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I've heard of a field, and I've heard of a non-commutative (or "not-necessarily commutative) rings. Do people ever study non-commutative fields?

For motivation, consider the set of all $n \times n$ invertible matrices over a (commutative) field.

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The invertible matrices are not closed under addition. –  JSchlather Mar 15 '13 at 7:46
    
As Robert Israel pointed out in his answer, what you're thinking of are division rings, of which perhaps the best-known noncommutative example is the division ring of quaternions. –  Branimir Ćaćić Mar 15 '13 at 7:47
    
@JSchlather Ohhh. Of course. –  goblin Mar 15 '13 at 7:47
    
The ring of quaternions may be the most well known example. They are used in computer graphics, among other things. –  Michael Hardy Mar 17 '13 at 14:54

3 Answers 3

Yes. They are called division rings or skew fields.

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Also, every finite divison ring is a field (there is a nice proof in "Proofs from the book") –  Stefan Mar 15 '13 at 10:15

Yes. A lot of interesting number theory is involved. The Brauer group classifies division algebras with a given center, and in class field theory that plays a big role (when the center field is a number field). See for example this question and this.

In addition to number theory, the topic is interesting on its own merits. Over more complicated center fields we no longer get all the division algebras as cyclic algebras. Hopefully some more knowledgable forumite can give you pointers to that theory.

My interest to skewfields (the number theoretic ones in particular) was awakened by the observation that lattices in skewfields yield interesting signal constellations in multi-antenna radio communications. Google for Golden code for the most widely known example.

Damien raised the study of roots of $x^2=-1$ in the ring of quaternions as a question with an unexpected answer. I have written up a shortish related answer. The root cause for the problem is explained in this exposition by Arturo Magidin.

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As said Robert Israel, they are called skew fields. But the study of skew fields is very different from commutative fields.

For example, if your field is the Quaternions $\mathbb H$, and you consider the polynomial with real coefficients $\rm X^2 + 1$, it has more than 2 roots in $\mathbb H$ !

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Excellent example. +1 –  user1551 Mar 15 '13 at 11:04
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@user18921 Actually the situation is much worse than Damien makes it sound... $X^2+1$ actually has infinitely many roots in $\Bbb H$! –  rschwieb Mar 15 '13 at 13:33

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