Archive for the ‘Division 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 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

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Here you can see part (2). We are now ready to prove a nice result.

Theorem. Let k be an algebraically closed field. Let A be a commutative k-domain and let B be a k-domain. Then A \otimes_k B is a k-domain.

Proof. Suppose that A \otimes_k B is not a domain. So there exist non-zero elements u, v \in A \otimes_k B such that uv=0. Let u = \sum_{i=1}^m a_i \otimes_k b_i and v = \sum_{i=1}^n a_i' \otimes_k b_i', where a_i,a_i' \in A, \ b_i,b_i' \in B and both sets \{b_1, \ldots , b_m\} and \{b_1', \ldots , b_n'\} are k-linearly independent. Let C=k[a_1, \ldots , a_m, a_1', \ldots , a_n'], which is a commutative domain because C \subseteq A. Also, note that u,v \in C \otimes_k B \subseteq A \otimes_k B. Now,  since u, v \neq0, there exist integers r,s such that a_r \neq 0 and a_s' \neq 0. Therefore, by the third part of the corollary in part (2), there exists a ring homomorphism \varphi : C \longrightarrow k such that \varphi(a_r) \neq 0 and \varphi(a_s') \neq 0. Let \psi = \varphi \otimes \text{id}_B. Then \psi : C \otimes_k B \longrightarrow k \otimes_k B \cong B is a ring homomorphism and hence \psi(u)\psi(v)=\psi(uv)=0. Therefore either \psi(u)=0 or \psi(v)=0, because B is a domain. But \psi(u)=\sum_{i=1}^m \varphi(a_i) \otimes_k b_i \neq 0, because \{b_1, \ldots , b_m \} is k-linearly independent and \varphi(a_r) \neq 0. Also, \psi(v)=\sum_{i=1}^n \varphi(a_i') \otimes_k b_i' \neq 0, because \{b_1', \ldots , b_n' \} is k-linearly independent and \varphi(a_s') \neq 0. This contradiction proves that A \otimes_k B is  a domain. \Box

Let k be an algebraically closed field. A trivial corollary of the theorem is a well-known result in field theory: if F_1,F_2 are two fields which contain k, then F_1 \otimes_k F_2 is a commutative domain. Another trivial result is this: if k is contained in both a field F and the center of a division algebra D, then F \otimes_k D is a domain.

Question. Let k be an algebraically closed field and let D_1,D_2 be two finite dimensional division k-algebras. Will D_1 \otimes_k D_2 always be a domain?

Answer. No! See the recent paper of Louis Rowen and David Saltman for an example.

Here you can see part (1). We are now going to prove a more interesting result than the one we proved in part (1). But we need to get prepared first. The following important result is known as Zariski’s lemma.

Lemma. (Zariski, 1946) Let k be a field and let A=k[a_1, \ldots , a_n] be a finitely generated commutative algebra. If A is a field, then A is algebraic over k  and thus \dim_k A < \infty.

Proof. The proof is by induction over n. If n=1, then A=k[a_1] and since A is a field, a_1^{-1} \in A. Thus a_1^{-1}=\sum_{i=0}^m \gamma_i a_1^i for some integer m \geq 0 and \gamma_i \in k. Then \sum_{i=0}^m \gamma_i a_1^{i+1}=0 and so a_1, and hence A, is algebraic over k. Now suppose that n \geq 2. If all a_i are algebraic over k, then A=k[a_1, \ldots , a_n] is algebraic over k and we are done. So we may assume that a_1 is transcendental over k. Since A is a field, K =k(a_1) \subseteq A and thus A=K[a_2, \ldots , a_n]. By the induction hypothesis, A is algebraic over K. So every a_i satisfies some monic polynomial f_i(x) \in K[x] of degree m_i. Let v_i \in k[a_1] be the product of the denominators of the coefficients of f_i(x) and put v=\prod_{i=1}^n v_i. Let m be the maximum of m_i. Then multiplying f(a_i)=0 through by v^m shows that each va_i is integral over k[a_1]. Note that since a_1 is transcendental over k, k[a_1] \cong k[x], the polynomial algebra over x. Thus I can choose an irreducible polynomial p(a_1) \in k[a_1] such that

\gcd(p(a_1), v)=1. \ \ \ \ \ \ \ \ \ (*)

Now (p(a_1))^{-1} \in A=k[a_1, \ldots , a_n], because A is a field. Thus for a large enough integer r, we have v^r(p(a_1))^{-1} \in k[va_1, \ldots , va_n] and hence v^r(p(a_1))^{-1} is integral over k[a_1]. But k[a_1] is a UFD and we know that every UFD is integerally closed (in its field of fraction). Therefore v^r(p(a_1))^{-1} \in k[a_1], which is absurd because then p(a_1) \mid v, contradicting (*). \ \Box

Corollary. Let k be a field and let A=k[a_1, \ldots , a_n] be a finitely generated commutative algebra.

1) If \mathfrak{m} is a maximal ideal of A, then \dim_k A/\mathfrak{m} < \infty.

2) If A is a field and k is algebraically closed, then A=k.

3) If A is a domain, k is algebraically closed and b_1, \ldots , b_m \in A \setminus \{0\}, then there exists a ring homomorphism \varphi : A \longrightarrow k such that \varphi(b_i) \neq 0 for all i=1, \ldots , m.

Proof. 1) Let \overline{a_i}=a_i + \mathfrak{m}, \ i=1, \ldots , n. Then A/\mathfrak{m} = k[\overline{a_1}, \ldots , \overline{a_n}] and we are done by the lemma.

2) By the lemma, A is algebraic over k and thus, since k is algebraically closed, A=k.

3) Let S be the set of all monomials in b_1, \ldots, b_m. Clearly S is multiplicatively closed and 0 \notin S because A is a domain. Consider B=S^{-1}A, the localization of A at S. Clearly B=k[a_1, \ldots , a_n, b_1^{-1}, \ldots , b_m^{-1}] and so B is finitely generated. Let \mathfrak{m} be a maximal ideal of B. Note that b_i \notin \mathfrak{m}, for all i, because each b_i is a unit in B. Now, we have B/\mathfrak{m} \cong k, by 2). Let f: B/\mathfrak{m} \longrightarrow k be a ring isomorphism. We also have the natural ring homomorphism g: B \longrightarrow B/\mathfrak{m} and an inejctive ring homomorphism h: A \longrightarrow B=S^{-1}A defined by f(a)=a/1. Let \varphi = fgh. Then \varphi : A \longrightarrow k and \varphi(b_i) \neq 0 for all i, because f is an isomorphism and b_i \notin \mathfrak{m} for all i. \ \Box

See part (3) here.

Definition. Let D be a division algebra with the center k. We denote by [D,D] the additive subgroup of D generated by the set \{ab-ba : \ a,b \in D \}. Also, we denote by D' the subgroup of D^{\times} generated by the set \{aba^{-1}b^{-1}: \ a,b \in D^{\times} \}.

Lemma. Let r, s be integers, a_1 , \ldots , a_r \in D and b_1, \ldots , b_r \in D'. There exists v \in D' such that (b_1a_1b_2a_2 \ldots b_ra_r)^s=(a_1a_2 \ldots a_r)^s v.

Proof. Apply the following repeatedly: if a \in D and b \in D', then ba = ac, where c = b(b^{-1}a^{-1}ba) \in D'. \ \Box

Theorem. Let D be a finite dimensional central division k-algebra of degree n. For every a \in D there exist u \in [D,D] and v \in D' such that \text{Trd}_D(a)=na + u and \text{Nrd}_D(a)=a^nv.

Proof. Let f(x) be the minimal polynomial of a over k and let m = \deg f(x). Let r = \frac{n}{m}. Then, by the theorem in this post, \text{Prd}_D(a,x)=(f(x))^r and by Wedderburn’s factorization theorem there exist non-zero elements c_1, c_2, \ldots , c_m such that

f(x) = (x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_m a c_m^{-1}) .

Let \alpha = \sum_{i=1}^m c_i a c_i^{-1} and \beta = \prod_{i=1}^m c_i a c_i^{-1}. Then

f(x)=x^m - \alpha x^{m-1} + \ldots + (-1)^m \beta

and so

\text{Prd}_D(a,x) = (x^m - \alpha x^{m-1} + \ldots + (-1)^m \beta)^r= x^n - r \alpha x^{n-1} + \ldots + (-1)^n \beta^r.

Therefore \text{Trd}_D(a) = r \alpha and \text{Nrd}_D(a) = \beta^r. Hence

\text{Trd}_D(a)=r \alpha = r \sum_{i=1}^m c_i ac_i^{-1} = na + r\sum_{i=1}^m(c_iac_i^{-1} - ac_i^{-1}c_i).

Let u =r\sum_{i=1}^m (c_iac_i^{-1} - ac_ic_i^{-1}). Then u \in [D,D] and \text{Trd}_D(a)=na+u. Now let b_i=c_iac_i^{-1}a^{-1}. Then the lemma gives us

\text{Nrd}_D(a)=\beta^r = (b_1ab_2a \ldots b_m a)^r = a^n v,

for some v \in D'. \ \Box

You can see part (1) in here.

Wedderburn’s Factorization Theorem. (Wedderburn, 1920) Let D be a division algebra with the center k. Suppose that a \in D is algebraic over k and let f(x) \in k[x] be the minimal polynomial of a over k. There exist non-zero elements c_1, c_2 , \ldots , c_n \in D such that

f(x)=(x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_nac_n^{-1}).

Proof. By Remark 2 in part (1), there exists g(x) \in D[x] such that f(x)=g(x)(x-a). Now let m be the largest integer for which there exist non-zero elements c_1, c_2 , \ldots , c_m \in D and p(x) \in D[x] such that

f(x)=p(x)(x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_m a c_m^{-1}).

Let

h(x) = (x - c_1ac_1^{-1})(x - c_2ac_2^{-1}) \ldots (x - c_mac_m^{-1}).

Claim. h(cac^{-1})=0 for all 0 \neq c \in D.

Proof of the claim. Suppose, to the contrary, that there exists 0 \neq c \in D such that h(cac^{-1}) \neq 0. Then, since f(x)=p(x)h(x) and f(cac^{-1})=0, there exists 0 \neq b \in D such that p(bcac^{-1}b^{-1})=0, by Lemma 1 in part (1). So, by Remark 2 in part (1), there exists q(x) \in D[x] such that p(x)=q(x)(x - bcac^{-1}b^{-1}). Hence f(x)=p(x)h(x)=q(x)(x - bca (bc)^{-1})(x-c_1ac_1^{-1}) \ldots (x - c_m a c_m^{-1}), contradicting the maximality of m. \ \Box

Therefore, by Lemma 2 in part (1), \deg h(x) \geq \deg f(x) and so h(x)=f(x) because f(x)=p(x)h(x) and both f(x) and h(x) are monic. \Box

In the next post, I will use Wedderburn’s factorization theorem to find an expression for the reduced trace and the reduced norm of an element in a finite dimensional central division algebra.

Let R be a ring and let R[x] be the ring of polynomials in the central indeterminate x. We will always write the coefficients of an element of R[x] on the left. Now, if R is commutative and f(x)=g(x)h(x) in R[x], then f(a)=g(a)h(a) for all a \in R. This is not true if R is noncommutative. For example, let g(x)=x-a and h(x)=x - b, where a,b \in R. Then f(x)=g(x)h(x)=x^2-(a+b)x+ab and hence f(a)=ab-ba. Thus if ab \neq ba, then f(a) \neq 0 but g(a)h(a) = (a-a)(a-b)=0.

Remark 1. Let R be a ring. If f(x)=g(x)h(x) in R[x] and if every coefficient of h(x) is in the center of R, then f(a)=g(a)h(a) for all a \in R.

Proof. Let g(x)=\sum_{i=0}^n b_ix^i and h(x)=\sum_{j=0}^m c_jx^j. Then, since a commutes with every c_j, we have

f(a)=\sum_{\ell=0}^{m+n}\sum_{i+j=\ell}b_ic_ja^{\ell}=\sum_{\ell=0}^{m+n}\sum_{i+j=\ell}b_ia^ic_ja^j=g(a)h(a).

Remark 2. Let R be a ring and f(x) \in R[x]. Then f(a)=0 if and only if f(x)=g(x)(x-a) for some g(x) \in R[x].

Proof. Let f(x)=\sum_{i=0}^n a_ix^i. If f(a)=0, then since x-a commutes with a, we have

f(x)=\sum_{i=0}^na_i(x-a +a)^i = \sum_{i=0}^n a_i \sum_{j=0}^i \binom{i}{j}a^{i-j}(x-a)^j =

g(x)(x-a) + f(a)=g(x)(x-a).

Conversely, if f(x)=g(x)(x-a) for some g(x) = \sum_{i=0}^m b_ix^i, then

f(x)=\sum_{i=0}^mb_ix^{i+1} - \sum_{i=0}^m b_iax^i

and thus f(a)=\sum_{i=0}^mb_ia^{i+1} - \sum_{i=0}^m b_ia^{i+1}=0. \ \Box

Lemma 1. Let D be a division algebra with the center k. Suppose that f(x)=g(x)h(x) in D[x]. If a \in D and b = h(a) \neq 0, then f(a) = g(bab^{-1})b. Thus if f(a)=0, then g(bab^{-1})=0.

Proof. Let g(x)=\sum_{i=0}^n b_ix^i and h(x)=\sum_{j=0}^m c_jx^j. Then

f(a)=\sum_{i,j} b_i c_j a^{i+j} = \sum_{i=0}^n b_i \left (\sum_{j=0}^m c_ja^j \right)a^i=\sum_{i=0}^n b_i h(a)a^i = \sum_{i=0}^n b_iba^i =

\sum_{i=0}^n b_i(bab^{-1})^ib=g(bab^{-1})b. \ \Box

Lemma 2. Let D be a division algebra with the center k. Suppose that a \in D is algebraic over k and f(x) \in k[x] is the minimal polynomial of a over k. If g(x) \in D[x] is non-zero and g(cac^{-1}) = 0, for all 0 \neq c \in D, then \deg g(x) \geq \deg f(x).

Proof. Let \deg f(x) = n and suppose that the lemma is false. Let m \geq 1 be the smallest integer for which there exists a non-zero polynomial g(x) \in D[x] of degree m such that m < n and g(cac^{-1})=0 for all 0 \neq c \in D. Note that since for every 0 \neq u \in D, the polynomial ug(x) has degree m and ug(cac^{-1})=0, for all 0 \neq c \in D, we may assume that g(x) is monic. Let g(x)=x^m + b_{m-1}x^{m-1} + \ldots + b_1x + b_0. To get a contradiction, we are going to find a non-zero polynomial h(x) \in D[x] such that \deg h(x) < m and h(cac^{-1})=0 for all 0 \neq c \in D. We first note that, since f(x) is the minimal polynomial of a over k, \ m < n and g(a)=0, we have g(x) \notin k[x]. So there exists 0 \leq j \leq m-1 such that b_j \notin k. Let b \in D be such that b_jb - bb_j \neq 0. So b \neq 0. Now we let

h(x) = g(x)b - bg(x).

Then h(x)= \sum_{i=0}^{m-1}(b_ib - bb_i)x^i. Hence \deg h(x) < m and h(x) \neq 0, because b_jb - bb_j \neq 0. Let 0 \neq c \in D. Then, by Lemma 1,

h(cac^{-1}) = g(bcac^{-1}b^{-1})b - bg(cac^{-1}) = g(bca(bc)^{-1})b = 0. \ \Box

Theorem. Let D be a finite dimensional central division k-algebra of degree n and let a \in D. Suppose that q(x)=x^d + \alpha_{d-1}x^{d-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x] is the minimal polynomial of a over k. Then \text{Prd}_D(a,x)=(q(x))^{n/d}.

Proof. We have \dim_k k(a) = \deg q(x) = d and hence \dim_{k(a)} D = n^2/d. Now, the set \{1,a, \ldots , a^{d-1} \} is a k-basis for k(a). Let \{u_1, u_2, \ldots , u_{n^2/d} \} be a k(a)-basis for D. Then the set

\mathfrak{B}=\{u_1, au_1, \ldots , a^{d-1}u_1, \ldots , u_{n^2/d}, au_{n^2/d} \ldots , a^{d-1}u_{n^2/d} \}

is a k-basis for D. Now define \phi : D \longrightarrow \text{End}_k(D) \cong M_{n^2}(k) by \phi(b)(c)=bc for all b,c \in D. Let p(x) be the characteristic polynomial of \phi(a). By the example in this post, we have

p(x)=(\text{Prd}_D(a,x))^n. \ \ \ \ \ \ \ \ \ \ (1)

We are now going to find p(x) in terms of q(x). To do so, we first find [\phi(a)]_{\mathfrak{B}}, the matrix of \phi(a) with respect to the ordered basis \mathfrak{B}. Notice that since q(a)=0 we have

a^d = -\alpha_0 - \alpha_1a - \ldots - \alpha_{d-1} a^{d-1}.

Now, since \phi(a)(a^iu_j)=a^{i+1}u_j for all i=0, \ldots , d-1 and j =1, \ldots , n^2/d, we have the following n^2 \times n^2 block matrix

[\phi(a)]_{\mathfrak{B}} = \begin{pmatrix} C & 0 & \ldots & 0 \\ 0 & C & \ldots & 0 \\ . & . & \ldots & . \\ . & . & \ldots & . \\ . & . & \ldots & . \\ 0 & 0 & \ldots & C \end{pmatrix}, \ \ \ \ \ \ \ \ \ \ \ (2)

where C = \begin{pmatrix} 0 & 0 & \ldots & 0 & -\alpha_0 \\ 1 & 0 & \ldots & 0 & -\alpha_1 \\ . & . & \ldots & . & . \\ . & . & \ldots & . & . \\ . & . & \ldots & . & . \\ 0 & 0 & \ldots & 1 & -\alpha_{d-1} \end{pmatrix}. It is easy to see that

\det(xI - C)=x^d + \alpha_{d-1}x^{d-1} + \ldots + \alpha_1x + \alpha_0 = q(x).

Thus (2) gives us

p(x)=(\det(xI - C))^{n^2/d}=(q(x))^{n^2/d}. \ \ \ \ \ \ \ \ (3)

Hence (q(x))^{n^2/d} = (\text{Prd}_D(a,x))^n, by (1) and (3). That means q(x) is the unique irreducible factor of \text{Prd}_D(a,x) and hence \text{Prd}_D(a,x)=(q(x))^r for some integer r \geq 1. Now, since \deg \text{Prd}_D(a,x)=n and \deg q(x)=d, we must have r = n/d. \ \Box

Example. (You should also see the example in this post!) Let \mathbb{H} be the division algebra of real quaternions and a = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H}. It is not hard to see that q(x)=x^2 - 2 \alpha x + \alpha^2 + \beta^2+ \gamma^2 + \delta^2 is the minimal polynomial of a over \mathbb{R}. Therefore, since \deg \mathbb{H} = \deg q(x)=2, we get from the above theorem that \text{Prd}_{\mathbb{H}}(a,x)=q(x).