted: December 23, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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We will assume that $C$ is a commutative ring with 1. Also, given an integer $n \geq 1,$  we will denote by $C \langle x_1, \cdots , x_n \rangle$ the ring of polynomials in non-commuting variables $x_1, \cdots , x_n$ and with coefficients in $C.$

Definition. A $C$-algebra $A$ is called a polynomial identity algebra (or a PI-algebra) if there exists an integer $n \geq 1$ and a non-zero polynomial

$f \in C \langle x_1, \cdots , x_n \rangle$

such that $f(a_1, \cdots , a_n)=0$ for all $a_j \in A.$ The smallest possible degree of such $f$ is called the PI-degree of $A.$

Remark 1. If an algebra $A$ satisfies a polynomial $f,$ then obviously every subring of $A$ and homomorphic image of $A$ will satisfy $f$ too.

Remark 2. It is known and easy to prove that if an algebra $A$ satisfies a non-zero polynomial of degree $n,$ then $A$ will also satisfy a non-zero multi-linear polynomial of degree at most $n.$

Example 1. Every commutative algebra is PI because it satisfies the polynomial $x_1x_2 - x_2x_1.$

Example 2. Let $C$ be a commutative ring and $A=M_2(C).$ We show that $A$ is a PI-algebra. Let $a_1,a_2 \in A.$ Then $\text{tr}(a_1a_2 - a_2a_1)=0$ and so, by Cayley-Hamilton theorem, $(a_1a_2 - a_2a_1)^2 = cI_2,$ for some $c \in C.$ Therefore $(a_1a_2-a_2a_1)^2$ commutes with every element of $A$ and so $A$ satisfies the polynomial

$f=(x_1x_2-x_2x_1)^2x_3 - x_3(x_1x_2 - x_2x_1)^2.$

Remark 2. Let $C$ be a commutative ring and suppose that $A=M_n(C)$ satisfies a non-zero polynomial of degree $\leq 2n-1.$ Then, by Remark 1, $A$ will also satisfy a non-zero multi-linear polynomial

$f(x_1, \cdots , x_k) = \sum_{\alpha \in S_k} c_{\alpha} x_{\alpha(1)} \cdots x_{\alpha (k)},$

for some $k \leq 2n-1$ and $c_{\alpha} \in C.$ Renaming the variables, if necessary, we may assume that $c=c_{\text{id}} \neq 0.$ Then $f(e_{11},e_{12},e_{22},e_{23}, \cdots )=c e_{1 \ell}=0,$ for some $\ell.$ Therefore $c e_{ij}=e_{i1}(c e_{1 \ell}) e_{\ell j} = 0,$ for all $i,j.$ Thus $c = 0,$ which is a contradiction. So $A$ does not satisfy any non-zero polynomial of degree $\leq 2n-1.$

Theorem. Let $A$  be an algebra. Suppose $A$ is left primitive, $M$ a faithful simple left $A$ module and $D=\text{End}_A(M).$ If $A$ satisfies a polynomial $f$ of degree $d,$ then $\dim_D M=n \leq [d/2]$ and $A \cong M_n(D).$

Proof. Suppose $\dim_D M > [d/2].$ Then there exists some $k > [d/2]$ such that either $A \cong M_k (D)$ or $M_k(D)$ is a homomorphic image of some subring of $A.$ In either case, by remark 1, $M_k (D)$ and hence $M_k(Z(D))$ satisfies $f$. Thus by remark 2: $d \geq 2k \geq 2([d/2] + 1) > d$, which is nonsense. $\Box$

## A theorem of Azumaya

Posted: December 22, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Theorem. Let $D$ be a division algebra with the center $Z$ and suppose that $K$ is a maximal subfield of $D.$ Let $R=D \otimes_Z K.$ Then

$1) \ R$ is a simple ring and thus left primitive. In fact:

$2) \ D$ is a faithful simple left $R$ module and $\text{End}_R(D) \cong K.$

Proof. 1) This part follows from the first part of this corollary but we will give a self-contained proof: let $\{ k_j : \ j \in J \}$ be a $Z$ basis for $K.$ Suppose that $R$ is not simple. Let $(0) \neq I$ be an ideal of $R.$ Choose $n$ to be the smallest integer for which there exists $0 \neq x=\sum_{j=1}^n d_j \otimes k_{t_j} \in I.$ Then $d_1 \neq 0$ and therefore, by replacing $x$ with $(d_1^{-1} \otimes 1)x$ if necessary, we may assume that $d_1 = 1.$ Now, for any $d \in D$ we have

$\sum_{j=2}^n (dd_j-d_jd) \otimes k_{t_j} = (d \otimes 1)x - x(d \otimes 1) \in I,$

which gives us $dd_j=d_jd,$ by minimality of $n.$ So $d_j \in Z,$ for all $j.$ Thus $x = 1 \otimes k$, for some $0 \neq k \in K.$  But then

$1_R = 1 \otimes 1 = (1 \otimes k^{-1})x \in I$

and so $I=R.$

2) Defining $(d_1 \otimes k)d_2=d_1d_2k,$ for all $d_1,d_2 \in D, \ k \in K,$ and extending it linearly will make $D$ a left $R$ module. Clearly $D$ is a faithful $R$ module because $\text{ann}_R D \neq R$ is an ideal of $R$ and $R$ is a simple ring by 1).  To prove that $D$ is a simple $R$ module let $d_1 \neq 0$ and $d_2 \in D.$ Then $(d_2d_1^{-1} \otimes 1)d_1=d_2$ and hence $Rd_1=D.$ Finally, $\text{End}_R D \cong K$ is just a special case of this theorem. Just note that, since $K$ is commutative, $K^{op}=K$ and since it is a maximal subfield, $C_D(K)=K. \ \Box$

Remark. By part 2) of the theorem and the density theorem, $R$ is a dense subring of $\text{End}_K(D).$ Also if $\dim_K D = n < \infty,$ then $R \cong \text{End}_K(D) \cong M_n(K).$

## Some applications of density theorem

Posted: December 19, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Remark. If $R$ is a left primitive ring, then, by the density theorem and remark 4 in this post, there exists a division ring $D$ such that  either $R \cong M_n(D),$ for some positive integer $n$ or for any positive integer $n$ there exists a subring $R_n$ of $R$ and an onto ring homomorphism $R_n \longrightarrow M_n(D).$

Example 1. A finite ring $R$ is left primitive if and only if $R \cong M_n(F),$ for some integer $n$ and some finite field $F.$

Proof. The “if” part is trivial because $M_n(F)$ is a simple ring and hence left primitive. Now suppose that $R$ is a finite left primitive ring. By the above remark we’ll have two possibilities.

Case 1. There exists a division ring $D$ such that $R \cong M_n(D).$ Then $D$ is finite and so a field.

Case 2. There exists a division ring $D$ such that for any positive integer $n$ there exists a subring $R_n$ and an onto ring homomorphism $R_n \longrightarrow M_n(D).$ But this case is not possible because $R_n$ is finite for all $n$ but the cardinality of $M_n(D)$ goes to infinity as $n$ goes to infinity.

Example 2. A ring $R$ has the property $(*)$ if for every $x \in R$ there exists an integer $n > 1,$ depending on $x,$ such that $x^{n}=x.$ A left primitive ring $R$ which has the property $(*)$ is a division ring.

Proof. Again, the same as example 1, we’ll consider two cases.

Case 1. $R \cong M_n(D),$ for some integer $n$ and division ring $D.$ But for $n \geq 2$ the ring $M_n(D)$ doesn’t have the property $(*)$ because for all $m \geq 2 : \ e_{12}^m = 0 \neq e_{12}.$ So $n = 1$ and thus $R \cong D.$

Case 2. Suppose that there exists an onto ring homomorphism from a subring $R_2$ of $R$ to $M_2(D).$ Then $R_2/I \cong M_2(D),$ for some ideal $I$ of $R_2.$ Then, since $R$ has the property $(*),$ the ring $R_2/I$ must have the property $(*).$ But we showed in Case 1 that $M_2(D)$ does not satisfy $(*).$ So this case is impossible.

Example 3. Let $R$ be a left primitive ring and suppose that $x(xy-yx)=(xy-yx)x,$ for all $x,y \in R.$ Then $R$ is a division ring.

Proof. The proof is the same as example 2. The only thing we’ll need is to note that for $n \geq 2$ the ring $M_n(D)$ doesn’t satisfy the above property because, for example, if we let $x=e_{11}, \ y=e_{12},$ then $x(xy-yx)=e_{12}$ but $(xy-yx)x=0.$

Example 4. Let $R$ be a left primitive ring such that $1+r^2$ is a unit for all $r \in R.$ Then $R$ is a division ring.

Proof. Let $M$ be a faithful simple left $R$-module and $D=\text{End}_R(M).$ So, by the above remark, we only need to prove that the dimension of $M,$ as a vector space over $D,$ is $1.$ Suppose that the dimension is at least $2$ and let $\{x,y \} \subset M$ be a $D$-linearly independent set. By the density theorem, there exists $r \in R$ such that $rx=y$ and $ry=-x.$ But then $r^2x=ry=-x$ and thus $(1+r^2)x=0,$ which gives us the contradiction $x=0$ because $1+r^2$ is a unit.

Example 5. (Wedderburn-Artin) Let $R$ be a left primitive ring. If $R$ is left Artinian, then $R \cong M_n(D),$ for some division ring $D$ and positive integer $n.$

Proof. Let $M$ be a faithful simple left $R$-module and $D=\text{End}_R(M).$ If $M$ is finite dimensional as a vector space over $D,$ we’re done by the above remark. Otherwise, choose an infinite $D$-linearly independent set $\{x_1,x_2, \cdots \} \subset M.$ For any positive integer $n$ let $I_n = \text{ann}_R \{x_1, \cdots , x_n \}.$ Clearly $I_{n+1} \subseteq I_n$ for all $n.$ Now, by the density theorem, there exists $r \in R$ such that $rx_1= \cdots = rx_n=0$ and $rx_{n+1}=x_1 \neq 0.$ Thus $r \in I_n$ but $r \notin I_{n+1},$ which means $I_n \supset I_{n+1}.$ So we have $I_1 \supset I_2 \supset \cdots ,$ which is a contradiction because $R$ is Artinian.

## Jacobson density theorem

Posted: December 19, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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We showed in the previous section that every dense subring of the ring of linear transformations of a vector space over a division ring is left primitive. Now, we’d like to prove the converse: every left primitive ring is a dense subring of the ring of linear transformations of some vector space over some division ring. We will assume that $R$ is a ring, $M$ is a simple left $R$ module and $D=\text{End}_R(M).$ As usual, $M$ is considered as a right vector space over $D.$

Remark. If $\varphi \in \text{Hom}_R (M^k, M),$ then there exist $d_i \in D$ such that for all $x_i \in M$ we have $\varphi (x_1, \cdots , x_k) = \sum_{i=1}^k x_i d_i.$

Proof. For any $1 \leq i \leq k$ let $\mu_i : M \longrightarrow M^k$ be the $i$-th injection map, i.e. $\mu_i(x)=(0, \cdots , x, \cdots , 0),$ where $x$ is the $i$-th coordinate. Clearly $d_i=\varphi \mu_i \in D.$ Now

$\varphi (x_1, \cdots , x_k)=\sum_{i=1}^k \varphi \mu_i(x_i)=\sum_{i=1}^k x_i d_i.$

Density Theorem. (Chevalley – Jacobson) Let $R$ be a left primitive ring, $M$ a faithful simple left $R$ module and $D=\text{End}_R(M).$ Then $R$ is a dense subring of $S=\text{End}_D(M).$

Proof. We have already showed that $R$ is a subring of $S.$ So we need to prove that for any $D$-linearly independent set $\{x_1, \cdots , x_n \} \subset M$ and any set $\{y_1, \cdots , y_n \} \subset M,$ there exists $f \in R$ such that $f(x_j)=y_j,$ for all $j.$ The proof is by induction over $n$: if $n = 1,$ then, since $M$ is simple and $x_1 \neq 0,$ we have $Rx_1=M$ and thus there exists $f \in R$ such that $fx_1=f(x_1)=y_1.$ Assuming that the result is true for $n-1,$ we will have $R(x_1, \cdots , x_{n-1})=M^{n-1}$  (density condition!). We now prove a claim:

Claim.  There exists $f \in R$ such that $f(x_n) \neq 0$ and $f(x_1) = \cdots = f(x_{n-1})=0.$

Proof of the claim. Suppose to the contrary that the claim is not true. Then $f(x_1)= \cdots f(x_{n-1})=0$ will imply that $f(x_n)=0,$ for any $f \in R.$ Define $\varphi : M^{n-1} \longrightarrow M$ by $\varphi (f(x_1), \cdots , f(x_{n-1}))=f(x_n), \ f \in R.$ See that $\varphi$ is well-defined, that is $\varphi \in \text{Hom}_R (M^{n-1},M).$ Therefore, if we put $f=1_R,$ then by the above remark there must exist $d_i \in D$ such that $x_n = \varphi (x_1, \cdots , x_{n-1}) = \sum_{i=1}^{n-1} x_i d_i,$ which contradicts $D$-linear independence of $x_1, \cdots , x_n. \ \Box$

So, using the above claim,  for any $1 \leq i \leq n,$ we can choose $f _i \in R$ such that $f_i(x_i) \neq 0$ and $f_i (x_j)=0,$ for all $j \neq i.$ Thus, since $M$ is simple, $Rf_i(x_i)=M,$ for all $1 \leq i \leq n.$ Hence there exist $g_i \in R$ such that $g_if_i(x_i)=y_i.$ Let $f = \sum_{i=1}^n g_i f_i.$ Then for any $1 \leq j \leq n$

$f(x_j)=\sum_{i=1}^n g_if_i(x_j)=g_jf_j(x_j)=y_j. \ \Box$

The Structure Theorem For Primitive Rings. Let $R$ be a left primitive ring, $M$ a faithful simple left $R$ module and $D=\text{End}_R(M).$
1) If $\dim_D M = n < \infty,$ then $R \cong M_n(D).$
2) If $\dim_D M = \infty,$ then for any positive integer $n$, there exists a subring $R_n$ of $R$ and an onto ring homomorphism $\varphi : R_n \longrightarrow M_n(D).$

Proof. It follows from the density theorem and remarks 3 and 4 in the previous post. $\Box$

## Dense subrings

Posted: December 18, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Throughout $D$ is a division ring, $M$ a right vector space over $D,$ and  $R$ is a subring of $S=\text{End}_D(M).$ Clearly $M$ can be viewed as a left $R$ module by defining $fx=f(x),$ for all $f \in R$ and $x \in M$.

Definition. $R$ is said to be a dense subring of $S$ if for every $D$-linearly independent set $\{x_1, \cdots, x_n \} \subset M$ and any set $\{y_1, \cdots, y_n \} \subset M,$ there exists $f \in R$ such that $f(x_j)=y_j,$ for all $1 \leq j \leq n.$

Remark 1. If $R$ is a dense subring of $S,$ then $R$ is left primitive.

Proof. Well, $M$ is clearly a faithful left $R$ module. To see why it is simple, let $0 \neq x \in M$ and $y \in M.$ By the density condition, there exists $f \in R$ such that $fx=f(x)=y.$ Thus $Rx=M.$

Remark 2. We proved in example 4 in here that $S$ itself is left primitive. In the above remark we showed that any dense subring of $S$ is also left primitive.

Remark 3. If $R$ is a dense subring of $S$ and $\dim_D M = n < \infty,$ then $R=S \cong M_n(D).$

Proof. Let $\{x_1, \cdots , x_n \}$ be a basis for $M,$ as a vector space over $D,$ and let $g \in S.$ Then, since $R$ is dense, there exists $f \in R$ such that $f(x_j)=g(x_j),$ for all $j.$. But then $g= f \in R$ and hence $S=R.$

Remark 4. If $R$ is a dense subring of $S$ and $\dim_D M = \infty,$ then for any positive integer $n$, there exists a subring $R_n$ of $R$ and an onto ring homomorphism $\varphi : R_n \longrightarrow M_n(D).$

Proof. Let $\{x_1, \cdots , x_n \}$ be a $D$-linearly independent subset of $M$ and let $N=\sum_{j=1}^n x_j D,$ i.e. the $D$-vector subspace of $M$ spanned by $x_1, \cdots , x_n.$ Now define $R_n = \{f \in R: \ f(N) \subseteq N \}.$ Clearly $R_n$ is a subring of $R$. Finally we define the map $\varphi : R_n \longrightarrow \text{End}_D(N) \cong M_n(D)$ by $\varphi(f)(x)=f(x),$ for all $f \in R_n$ and $x \in N.$ It’s easy to see that $\varphi$ is a well-defined ring homomorphism. To prove that $\varphi$ is onto, let $g \in \text{End}_D(N).$ Then, by the density condition, there exists $f \in R$ such that $f(x_j)=g(x_j),$ for all $j.$ Thus $f \in R_n,$ because $g \in \text{End}_D(N),$ and clearly $\varphi(f)=g.$