Archive for the ‘Primitive Rings’ Category

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

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

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

We proved part 1) of the theorem in the previous section. So we just need to prove 2). 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 2). 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 1) 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

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 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 the coefficient of at least one of the monomials of the highest degree in f is 1 and f(a_1, \cdots , a_n)=0, for all a_j \in A. Then 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). Then for any a_1,a_2 \in A we have \text{tr}(a_1a_2 - a_2a_1)=0. Thus 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 any element of A. So we’ve proved that A is PI because it 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

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).

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.

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

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.

Fact 1. Let R be a left primitive ring and M a faithful simple left R module. By Schur’s lemma D=\text{End}_R(M) is a division ring and M can be considered as a right vector space over D in the usual way. Let S=\text{End}_D(M) and define \varphi : R \longrightarrow S by \varphi(r)(x)=rx, for all r \in R and x \in M. Then \varphi is a well-defined ring homomorphism. Also \varphi is one-to-one because M is faithful. So R can be viewed as a subring of S.

Fact 2. Every left primitive ring R is prime. To see this, suppose M is a faithful simple left R module and I,J be two non-zero ideals of R with IJ=(0).  Now JM is a submodule of M and M is simple. Therefore either JM=(0) or JM=M. If JM=(0), then we get (0) \neq J \subseteq \text{ann}_R M = (0), which is nonsense. Finally, if JM=M, then we will have (0)=(IJ)M=I(JM)=IM. Thus I \subseteq \text{ann}_R M = (0) and so I=(0), a contradiction!

Fact 3. A trivial result of Fact 2 is that the center of a left primitive ring is a commutative domain. A non-trivial fact is that every commutative domain is the center of some left primitive ring. For a proof of this see: T. Y. Lam,  A First Course in Noncommutative Ring Theory, page 195.

Fact 4. Let R be a prime ring and M a faithful left R module of finite length. Then R is left primitive. To see this, let (0)=M_0 \subset M_1 \subset \cdots \subset M_n=M be a composition series of M. Therefore M_k/M_{k-1} is a simple left R module for every 1 \leq k \leq n. We also let I_k=\text{ann}_R (M_k/M_{k-1}). Then each I_k is an ideal of R and it’s easy to see that I_1I_2 \cdots I_nM = (0). Thus I_1I_2 \cdots I_n = (0), because M is faithful. Hence I_{\ell} = (0), for some \ell, because R is prime. Therefore M_{\ell}/M_{\ell - 1} is a faithful simple left R module.

Fact 5. Every left primitive ring R is semiprimitive. This is easy to see: let M be a faithful simple left R module and J=J(R), as usual, be the Jacobson radical of R. The claim is that J=(0). So suppose that J \neq \{0\}  and choose 0 \neq x \in M. Then Rx=M, because M is simple, and so JM=Jx. Also either JM=(0), which is impossible because then J \subseteq \text{ann}_R M=(0), or JM=M. If Jx=JM=M, then ax =x, for some a \in J. Thus (1-a)x=0, which gives us the contradiction x = 0, because 1-a is invertible in R.