Posts Tagged ‘Wedderburn’s little theorem’

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. $\Box$

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. $\Box$

For notations and the results we have already proved, see part (1).

Lemma 3.  Let $q \geq 1, \ n \geq 2$ be integers. Then $|\Phi_n(q)| > q-1.$

Proof. By definition of $\Phi_n,$ it suffices to prove that $|q - \alpha| > q- 1$ for any $n$-th root of unity $\alpha \neq 1.$ So if $\alpha = \cos \theta + i \sin \theta,$ then $\cos \theta < 1$ and thus

$|q - \alpha|^2=q^2-(2\cos \theta)q + 1 > (q-1)^2 \geq q-1. \ \Box$

Wedderburn’s Little Theorem. (J. M. Wedderburn, 1905) Every finite division ring is a field.

Proof. Let $D$ be a finite division ring with the center $Z.$ Then $Z$ is a (finite) field and $D$ is a finite dimensional vector space over $Z.$ So if $|Z|=q$ and $\dim_Z D=n,$ then $|D|=q^n.$ If $n = 1,$ then $D=Z$ and we are done. So we will assume that $n \geq 2$ and we will get a contradiction. Let $D^{\times}=D \setminus \{0\},$ as usual, be the multiplicative group of $D.$ Clearly $Z^{\times}=Z \setminus \{0\}$ is the center of $D^{\times}.$ Also, for any $a \in D,$ let $C(a)$ be the centralizer of $a$ in $D.$ Then $C(a)$ is also a finite division ring and thus a finite dimensional vector space over $Z.$ Let $\dim_Z C(a)=n_a.$ Then $|C(a)|=q^{n_a}.$ It is clear that the centralizer of $a$ in $D^{\times}$ is $C(a)^{\times}=C(a) \setminus \{0\}.$ So the class equation of $D^{\times}$ gives us

$\displaystyle q^n-1=|D^{\times}|=|Z^{\times}| + \sum_{a}[D^{\times}:C(a)^{\times}]=q-1 + \sum_{a} \frac{q^n-1}{q^{n_a}-1}. \ \ \ \ \ \ (\dagger)$

By Lemma 1 and Lemma 2 in part (1), $|\Phi_n(q)|$ divides both $q^n-1$ and $\sum_a \displaystyle \frac{q^n-1}{q^{n_a}-1}.$ So $|\Phi_n(q)| \mid q-1,$ by $(\dagger ).$ Thus $|\Phi_n(q)| \leq q-1,$ contradicting Lemma 3. $\Box$

Wedderburn’s little theorem (1)

Posted: September 24, 2011 in Division Rings, Noncommutative Ring Theory Notes
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In this two-part note I’m going to prove that every finite division ring is a field. This result is called Wedderburn’s little theorem. The proof we are going to give is due to Ernst Witt and that’s probably the best proof available. But before getting into the proof, we need to know a little bit about cyclotomic polynomials.

Notation. For any integer $n \geq 1$ we have the $n$-th root of unity $\zeta_n = e^{2 \pi i/n}.$

Definition. The $n$-th cyclotomic polynomial is defined by $\Phi_n(x) = \prod_{1 \leq k \leq n, \ \gcd(k,n)=1}(x - \zeta_n^k), \ n \geq 1.$

Lemma 1. $x^n - 1 = \prod_{d \mid n} \Phi_d(x).$ In particular, $\Phi_n(x) \mid x^n -1.$

Proof. We have $x^n-1 = \prod_{j=1}^n (x - \zeta_n^j)=\prod_{d \mid n} \prod_{\gcd(j,n)=d}(x-\zeta_n^j).$ But

$\prod_{\gcd(j,n)=d} (x - \zeta_n^j) = \prod_{\gcd(k, n/d)=1}(x - \zeta_n^{kd})$

and obviously $\zeta_n^d = \zeta_{n/d}.$ Hence

$x^n-1 = \prod_{d \mid n} \prod_{\gcd(k,n/d)=1}(x - \zeta_{n/d}^k)=\prod_{d \mid n} \Phi_{n/d}(x)=\prod_{d \mid n} \Phi_d(x). \ \Box$

Corollary. For every $n \geq 1 : \ \Phi_n(x) \in \mathbb{Z}[x].$

Proof. By induction over $n.$ There is nothing to prove if $n=1$ because $\Phi_1(x)=x-1.$ Now let $n \geq 2$ and suppose the $\Phi_m(x) \in \mathbb{Z}[x]$ for all $m < n.$ Note that cyclotomic polynomials are all monic. Thus, by Lemma 1

$x^n-1 = g(x) \Phi_n(x), \ \ \ \ \ \ \ (*)$

for some monic polynomial $g(x) \in \mathbb{Z}[x].$ Since $x^n-1$ is monic too, it follows from $(*)$ that $\Phi_n(x) \in \mathbb{Z}[x]. \ \Box$

Lemma 2. If $1 \leq d < n$ and $d \mid n,$ then $\displaystyle \Phi_n(x) \mid \frac{x^n - 1}{x^d - 1}.$

Proof. By Lemma 1 we have

$x^n - 1 = \prod_{m \mid n} \Phi_m(x)=\Phi_n(x) \prod_{m

Therefore, again by Lemma 1, $x^n-1 = \Phi_n(x) (x^d -1) f(x),$ where $f(x) =\prod_{m \nmid d, \ m \mid n} \Phi_m(x). \ \Box$

We will continue our discussion in part (2).

Finite subgroups of the multiplicative group of a division ring

Posted: September 20, 2011 in Division Rings, Noncommutative Ring Theory Notes
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It is a well-known fact that a finite subgroup of the multiplicative group of a field is cyclic. We will prove this result shortly.  We will also extend it to any division ring of non-zero characteristic. Note that this result is not necessarily true in a division ring of zero characteristic. For example, in the division ring of real quaternions, the subgroup $\{\pm 1, \pm i, \pm j, \pm k \}$ is not even abelian let alone cyclic. We will also prove that finite abelian subgroups of the multiplicative group of any division ring are cyclic.

Theorem. Every finite subgroup of the multiplicative group of a field is cyclic.

Proof. Let $F$ be a field and let $G$ be a finite subgroup of $F^{\times}.$ Let $|G|=n$ and, for any divisor $d$ of $n,$ let $f(d)$ be the number of elements of $G$ of order $d.$ Obviously

$\sum_{d \mid n} f(d) = n. \ \ \ \ \ \ (1)$

Let $\varphi$ be the Euler’s totient function. Recall from number theory that

$\sum_{d \mid n} \varphi(d)=n. \ \ \ \ \ \ (2)$

Claim. If $d \mid n$ and $f(d) \neq 0,$ then $f(d)=\varphi(d).$

Proof of the claim. Since $f(d) \neq 0,$ there exists $g \in G$ such that $o(g)=d.$ Let $H = \langle g \rangle$ and $p(x)=x^d - 1.$ Then every element of $H$ is a root of $p(x).$ But $p(x)$ has at most $d$ roots in $F.$ Thus $H$ is exactly the set of roots of $p(x).$ Finally, the fact that an element $g^m \in H$ has order $d$ if and only if $\gcd(m,d)=1$ implies $f(d)=\varphi(d). \ \Box$

It now follows from $(1), \ (2)$ and the claim that $f(d)=\varphi(d)$ for all $d \mid n.$ In particular, $f(n)=\varphi(n) \neq 0$ and so $o(g)= n = |G|$ for some $g \in G. \ \Box$

Corollary 1. Every finite abelian subgroup of the multiplicative group of a division ring is cyclic.

Proof. Let $D$ be a division ring with the center $Z.$ Let $G$ be a finite abelian subgroup of $D^{\times}$ and put $F=\sum_{g \in G} Zg.$ It is obvious that $F$ is a commutative domain and $G \subset F.$ Also,  since $G$ is finite, $F$ is a finite dimensional vector space over $Z$ and thus every element of $F$ is algebraic over $Z.$ Let $0 \neq c \in F$ and suppose that $q(x)=x^m + \ldots + a_1x + a_0 \in Z[x]$ is the minimal polynomial of $c$ over $Z.$ Then $a_0 \neq 0$ and so $c(c^{m-1} + \ldots + a_1)(-a_0^{-1})=1.$ Therefore $F$ is a field and we are done by the above theorem. $\Box$

Corollary 2.  Every finite subgroup of the multiplicative group of a division ring of non-zero characteristic is cyclic.

Proof. Let $D$ be a division ring with $\text{char}(D)=p > 0.$ Let $\mathbb{F}_p$ be the prime subfield of $D$ and suppose that $G$ is a finite subgroup of $D^{\times}.$ Let $F = \sum_{g \in G} \mathbb{F}_p g.$ Clearly $F$ is a finite subring of $D$ and $F$ contains $G.$ Let $0 \neq c \in F.$ Since $D$ is a domain, $F$ is a domain too. Thus $\{cx : \ x \in F\}=F$ and so $cx = 1$ for some $x \in F.$ Therefore $F$ is a division ring. But, by Wedderburn’s little theorem, a finite division ring is a field. So $F$ is a field and we are done by the above theorem. $\Box$