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.  $\Box$