Division Rings

In this post, we study in earnest rings in which every nonzero element has a multiplicative inverse, i.e., division rings.

Unless otherwise stated, a ring is assumed to have an identity $1\neq 0$, and a division ring is not assumed to be commutative.

Division Rings and Zero Divisors

The definition of a division ring has already been given, but we restate it to begin this post. ([Algebraic Structures] §Field of Fractions, ⁋Definition 3)

Definition 1 A ring $D\neq 0$ is called a division ring or skew field if every nonzero element of $D$ has a two-sided multiplicative inverse. A commutative division ring is called a field.

We have already verified in §Units and Zero Divisors, ⁋Definition 1 that the unit group $D^\times$ is a group under multiplication, and by definition $D^\times=D\setminus\{0\}$ in a division ring, so this is a group under multiplication. We call this group the multiplicative group of $D$.

The first property of a division ring is as follows.

Proposition 2 A division ring $D$ has no nonzero zero divisors. That is, if $ab=0$ then $a=0$ or $b=0$. In particular, every field is an integral domain.

Proof

Let $a,b\in D$ with $ab=0$ and $a\neq 0$. Since $D$ is a division ring, $a$ has an inverse $a^{-1}$, and multiplying both sides on the left by $a^{-1}$ gives

\[b=1\cdot b=(a^{-1}a)b=a^{-1}(ab)=a^{-1}\cdot 0=0\]

Thus if $a\neq 0$ then $b=0$, which means that $a=0$ or $b=0$ whenever $ab=0$. Hence $D$ has no nonzero zero divisors. If $D$ is a field, then it is additionally commutative and $0\neq 1$, so it is an integral domain ([Algebraic Structures] §Field of Fractions, ⁋Definition 5).

Wedderburn’s Little Theorem

The above Proposition 2 can in fact be obtained immediately from §Units and Zero Divisors, ⁋Proposition 4, since in a division ring every nonzero element is a unit, so there is no possibility of a nonzero zero divisor existing. However, the converse is not generally true; for instance, $\mathbb{Z}$ is such an example, as we already observed right after the above proposition.

Moreover, in that post we already examined a partial converse of the above Proposition 2, namely that for finite rings, an integral domain is always a field. (§Units and Zero Divisors, ⁋Corollary 6) The proof of this corollary does not essentially use the commutativity of the ring, yet the reason commutativity was assumed in that corollary is that the finite non-commutative zero-divisor-free rings that would be dealt with when commutativity is dropped do not exist.

To examine this phenomenon, we first organize properties of the center $Z(D)$ of a division ring $D$. This is a commutative subring of $D$ ([Algebraic Structures] §Definition of a Ring, ⁋Definition 8), and moreover it is a field. This is because any nonzero $z\in Z(D)$ has an inverse $z^{-1}$ in $D$, and for any $x\in D$ we have

\[z^{-1}x=z^{-1}xzz^{-1}=z^{-1}zxz^{-1}=xz^{-1}\]

so $z^{-1}\in Z(D)$. Hence if $D$ is finite, then $Z(D)$ is a finite field, and letting $q$ be its number of elements, we have $q\geq 2$.

Still assuming the above finiteness, if we view $D$ as a vector space over $Z(D)$, then for $n=\dim_{Z(D)} D$ we have $\lvert D\rvert=q^n$. More generally, a sub-division ring $D'$ of $D$ containing $Z(D)$ is also a vector space over $Z(D)$, so its number of elements is of the form $q^d$, and if $D$ is an $m$-dimensional vector space over $D'$ then from $q^n=(q^d)^m$ we obtain $d\mid n$. In a similar vein, we need the following cyclotomic polynomial to prove the main result of this section.

Definition 3 For a positive integer $n$, the $n$th cyclotomic polynomial $\Phi_n(\x)$ is defined by

\[\Phi_n(\x)=\prod_{\substack{1\leq m\leq n\\ \gcd(m,n)=1}}\bigl(\x-\zeta^m\bigr),\qquad \zeta=e^{2\pi i/n}\]

That is, $\Phi_n(\x)$ is the monic polynomial whose roots are the primitive $n$th roots of unity.

Since a detailed analysis of cyclotomic polynomials is not our current concern, we postpone this to [Field Theory] §Cyclotomic Extensions, ⁋Definition 1 and use only two basic properties. First, since every root of $\x^n-1$ is, for some $d\mid n$, a primitive $d$th root of unity,

\[\x^n-1=\prod_{d\mid n}\Phi_d(\x)\]

holds, and therefore by performing the division from §Polynomial Rings, ⁋Proposition 5 inside $\mathbb{Z}[\x]$, we inductively know that each $\Phi_d(\x)$ has integer coefficients. Second, for a proper divisor of $n$, $d$, since the right-hand side of

\[\x^n-1=(\x^d-1)\cdot\prod_{e\mid n\text{ but }e\nmid d}\Phi_e(\x)\]

contains the factor $\Phi_n(\x)$, $\Phi_n(\x)$ divides, in $\mathbb{Z}[\x]$, $(\x^n-1)/(\x^d-1)$. In particular, substituting an integer $q$ into these integer polynomials, we know that $\Phi_n(q)$ divides both $q^n-1$ and $(q^n-1)/(q^d-1)$ as integers.

Finally, we isolate one analytic inequality that we will need.

Proposition 4 For integers $q\geq 2$ and $n\geq 2$, we have $\lvert\Phi_n(q)\rvert>q-1$.

Proof

Substituting $q$ into the cyclotomic polynomial, by Definition 3 we have

\[\Phi_n(q)=\prod_{\substack{1\leq m\leq n\\ \gcd(m,n)=1}}(q-\zeta^m)\]

We estimate the absolute value of each factor from below. If we write $\zeta=\cos\theta+i\sin\theta$ ($\theta\neq 0$), then

\[\lvert q-\zeta^m\rvert^2=(q-\cos m\theta)^2+\sin^2 m\theta=q^2-2q\cos m\theta+1\]

and therefore

\[\lvert q-\zeta^m\rvert^2-(q-1)^2=q^2-2q\cos m\theta+1-(q^2-2q+1)=2q(1-\cos m\theta)\geq 0\]

For every primitive root $\zeta^m$, we have $\lvert q-\zeta^m\rvert\geq q-1\geq 1$, and in particular

\[\lvert\Phi_n(q)\rvert=\prod_{\substack{1\leq m\leq n\\ \gcd(m,n)=1}}\lvert q-\zeta^m\rvert\geq\lvert q-\zeta\rvert\geq q-1\]

where, for the last inequality, using $\zeta\neq 1$, that is, $\cos\theta\neq 1$, this inequality is strict, so we obtain the strict inequality of the proposition.

We now prove the theorem.

Theorem 5 (Wedderburn) Every finite division ring is a field. That is, every finite division ring is commutative.

Proof

Let $D$ be a finite division ring and $Z=Z(D)$ its center. As we saw earlier, $Z$ is a finite field, and if we let its number of elements be $q\geq 2$, then $D$ is a finite-dimensional vector space over $Z$, so its number of elements is of the form $\lvert D\rvert=q^n$. Our claim is that $n=1$, so that $D=Z$ is commutative.

To this end, we write the class equation of the multiplicative group $D^\times=D\setminus\{0\}$. ([Algebraic Structures] §Group Action, ⁋Theorem 14 (Orbit-stabilizer theorem)) The class equation of $D^\times$ for the conjugation action from [Algebraic Structures] §Group Action, ⁋Proposition 9 is

\[\lvert D^\times\rvert=\lvert Z(D^\times)\rvert+\sum_{x}\bigl[D^\times:C_{D^\times}(x)\bigr]\]

where $C_{D^\times}(x)$ is the centralizer of $x$ defined right after [Algebraic Structures] §Group Action, ⁋Definition 12, and the sum is over all representatives not belonging to $Z(D^\times)$. Also, since $Z(D^\times)=Z^\times=Z\setminus\{0\}$, we have $\lvert Z(D^\times)\rvert=q-1$.

Now for each $x\in D^\times$, $C_D(x)=\{y\in D\mid xy=yx\}$ is a sub-division ring of $D$ and contains $Z$. In this case, we have seen that $C_D(x)$ is a $Z$-vector space, and since $\lvert Z\rvert=q$, it is of the form $\lvert C_D(x)\rvert=q^{d(x)}$. Also, since $D$ is a vector space over $C_D(x)$, we have $d(x)\mid n$. Since $C_{D^\times}(x)=C_D(x)\setminus\{0\}$, we have

\[\bigl[D^\times:C_{D^\times}(x)\bigr]=\frac{q^n-1}{q^{d(x)}-1}\]

and for this number to be an integer, we must have $d(x)\mid n$. Now if $x$ does not belong to the center, then $C_D(x)\neq D$, so $d(x)<n$. Hence the class equation takes the form

\[q^n-1=(q-1)+\sum_{x}\frac{q^n-1}{q^{d(x)}-1}\tag{$\ast$}\]

where each $d(x)$ in the sum is a proper divisor of $n$.

Now assume $n\geq 2$ and derive a contradiction. The cyclotomic polynomial $\Phi_n(q)$ divides $q^n-1$, and for each proper divisor $d=d(x)<n$, it also divides $\frac{q^n-1}{q^d-1}$. Therefore, in $(\ast)$ above, $\Phi_n(q)$ must also divide $q-1$. That is, $\Phi_n(q)\mid q-1$, and since $q-1\geq 1$, we have $\lvert\Phi_n(q)\rvert\leq q-1$. However, by Proposition 4, if $n\geq 2$, then $\lvert\Phi_n(q)\rvert>q-1$, a contradiction.

The first consequence of this theorem is to reconfirm the result about finite integral domains.

Corollary 6 If $0\neq 1$ and a finite ring $A$ has no nonzero zero-divisors, then $A$ is a field.

Proof

Let $A$ be a finite ring with no zero divisors other than $0$. For any nonzero $a\in A$, consider the left multiplication morphism $\lambda_a:A\rightarrow A$, $\lambda_a(x)=ax$. If $ax=ay$, then $a(x-y)=0$, and since $a$ is not a zero divisor, $x=y$, that is, $\lambda_a$ is injective. Since $A$ is a finite set, $\lambda_a$ is surjective, and $av=1$ for some $v$. Applying the same argument to right multiplication, $wa=1$ for some $w$, and since $w=w(av)=(wa)v=v$, $v$ is a two-sided inverse of $a$. Therefore, every nonzero element is a unit and $A$ is a division ring. A finite division ring is a field by Theorem 5 (Wedderburn).

What makes this essentially different from §Units and Zero Divisors, ⁋Corollary 6 is that we do not assume the commutativity of $A$. If we had assumed commutativity from the outset, then Theorem 5 (Wedderburn) would not have been needed to prove this corollary. The power of this theorem lies in achieving the same result using only the assumption of finiteness, without commutativity.

Quaternions

By Theorem 5 (Wedderburn), a non-commutative division ring is necessarily infinite, so one must look for examples among infinite rings. The most classical one is the space of quaternions defined by Hamilton, which is a vector space over the real field $\mathbb{R}$ of dimension $4$ equipped with a multiplication.

Definition 7 The quaternion algebra $\mathbb{H}$ is the vector space with basis $1,i,j,k$ over $\mathbb{R}$ of dimension $4$, whose elements are of the form

\[q=a+bi+cj+dk\qquad(a,b,c,d\in\mathbb{R})\]

and whose multiplication has $1$ as the identity and is defined by extending the relations

\[i^2=j^2=k^2=-1,\qquad ij=k,\quad jk=i,\quad ki=j,\qquad ji=-k,\quad kj=-i,\quad ik=-j\]

on the basis elements $\mathbb{R}$-bilinearly.

That this multiplication is associative does not follow merely from extending the relations bilinearly, so this must be checked separately. The simplest method is to realize $\mathbb{H}$ inside the ring of $2\times 2$ complex matrices $\Mat_2(\mathbb{C})$. For a given quaternion $q=a+bi+cj+dk$, set $z=a+bi$, $w=c+di$ and define the $\mathbb{R}$-linear map $\varphi:\mathbb{H}\rightarrow\Mat_2(\mathbb{C})$ by

\[\varphi(q)=\begin{pmatrix}z&w\\ -\bar{w}&\bar{z}\end{pmatrix}\]

Then the images of the basis elements are

\[\varphi(1)=I,\qquad\varphi(i)=\begin{pmatrix}i&0\\ 0&-i\end{pmatrix},\qquad\varphi(j)=\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\qquad\varphi(k)=\begin{pmatrix}0&i\\ i&0\end{pmatrix}\]

and that these four matrices satisfy all the relations of Definition 7 is verified by direct computation. Since both sides are bilinear in $q,q'$, it follows that $\varphi(qq')=\varphi(q)\varphi(q')$ for all quaternions, and since the four matrices above are linearly independent over $\mathbb{R}$, $\varphi$ is injective. Thus the associativity of multiplication in $\Mat_2(\mathbb{C})$ carries over to $\mathbb{H}$, and $\mathbb{H}$ becomes a ring isomorphic to the subring $\varphi(\mathbb{H})\subseteq\Mat_2(\mathbb{C})$. On the other hand, the determinant of this matrix

\[\lvert z\rvert^2+\lvert w\rvert^2=a^2+b^2+c^2+d^2\]

defines the norm of the quaternion.

Definition 8 For a quaternion $q=a+bi+cj+dk$, we define its conjugate by

\[\bar{q}=a-bi-cj-dk\]

and its norm by

\[N(q)=q\bar{q}\]

For any quaternion $q=a+bi+cj+dk$, multiplying by the conjugate $\bar{q}$, one can verify that all coefficients of the $i,j,k$ terms indeed cancel out by the relations of Definition 7, so that

\[N(q)=q\bar{q}=a^2+b^2+c^2+d^2\in\mathbb{R}\]

holds. In particular, $N(q)=0$ is equivalent to $a=b=c=d=0$, that is, $q=0$. Also, since the conjugate of $\bar{q}$ is again $q$, we have $\bar{q}q=\bar{q}\bar{\bar{q}}=N(\bar{q})=a^2+(-b)^2+(-c)^2+(-d)^2=N(q)$, and therefore both $q\bar{q}$ and $\bar{q}q$ are equal to $N(q)$.

Another property of the norm is that it preserves multiplication. Indeed, it is easily verified that the conjugate satisfies $\overline{q_1q_2}=\bar{q}_2\bar{q}_1$, and using this,

\[N(q_1q_2)=q_1q_2\overline{q_1q_2}=q_1q_2\bar{q}_2\bar{q}_1=q_1N(q_2)\bar{q}_1=N(q_2)q_1\bar{q}_1=N(q_1)N(q_2)\]

holds. Writing this multiplicativity in coordinates gives

\[(a_1^2+b_1^2+c_1^2+d_1^2)(a_2^2+b_2^2+c_2^2+d_2^2)=(\cdots)^2+(\cdots)^2+(\cdots)^2+(\cdots)^2\]

which is Euler’s four-square identity, stating that the product of two sums of four squares is again a sum of four squares. In any case, what is important for us is that we can use this to prove that $\mathbb{H}$ is a division ring.

Proposition 9 The quaternion algebra $\mathbb{H}$ is a noncommutative division ring.

Proof

That $\mathbb{H}$ is a ring with $1\neq 0$ was verified above, and that it is not commutative is obvious from $ij=k\neq -k=ji$. It remains to show that every nonzero $q\in\mathbb{H}$ has a two-sided multiplicative inverse.

Let $q=a+bi+cj+dk\neq 0$. We have seen above that $N(q)=a^2+b^2+c^2+d^2$ is a positive real number. Then this can be viewed as an element of $\mathbb{H}$, and moreover it commutes with every element of $\mathbb{H}$. The same holds for the inverse $N(q)^{-1}$ of $N(q)$, and therefore

\[q\cdot\bigl(N(q)^{-1}\bar{q}\bigr)=N(q)^{-1}(q\bar{q})=N(q)^{-1}N(q)=1\]

and in the same way, one can verify that

\[\bigl(N(q)^{-1}\bar{q}\bigr)\cdot q=N(q)^{-1}(\bar{q} q)=N(q)^{-1}N(q)=1\]

holds. That is, $q^{-1}=N(q)^{-1}\bar{q}$ is a two-sided inverse of $q$, from which we obtain the desired claim.

Endomorphism Rings of Simple Modules

Division rings are useful when dealing with endomorphisms of modules. A nonzero module $M$ defined over a ring $A$ is called a simple module if $M$ has no submodules other than $0$ and itself. For convenience of discussion, let us fix $M$ to be a left module. Then the following holds.

Lemma 10 (Schur) For simple modules $M,N$ over a ring $A$, the following hold.

Any $A$-module homomorphism $f:M\rightarrow N$ is either the zero map or an isomorphism.
In particular, the endomorphism ring $\End_A(M)$ of a simple module $M$ is a division ring.

Proof

Let $f:M\rightarrow N$ be a nonzero $A$-module homomorphism. The kernel $\ker f$ is a submodule of $M$, and since $f\neq 0$, we have $\ker f\neq M$. Since $M$ is simple, $\ker f=0$, i.e. $f$ is injective. Also $\im f$ is a nonzero submodule of $N$, and since $N$ is simple, $\im f=N$, i.e. $f$ is surjective. Hence $f$ is an isomorphism. This establishes the first result.

Now consider the case $M=N$. Then $\End_A(M)$ is a ring with composition of morphisms as multiplication and the identity map $\id_M$ as identity. Since $M$ is nonzero, $\id_M\neq 0$, i.e. this ring is not $0$. By the first result, any nonzero $f\in\End_A(M)$ is an isomorphism, and its inverse $f^{-1}$ is also an $A$-module homomorphism, hence an element of $\End_A(M)$. Moreover $f\circ f^{-1}=f^{-1}\circ f=\id_M$, so $f$ is a unit. That is, every nonzero element is a unit, and $\End_A(M)$ is a division ring.

This lemma supplies division rings in abundance in the form of endomorphism rings of simple modules. Conversely, viewing a division ring itself as a vector space over a smaller field, it is represented by matrices inside the ring of linear endomorphisms over that field, and the matrix representation of the quaternions $\mathbb{H}$ written by hand right after Definition 7 is also obtained in this way. Viewing $\mathbb{H}$ as a $\mathbb{C}$-vector space by multiplying the subfield $\mathbb{C}=\mathbb{R}+\mathbb{R}i$ of $\mathbb{H}$ on the left, a quaternion $q=a+bi+cj+dk$ can be written uniquely as

\[q=z+wj\]

for $z=a+bi$, $w=c+di$, so $\{1,j\}$ is a basis of this vector space and $\dim_{\mathbb{C}}\mathbb{H}=2$. Now considering right multiplication $\rho_q(x)=xq$ for each $q\in\mathbb{H}$, we have

\[\rho_q(ux)=uxq=u\rho_q(x)\qquad(u\in\mathbb{C})\]

so $\rho_q$ is a $\mathbb{C}$-linear map, i.e. an element of $\End_{\mathbb{C}}(\mathbb{H})$, and if $\rho_q=0$ then $q=\rho_q(1)=0$, so $q\mapsto\rho_q$ is injective. By associativity, $\rho_{qq'}=\rho_{q'}\circ\rho_q$, so this correspondence reverses the order of multiplication, but if we write coordinates as row vectors and represent a $\mathbb{C}$-linear map by a matrix $M_q$ acting by right multiplication, the order is reversed once more, so

\[\mathbb{H}\rightarrow\Mat_2(\mathbb{C});\quad q\mapsto M_q\]

is an injective ring homomorphism. Here the two rows of $M_q$ are the coordinates of $\rho_q(1)$ and $\rho_q(j)$, respectively, and since $ji=-ij$ from the relations of Definition 7, using the fact that $ju=\bar{u}j$ holds for any $u\in\mathbb{C}$, from $\rho_q(1)=q=z+wj$ and

\[\rho_q(j)=jq=jz+jwj=\bar{z}j+\bar{w}j^2=-\bar{w}+\bar{z}j\]

we obtain

\[M_q=\begin{pmatrix}z&w\\ -\bar{w}&\bar{z}\end{pmatrix}\]

This is the matrix representation written above, and its determinant $\lvert z\rvert^2+\lvert w\rvert^2$ is precisely the norm $N(q)$, so the fact that $M_q$ is an invertible matrix for nonzero $q$ is the same content as what Proposition 9 showed.

References

[DF] D. S. Dummit and R. M. Foote, Abstract algebra, 3rd ed., Wiley, 2004.
[Her] I. N. Herstein, Noncommutative rings, Carus Mathematical Monographs 15, Mathematical Association of America, 1968.
[Lam] T. Y. Lam, A first course in noncommutative rings, 2nd ed., Graduate Texts in Mathematics 131, Springer, 2001.

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Division Rings

Division Rings and Zero Divisors

Wedderburn’s Little Theorem

Quaternions

Endomorphism Rings of Simple Modules

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