Posts Tagged ‘Artin-Wedderburn theorem’

Throughout this post, U(R) and J(R) are the group of units and the Jacobson radical of a ring R. Assuming that U(R) is finite and |U(R)| is odd, we will show that |U(R)|=\prod_{i=1}^k (2^{n_i}-1) for some positive integers k, n_1, \ldots , n_k. Let’s start with a nice little problem.

Problem 1. Prove that if U(R) is finite, then J(R) is finite too and |U(R)|=|J(R)||U(R/J(R)|.

Solution. Let J:=J(R) and define the map f: U(R) \to U(R/J)) by f(x) = x + J, \ x \in U(R). This map is clearly a well-defined group homomorphism. To prove that f is surjective, suppose that x + J \in U(R/J). Then 1-xy \in J, for some y \in R, and hence xy = 1-(1-xy) \in U(R) implying that x \in U(R). So f is surjective and thus U(R)/\ker f \cong U(R/J).
Now, \ker f = \{1-x : \ \ x \in J \} is a subgroup of U(R) and |\ker f|=|J|. Thus J is finite and |U(R)|=|\ker f||U(R/J)|=|J||U(R/J)|. \Box

Problem 2. Let p be a prime number and suppose that U(R) is finite and pR=(0). Prove that if p \nmid |U(R)|, then J(R)=(0).

Solution. Suppose that J(R) \neq (0) and 0 \neq x \in J(R). Then, considering J(R) as an additive group, H:=\{ix: \ 0 \leq i \leq p-1 \} is a subgroup of J(R) and so p=|H| \mid |J(R)|. But then p \mid |U(R)|, by Problem 1, and that’s a contradiction! \Box

There is also a direct, and maybe easier, way to solve Problem 2: suppose that there exists 0 \neq x \in J(R). On U(R), define the relation \sim as follows: y \sim z if and only if y-z = nx for some integer n. Then \sim is an equivalence relation and the equivalence class of y \in U(R) is [y]=\{y+ix: \ 0 \leq i \leq p-1 \}. Note that [y] \subseteq U(R) because x \in J(R) and y \in U(R). So if k is the number of equivalence classes, then |U(R)|=k|[y]|=kp, contradiction!

Problem 3. Prove that if F is a finite field, then |U(M_n(F))|=\prod_{i=1}^n(|F|^n - |F|^{i-1}). In particular, if |U(M_n(F))| is odd,  then n=1 and |F| is a power of 2.

Solution. The group U(M_n(F))= \text{GL}(n,F) is isomorphic to the group of invertible linear maps F^n \to F^n. Also, there is a one-to-one correspondence between the set of invertible linear maps F^n \to F^n and the set of (ordered) bases of F^n. So |U(M_n(F))| is equal to the number of bases of F^n. Now, to construct a basis for F^n, we choose any non-zero element v_1 \in F^n. There are |F|^n-1 different ways to choose v_1. Now, to choose v_2, we need to make sure that v_1,v_2 are not linearly dependent, i.e. v_2 \notin Fv_1 \cong F. So there are |F|^n-|F| possible ways to choose v_2. Again, we need to choose v_3 somehow that v_1,v_2,v_3 are not linearly dependent, i.e. v_3 \notin Fv_1+Fv_2 \cong F^2. So there are |F|^n-|F|^2 possible ways to choose v_3. If we continue this process, we will get the formula given in the problem. \Box

Problem 4. Suppose that U(R) is finite and |U(R)| is odd. Prove that |U(R)|=\prod_{i=1}^k (2^{n_i}-1) for some positive integers k, n_1, \ldots , n_k.

Solution. If 1 \neq -1 in R, then \{1,-1\} would be a subgroup of order 2 in U(R) and this is not possible because |U(R)| is odd. So 1=-1. Hence 2R=(0) and \mathbb{Z}/2\mathbb{Z} \cong \{0,1\} \subseteq R. Let S be the ring generated by \{0,1\} and U(R). Obviously S is finite, 2S=(0) and U(S)=U(R). We also have J(S)=(0), by Problem 2. So S is a finite semisimple ring and hence S \cong \prod_{i=1}^k M_{m_i}(F_i) for some positive integers k, m_1, \ldots , m_k and some finite fields F_1, \ldots , F_k, by the Artin-Wedderburn theorem and Wedderburn’s little theorem. Therefore |U(R)|=|U(S)|=\prod_{i=1}^k |U(M_{m_i}(F_i))|. The result now follows from the second part of Problem 3. \Box


Throughout k is a field, K is the algebraic closure of k and A is a finite dimensional central simple k-algebra.

Lemma. A \otimes_k K \cong M_n(K), for some integer n.

Proof. Let S:=A \otimes_k K. By the first part of the corollary in this post we know that S is simple. We also have

Z(S) = Z(A) \otimes_k K = k \otimes_k K \cong K.

It is easy to see that if \{a_i \} is a k-basis for A, then \{a_i \otimes_k 1 \} is an K-basis for S. Thus \dim_K S = \dim_k A. So S is a finite dimensional central simple K-algebra and hence, since K is algebraically closed, S \cong M_n(K), for some n, by Remark 2 in this post. \Box

Theorem. If A is a finite dimensional central simple k-algebra, then \dim_k A is a perfect square.

Proof.  By the lemma, there exists an integer n such that A \otimes_k K \cong M_n(K). Thus

\dim_k A = \dim_K A \otimes_k K = \dim_K M_n(K) = n^2. \ \Box

Definition. The degree of A is defined by \deg A = \sqrt{\dim_k A}.

Remark. Let R be a finite dimensional k-algebra. Then R is reduced if and only if R is a finite direct product of finite dimensional division k-algebras. In this case, \dim_k R=n^2\dim_k Z(R) for some integer n \geq 1.

Proof. obviously R is (left) Artinian because \dim_k R < \infty and so J(R) is nilpotent. Thus J(R)=(0) because R is reduced and so R is semisimple. The result now follows from the Artin-Wedderburn theorem. The converse is trivial. Finally, the fact that, by the above theorem, \dim_{Z(D)}D is a perfect square for any finite dimensional division algebra D, proves the last part of the remark.  \Box