When I was studying linear algebra in the first year, from what I remember, vector spaces were always defined over a field, which was in every single concrete example equal to either $\mathbb{R}$ or $\mathbb{C}$.

In Associative Algebra course, we sometimes mentioned (when talking about $R$-modules) that if $R$ is a division ring, everything becomes trivial and known from linear algebra.

During the summer, I'm planning to revisit my notes of linear algebra, write them in tex, and try to prove as much as possible in a general setting.

Are there any theorems in linear algebra, that hold for vector spaces over a field and not over a division ring? How much linear algebra can be done over a division ring?

Also, what are some examples of division rings, that aren't fields? $\mathbb{H}$ is the only one that comes to mind. I know there aren't any finite ones (Wedderburn). Of course, I'm looking for reasonably nice ones...

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    $\begingroup$ Let me state a result (in case you do not know): every finite dimensional associative division algebra over the real field is isomorphic to either the real field, the complex field, or the quaternions. In particular, it is not possible to obtain examples of division rings (that are not fields) by somehow "generalizing" the quaternions along these lines. $\endgroup$ Commented Jun 13, 2011 at 7:29
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    $\begingroup$ Yes, that's the Theorem of Frobenius. But nobody said anything about algebras. I imagine being an $\mathbb{R}$-algebra is quite a restrictive condition, and omitting it, opens the door to many division rings... Thanks for the info though. $\endgroup$
    – Leo
    Commented Jun 13, 2011 at 7:38
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    $\begingroup$ The whole theory of eigenvalues breaks down over division rings. $\endgroup$ Commented May 29, 2013 at 10:34
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    $\begingroup$ What happened in the end? what are the consequences? i would like to have a copy of this article if you may. $\endgroup$
    – Ido4848
    Commented Dec 25, 2015 at 15:12
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    $\begingroup$ See Gertrude Ehrlich's Fundamental Concepts of Abstract Algebra book. The author starts assuming vector spaces over a division ring, and uses field only when it is necessary. $\endgroup$
    – Our
    Commented Jan 6, 2018 at 12:08

5 Answers 5


In my experience, when working over a division ring $D$, the main thing you have to be careful of is the distinction between $D$ and $D^{op}$.

E.g. if $F$ is a field, then $End_F(F) = F$ ($F$ is the ring of $F$-linear endomorphisms of itself, just via multiplication), and hence $End(F^n) = M_n(F)$; and this latter isomorphism is what links matrices and the theory of linear transformations.

But, for a general division ring $D$, the action of $D$ by left multiplication on itself is not $D$-linear, if $D$ is not commutative. Instead, the action of $D^{op}$ on $D$ via right multiplication is $D$-linear, and so we find that $End_D(D) = D^{op}$, and hence that $End_D(D^n) = M_n(D^{op}).$

As for examples of division algebras, they come from fields with non-trivial Brauer groups, although this may not help particularly with concrete examples.

A standard way to construct examples of central simple algebra over a field $F$ is via a crossed product. (Unfortunately, there does not seem to be a wikipedia entry on this topic.)

What you do is you take an element $a\in F^{\times}/(F^{\times})^n$, and a cyclic extension $K/F$, with Galois group generated by an element $\sigma$ of order $n$, and then define a degree $n^2$ central simple algebra $A$ over $F$ as follows:

$A$ is obtained from $K$ by adjoining a non-commuting, non-zero element $x$, which satisfies the conditions

  1. $x k x^{-1} = \sigma(k)$ for all $k \in K$, and
  2. $x^n = a$.

This will sometimes produce division algebras.

E.g. if we take $F = \mathbb R$, $K = \mathbb C$, $a = -1$, and $\sigma =$ complex conjugation, then $A$ will be $\mathbb H$, the Hamilton quaternions.

E.g. if we take $F = \mathbb Q_p$ (the $p$-adic numbers for some prime $p$), we take $K =$ the unique unramified extension of $\mathbb Q_p$ of degree $n$, take $\sigma$ to be the Frobenius automorphism of $K$, and take $a = p^i$ for some $i \in \{1,\ldots,n-1\}$ coprime to $n$, then we get a central simple division algebra over $\mathbb Q_p$, which is called the division algebra over $\mathbb Q_p$ of invariant $i/n$ (or perhaps $-i/n$, depending on your conventions).

E.g. if we take $F = \mathbb Q$, $K =$ the unique cubic subextension of $\mathbb Q$ contained in $\mathbb Q(\zeta_7)$, and $a = 2$, then we will get a central simple division algebra of degree $9$ over $\mathbb Q$. (To see that it is really a division algebra, one can extend scalars to $\mathbb Q_2$, where it becomes a special case of the preceding construction.)

See Jyrki Lahtonen's answer to this question, as well as Jyrki's answer here, for some more detailed examples of this construction. (Note that a key condition for getting a division algebra is that the element $a$ not be norm from the extension $K$.)

Added: As the OP remarks in a comment below, it doesn't seem to be so easy to find non-commutative division rings. Firstly, perhaps this shouldn't be so surprising, since there was quite a gap (centuries!) between the discovery of complex numbers and Hamilton's discovery of quaternions, suggesting that the latter are not so easily found.

Secondly, one easy way to make interesting but tractable non-commutative rings is to form group rings of non-commutative finite groups, and if you do this over e.g. $\mathbb Q$, you can find interesting division rings inside them. The one problem with this is that a group ring of a non-trivial group is never itself a division ring; you need to use Artin--Wedderburn theory to break it up into a product of matrix rings over division rings, and so the interesting division rings that arise in this way lie a little below the surface.

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    $\begingroup$ Excellent answer, needless to say! @Leon If you are not familiar with some of the terms in Matt E's answer, a good source is "Noncommutative Algebra" by Benson Farb and R. Keith Dennis. $\endgroup$ Commented Jun 13, 2011 at 7:32
  • $\begingroup$ Hmm, got more than I bargained for :). Thanks @Amitesh, I am not familiar with the second part of the answer, I guess it's not easy to find nice noncommutative division rings. Also, I was not asking for division algebras, but it seems they are the most popular type of division rings. In any case, thank you @Matt. $\endgroup$
    – Leo
    Commented Jun 13, 2011 at 7:45
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    $\begingroup$ @Leon: If $D$ is a division ring, then its centre is a field, and (in practice and in terms of constructions) it is easiest to consider the case when $D$ is finite-dimensional over its centre; thus division algebras are natural candidates to consider when thinking about division rings. And you are right that in some sense it is a bit tricky to find nice examples; I have added one more philosophical comment about that to my answer. Regards, $\endgroup$
    – Matt E
    Commented Jun 13, 2011 at 7:57
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    $\begingroup$ @Chandru: Dear Chandru, $D^{op}$ is the opposite ring, i.e. same underlying set, same addition, but $a \cdot b$ (in $D^{op}$) is $b a$ (computed in $D$). Regards, $\endgroup$
    – Matt E
    Commented Jun 13, 2011 at 8:05
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    $\begingroup$ I want to emphasize that the norm condition really says that no power $a^k, 0<k<n$ should be a norm. Of course, it suffices to check this, when $k$ is a proper divisor of $n$. So when $n$ is a prime itself, it suffices to check that $a$ itself is not a norm. $\endgroup$ Commented Jun 13, 2011 at 13:04

Let me quote a relevant paragraph in the Wikipedia article on "Division ring":

Much of linear algebra may be formulated, and remains correct, for (left) modules over division rings instead of vector spaces over fields. Every module over a division ring has a basis; linear maps between finite-dimensional modules over a division ring can be described by matrices, and the Gaussian elimination algorithm remains applicable. Differences between linear algebra over fields and skew fields occur whenever the order of the factors in a product matters. For example, the proof that the column rank of a matrix over a field equals its row rank yields for matrices over division rings only that the left column rank equals its right row rank: it does not make sense to speak about the rank of a matrix over a division ring.

I hope this helps!

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    $\begingroup$ @Myself: Thats true, and I have acted accordingly. $\endgroup$
    – user9413
    Commented Jun 13, 2011 at 9:23
  • $\begingroup$ Dear Chandru, what do you mean? I notice you deleted your answer ... $\endgroup$ Commented Jun 13, 2011 at 9:38
  • $\begingroup$ Myself informed there is not use of keeping 2 same answers, so i have deleted mine. :) $\endgroup$
    – user9413
    Commented Jun 13, 2011 at 9:39
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    $\begingroup$ I find the last sentence of the quoted article very misleading, and I recently explained why int this MathOverflow answer. In short, the left column rank and the right row rank are both misguided notions: over a division ring the column rank is naturally the right column rank, and the row rank is the left row rank, since only right-linear combinations of columns and left-linear combinations of rows can be expressed using matrix multiplication. So it does make sense to speak of the rank of a matrix over a division ring. $\endgroup$ Commented May 19, 2014 at 11:20

The others have outlined the basic differences. Since you asked for reasonably nice examples let me give you the following. Let $F=\mathbf{Q}(i)$, $E=F(\sqrt5)$ and $\sigma$ be the non-trivial element of $Gal(E/F)$ determined by $\sigma(i)=i, \sigma(\sqrt5)=-\sqrt5$. Then the set of matrices of the form $$ A(x_0,x_1)=\left(\begin{array}{rr} x_0&\sigma(x_1)\\ ix_1&\sigma(x_0) \end{array}\right), $$ where $x_0$ and $x_1$ range over $E$, is a division algebra. It is easy to see that it is closed under multiplication. Whenever $(x_0,x_1)\neq(0,0)$ the determinant of $A(x_0,x_1)\neq0$. This is easiest to see as follows: $\det A(x_0,x_1)=N(x_0)-iN(x_1)$, where $N$ is the relative norm map $N:E\rightarrow F, x\mapsto x\sigma(x)$. For the determinant to vanish, we must have $i=N(x_0/x_1)$ by the multiplicativity of the norm. But a 5-adic study of this equation shows that $i$ does not belong to the image of the norm map. From the above we also see that $\det A\in F$, so the usual formula for the inverse of a 2x2 matrix shows that $A(x_0,x_1)$ has an inverse of the same form.

In the language of Matt E's answer this division algebra represents an element of the Brauer group $Br(F)$. It is of order two. To the initiated I will tell that its non-trivial Hasse invariants lie above the primes $2+i$ and $2-i$.

This algebra (and other cyclic division algebras) have found their way to multiantenna radio communications. The discoverers call it the Golden code. IIRC this code is in the hyperWLAN standard, but I won't trust my memory 100% here.

Another division algebra that reasonably often shows up in lattice constructions (see Conway & Sloane) is the ring of Icosians. It is really a subalgebra of the usual Hamiltonian quaternions, where in place of $\mathbf{R}$ we restrict the coordinates to the field $\mathbf{Q}(\sqrt5)$. The resulting algebra has its non-trivial Hasse invariants only at the infinite places, so its discriminant is small, and hence it is destined to yield dense lattices and such.

  • $\begingroup$ Thanks for the info, and very interesting links. I'm sorry to say that I don't yet understand much of this, will have to study more, but I never thought abstract algebra had that much contact with non-mathematics. appreciated, +1 $\endgroup$
    – Leo
    Commented Jun 13, 2011 at 11:48
  • $\begingroup$ The link to Golden Code is broken (Emanuele moved to a different university, and they have taken his page down). Ping me, if you need more information about that. $\endgroup$ Commented Apr 23, 2012 at 5:07
  • $\begingroup$ Question: in the example above, is it possbile to use sqrt(3) instead of sqrt(5)? $\endgroup$
    – user34846
    Commented Jul 1, 2012 at 14:29
  • $\begingroup$ @gshmelev: Certainly there exists cyclic division algebras, where the field $\mathbf{Q}(i,\sqrt3)$ and its automorphism $\sigma:i\mapsto i,\sqrt3\mapsto-\sqrt3$ play a similar role. But the entries $ix_1$ in the bottom left corner of the matrices need to be something else. This is because (barring a mistake) the element $$z=\frac{1-i}2(1+\sqrt3)$$ has relative norm equal to $i$. We would need to replace that entry with $\gamma x_1$ for some other number $\gamma$ that is not a norm. Finding one is not too difficult, but perhaps we should discuss that in another question? Doesn't fit here. $\endgroup$ Commented Jul 1, 2012 at 16:57
  • $\begingroup$ Fixed a typo and the dead links. Sorry about the bump. $\endgroup$ Commented Nov 3, 2015 at 18:41

Anything that involves general homothecies (scalar multiples of the identity viewed as linear operators), or multi-linearity (in particular determinants) will not go through over division rings (or maybe only in some some much weakened form; for instance homothecies by central scalars can still be considered). Anything involving eigenvalues and eigenspaces falls into this category (an eigenspace is a subspace where a linear operator acts as a homothecy). As an example of a theorem that won't be available, the Cayley-Hamilton theorem (indeed the characteristic polynomial of a finite-dimensional linear operator over a division ring is not even defined).


By Schur's lemma, whenever $M$ is an irreducible module over some ring $R$, then $\operatorname{End}(M)$ is a division ring.


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