I've always wondered about what a field is meant to represent. For example, group automorphisns naturally represent symmetry in many areas.

I'm not looking for a solid answer, just an idea.

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    $\begingroup$ On some level, I think you'd have to first come up with an intuition for what a ring is. Rings are more primitive in a number of ways. If "group theory" studies certain nice subsets of permutations of a set, rings study subsets of homomorphisms from an abelian group to itself. $\endgroup$ – Thomas Andrews Jan 3 '12 at 18:03
  • $\begingroup$ See this answer, which shows you how to view a field as the "symmetries" of its additive group, considered as a vector space. $\endgroup$ – Bill Dubuque Jan 3 '12 at 19:47

I'm not sure if this answers your question, but a nice line that I'm paraphrasing from an answer from Greg Stevenson over at MO is that if you buy that groups are important for representing symmetries on sets, then you should naturally buy that rings are important for being the things that represent symmetries in abelian groups!

Namely, the set of endomorphisms of an abelian group always forms a ring, and so now if you're looking for automorphisms of an abelian group, you're naturally led to looking at asking question about inverses of elements in rings. The nicest possible world is one in which all of the elements have inverses, i.e., fields.

Of course, this a very narrow response, designed to mirror the question. Fields arise very naturally in very many areas of mathematics (rationals, reals, complex numbers, finite fields, $p$-adics, meromorphic functions, etc.), not all with the intent of modeling the symmetry of an abelian group.

  • $\begingroup$ Also, note that every ring is of this form, since the ring acts on its own additive group, which is very similar to the way we represent every group as permutations of itself. $\endgroup$ – Thomas Andrews Jan 3 '12 at 18:05
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    $\begingroup$ I mean, ever ring is some subring of the set of endomorphisms of an abelian group, not necessarily the full endomorphism ring. $\endgroup$ – Thomas Andrews Jan 3 '12 at 18:12
  • $\begingroup$ Okay so the natural numbers form a ring with multiplication and addition. What is the abelian group they are representing the endo- and auto- morphisms of? $\endgroup$ – Cris Stringfellow Jan 3 '12 at 19:02
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    $\begingroup$ @DustanLevenstein Given any ring, $(R,+,\times)$, it can be seen as isomorphic to a sub-ring of the ring of endomorphisms of the abelian group $(R,+)$. $\endgroup$ – Thomas Andrews Jan 3 '12 at 19:22
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    $\begingroup$ got it, thanks, Cam. $\endgroup$ – Dustan Levenstein Jan 3 '12 at 19:27

Fields allow us to generalise much of what we take for granted when working over the real, rational and complex numbers: in particular, the operations of addition, subtraction, multiplication and division. Division in particular is what makes a field special, separating it from, say, a ring.

So the short answer to your question is: a field is an algebraic structure on a set which allows us to make sense of addition, subtraction, multiplication and division. These operations are tied together using the underlying group structure and the distributivity law, and what we get turns out to be very useful.

Finite fields, for example, are incredibly useful in cryptography: they allow us to take a finite set of integers and divide one integer by another to get another integer, by using modular arithmetic. There are numerous other important types of fields: function fields, cyclotomic fields, number fields, etc. And furthermore, without fields, we wouldn't have vector spaces!

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    $\begingroup$ I thought Monoids let us generalize binary operations, what do Fields have to do with Monoids? Is a field a tightly coupled collection of 4 Monoids(tries to do too much), specifically (+), (-), (/), (*)? $\endgroup$ – Dmitry Dec 16 '16 at 5:38

From a semi-category-theoretic point of view, fields are to commutative rings as simple groups are to groups.

Specifically, a group, $G$, is simple if any only if any homomorphism $G\rightarrow H$ is either a monomorphism or trivial.

In the category of commutative rings, $F$ is a field if and only if any ring homomorphism $F\rightarrow R$ is either a monomorphism or trivial.

  • $\begingroup$ In other words, fields are to simple groups as ideals in commutative rings are to normal subgroups. $\endgroup$ – Alex Becker Jan 4 '12 at 0:40
  • $\begingroup$ Yes. That's less category-theoretical way of saying the same thing. @AlexBecker $\endgroup$ – Thomas Andrews Jan 4 '12 at 1:32

Like many things in mathematics, a field is a generalization rather than representation. But, also like many things in mathematics, fields have certain examples that inspired them and form the classical example. The real numbers are the classical example of a field, but certainly not the only one as Complex Numbers, Surreal Numbers, and many others are also prominent examples of fields.

When I first started learning about fields, I thought of all fields as "like real numbers" Obviously that is an oversimplification, but I found it a useful model to get something of an intuitive grasp rather than just trying to deal with naked axioms.

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    $\begingroup$ This is a good place to start. Then you can ask the question "How are fields not like the Real Numbers?" $\endgroup$ – Joshua Shane Liberman Jan 3 '12 at 18:31
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    $\begingroup$ I would start with the rationals, rather than the reals - as a purely algebraic structure, the reals are a monster. $\endgroup$ – Thomas Andrews Jan 3 '12 at 18:44
  • $\begingroup$ @ThomasAndrews You have a great point, but I had a strong computer background and ran into Pi and e frequently, so the reals were more comfortable for me personal at that point. For many others, the rationals might be better, and they are certainly simpler. $\endgroup$ – TimothyAWiseman Jan 3 '12 at 20:33
  • $\begingroup$ Of course, on the computer you don't really have $\pi$ and e … $\endgroup$ – celtschk Aug 24 '12 at 15:28
  • $\begingroup$ @celtschk This is true, but only to a degree. When I wanted to deal with a circle using a computer, I thought about pi and e, the computer worked with binary approximations that were close enough for the pixels on the screen. Of course, with packages like sympy the computer could deal with them simply as symbols without numeric values that obeyed certain rules as well, but even then it wasn't dealing with pi and e the way I was and I probably wasn't dealing with them the way a real expert would. $\endgroup$ – TimothyAWiseman Aug 27 '12 at 15:23

There is one thing that always fascinated me:

Say one day you come across math written in alien symbols, and they all look something like this

$$\blacklozenge \boxtimes\ \blacksquare\ \triangle\ \bigstar.$$

After hours of studying these you come to the conclusion that $\triangle$ denoes ''$=$'' and $\boxtimes$ must be denoting either plus or times. What is it? Well if you find that

$$\blacklozenge \boxtimes \blacktriangledown\ \triangle\ \blacktriangledown$$


$$\blacksquare \boxtimes \blacktriangledown\ \triangle\ \blacktriangledown$$

and they don't just use multiple symbols for the same thing, then know that $\boxtimes$ must be times and $\blacktriangledown$ is the zero. Because in addition of the abelian group with ''+'' you always have the unique inverse and the two symbols $\blacklozenge, \blacksquare$ can't both be the neutral element. I think it's pretty cool that these two operations are seperated by such a simple line.

If you have a field and $a$ is an element, then there is also $-a$. From

$$0=b\times a-b\times a=b\times(a-a)\ \ \ \forall b,$$

you can always find this object $0$ in a field by adding any element and its inverse. A field is a set of things you can add with the abelian plus operator like you expect things in the universe to behave if you add them. But there is also this mysterious thing ''$0$'' with

$$0\times a=a\ \ \ \forall a$$

The zero thing. Isn't that object weird?

  • $\begingroup$ Unfortunately it turns out that the aliens include $\pm\infty$ in their numbers, $\blacktriangledown$ is $-\infty$ (while $\blacktriangle$ is $+\infty$) and $\boxtimes$ is plus. :-) $\endgroup$ – celtschk Aug 24 '12 at 15:41

I think that a field is not meant to represent something in particular, but it is just a tool. A field is essentialy a set with two operation where you are always garantee to find an inverse and an opposite for every element. Such property is really usefull in some application like cryptography, data encoding, etc..


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