# In plain language, what's the significance of a field?

I just started Linear Algebra. Yesterday, I read about the ten properties of fields. As far as I can tell a field is a mathematical system that we can use to do common arithmetic. Is that correct?

• Fields can have pretty uncommon properties, though, such as $1+1=0$. So beware of trusting your intuition too much. Sep 11, 2014 at 14:45
• Are you asking what the significance of a field is for linear algebra, or in general? Sep 11, 2014 at 15:15
• @Hal Not necessarily, e.g. do any of the current answers help you understand the significance of the field axiom that nonzero elements are invertible, i.e. why not define vector spaces over rings (modules) vs. fields? If that is the sort of thing you are interested in then you should clarify that in your question. Sep 11, 2014 at 15:23
• Were no examples given? If you learn that the integers modulo $7$ form a field and the integers modulo $6$ or modulo $9$ do not, then you're beginning to understand something. Sep 11, 2014 at 16:28
• It might be worth (at some point) finding a good, perhaps informal, book on Galois theory, which is all about extending fields to bigger fields and can be used to prove all sorts of things such as that there is no general solution in radicals to the general quintic and that angles can't be trisected with rule and compass. Unlike many of the examples given (below) you really need the abstract concept of a field to do this tidily and might be motivating? Sep 12, 2014 at 7:46

More or less, yes. You can do arithmetic already with natural numbers (in $$\Bbb N$$), but you have to be careful with some operations:

The equation $$x+a=b$$ does not necessarily have a solution.

So one would prefer working in a ring, like the integers ($$\Bbb Z$$). The preceding equation has always one solution, $$x=b-a$$.

But then, the equation $$ax=b$$ does not have always a solution, in a ring.

So one would prefer working in a field, like rational numbers ($$\Bbb Q$$). The preceding equation has always one solution, $$x=\frac ba$$, if $$a\neq0$$.

So, working in a field is much more friendly. However, there are strange fields too, like finite fields. Or quaternions, for which commutativity does not hold anymore ($$ab\neq ba$$ usually): they form what is called a skew-field.

Just to develop this way of building larger tools a bit more:

But even then some equations have no solutions, like $$x^2-2=0$$ or $$x^2+1=0$$. So you may build larger fields: real algebraic numbers, complex algebraic numbers, real numbers ($$\Bbb R$$), and finally complex numbers ($$\Bbb C$$).

The field of real numbers has a valuable property, very important in analysis: every Cauchy sequence has a limit. That is, $$\Bbb R$$ is complete. It's for example easy to prove that the sequence of rationals $$u_n=1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}$$ is a Cauchy sequence, but does not converge to a rational.

The field of complex numbers is also complete, and have the extra feature that every nonconstant polynomial can be factored in factors of degree $$1$$. It's said to be an algebraically closed field. However, you lose something too, since, unlike the reals, it's not an ordered field: there are total orders on complex numbers, but none is compatible with the arithmetic operations.

As you can guess, while building larger and larger sets, you are more and more generalizing what you mean by a number.

Other genralizations of numbers that form a field include hyperreal numbers and surreal numbers.

The story does not end here, and many mathematical objects that form a field do not resemble numbers very much, apart from having the basic arithmetic rules of a field.

Given any integral domain, you can build its field of fractions:

Another example of a field not related to the notion of number: Hardy fields are fields of equivalence classes of functions.

• Good answer, but I think its a mistake to call the quaternions a field. They're a skew-field (what I would call a division ring), which is not a kind of field. Sep 12, 2014 at 4:14
• @goblin Mostly agreed, and I called it a skew-field too. In France, it's called a "noncommutative field" (corps non commutatif), that is, a field may or may not be commutative. Just a matter of convention. For instance, Wedderburn's theorem states that: every finite field is commutative, whereas in the english Wikipedia it's every finite domain is a field. Sep 12, 2014 at 6:08
• @Jean-ClaudeArbaut It's actually called "body" in French, right? There is "body", and there is "commutative body" (field). Such term usage used to exist in Russian as well, but it disappeared as English terms became more influential: now there're fields and non-commutative fields. Sep 12, 2014 at 6:38
• @Joker_vD Speaking of "body," this also explains why $k$ is a popular letter for fields: it comes from the German Körper. Sep 12, 2014 at 13:08

Yes, at its most basic level, a field is a generalization of the rational numbers. In a field, you can do addition, subtraction, multiplication and division as you could in $\Bbb Q$.

At a deeper level, fields have geometric significance. If you've ever studied a little geometry, then you'd know that there are at least two famous ways to approach geometry: with axioms akin to Euclid's axioms (the synthetic approach) and another way by using vector spaces and equations (the linear algebra approach).

We know that $\Bbb R\times \Bbb R$ (an $\Bbb R$ vector space) can be interpreted as a model of Euclidean geometry, and how its 1-d subspaces represent lines, it's elements represent points etc, and that it satisfies the synthetic axioms of Euclidean geometry.

But what about the other direction? Why can't we start with synthetic axioms and get vector spaces? Well, that's the thing: you can (if you have enough axioms.)

It turns out that if you adopt Hilbert's axioms groups $I-IV$ for plane geometry, then you can systematically build a field $F$, such that $F\times F$ models that plane exactly when the plane satisfies Pappus's theorem.

Another way to ensure the existence of the field is to adopt Hilbert's continuity axiom $V$ called "Archimedes's axiom." It's known that this axiom, in the presence of the others, implies Pappus's theorem, and the resulting field will be an Archimedian ordered field.

You can, of course, do higher dimensional geometry and get vector spaces $F^n$ and so on, as long as you have something like Pappus's theorem or Archimedes's axiom in your axioms.

If you asked me for a rough description of how the axioms of fields translate into geometric ideas for vector spaces, then this is how I would start. Since $F$ is an additive group, $F^n$ is also an additive group, and you can translate any point to another point using addition of vectors. For multiplication, you can use it to scale any vector to another vector in the same 1-dimensional subspace.

Now, this is just the first hint at the geometric nature of fields. Galois theory and then algebraic geometry really take the connection to more extreme altitudes!

• "It turns out that if you adopt Hilbert's axioms groups I−IV for plane geometry, then you can systematically build a field F" Awesome, I did not know any of that. This stuff really ought to get taught at universities. Sep 11, 2014 at 14:59
• @goblin There is a nice book on this: Geometric Algebra, by Emil Artin. Sep 11, 2014 at 15:00
• @goblin Yes, I thought the same thing too, when I learned it :) Hilbert does it himself in Grundlagen der geometrie, and I would definitely second Jean-Claude's suggestion of Geometric algebra. The latter is fresher than the former. Sep 11, 2014 at 15:02
• @goblin I can even argue that division rings are more fundamental: you can make the division ring exactly when the plane satisfies Desargues's theorem. Desargues theorem is automatically satisfied for dimensions $3$ and greater. Sep 11, 2014 at 15:03
• There's another nice book on it -- Hartshorne's Introduction to Projective Geometry (or something like that). Suitable for any math undergraduate who is used to doing proofs, and knows a bit of linear algebra. Sep 11, 2014 at 18:13

Certain items keep showing up in mathematics. When they show up often enough, we give them names.

"Field" is an example of this. You know about the real numbers: they're things that you can add and subtract, multiply and divide. Addition is commutative. So is multiplication. And multiplication distributes over addition.

It turns out that there are other sets that have all these properties, like the Complex Numbers.

There are some other sets that are kind of interesting, like the set {O, I} (those are the letters "O" and "I"), on which I can define addition by the rules $O + O = O, I + I = O, O + I = I + O = I$. That might look silly, but you can check that there's an identity (O), negation (the "negation" of each element is itself!), and addition is commutative. I can then define multiplication: $O*O = O*I = I*O = O; I*I = I$. Together, these have all the properties I mentioned above, and this is called "the finite field with two elements".

Because these properties let you manipulate objects using the rules of algebra you learned in high-school, they're pretty convenient to work with. And if you ONLY use those rules, then whatever you do will work not just for the reals, but for the complexes and various finite fields, and lots of others. So proving things about general fields gives you more power than proving things just about the real numbers.

(Fields generally DON'T have to have a notion of "less than" and "greater than", so when you're working with them, you have to forget that part of high-school. :) )

• I learned a lot from that answer - thank you. About the last paragraphs in parenthesis. Do they not have less than/greater than because the numbers are conceived of modularly (e.g. 1,2,3,1,2,3)? I've seen modular arithmetic arrayed around a circle (like a clock). But in HS we learn number lines - an infinitely large circle appears as a line - so I suppose any part of the circumference of a 'clock' of infinitely many numbers would appear so as well... Saying that to ask - should I think of arithmetic with the reals as modular arithmetic with an infinite quantity of numbers?
– Hal
Sep 11, 2014 at 15:15
• You should probably NOT think of the reals (or even integers) as a "limit" of modular arithmetic, because many important qualiteis do not pass to the limit. For instance, in modular arithmetic, adding any number to itself enough times always gives zero, but in integer arithmetic, that only works for the single number "zero". Sep 11, 2014 at 18:12
• @Hal: Not always. Another thing that would keep a field from having less/greater than is the numbers being conceived of two-dimensionally. Think of $\mathbb{C}$.
– Dan
Sep 12, 2014 at 5:31
• @Hal, the clock model doesn't give us an order, because we would want to say $-1 < 0 < 1$, but for integers mod 2, $-1 = 1$ so $1 < 0 < 1$, and in fact, we'd probably end up saying $a < b < a$ for every $a, b$ and that makes our order absolutely useless, and not even an order.
– jxnh
Sep 12, 2014 at 12:19

Have you discovered yet that finite fields also satisfy the field axioms? Then the arithmetic is rather uncommon. For example $F_p$ is the field of integers modulo $p$, a prime. In $F_7$ we have things like $5\times 4=-1$. But it still has many nice properties.

However, there is a theorem, basically easy, but tedious, to prove that if you also add an order axiom (we have things like $x>0$ that behaves the way you would expect) and a "completeness" axiom (that every let has a least upper bound) then the field must be essentially the ordinary real numbers (ie just the reals relabeled).

Note, for example, that any sensible order must have the property that if you start with 0 and keep adding 1, you get bigger and bigger numbers. That must be false in a finite field.

• Ah that's kind of neat (5 x 4 = -1); I get it.
– Hal
Sep 11, 2014 at 14:51
• But doesnt modular arithmetic have an order too? (I don't know much about it.) I mean 1,2,3,1,2,3 is still ordered in some way. Even though 3 precedes and proceeds 1.
– Hal
Sep 11, 2014 at 14:53
• But the order does not have the nice properties that we want it to have. Just as you cannot order the complex numbers. The most basic property you want is that if you keep adding 1 to itself you get bigger and bigger numbers. But that must be false in a finite field. Sep 11, 2014 at 14:57
• @Hal I'd like to recommend you take a look at this section of the wiki article. It mentions ordered fields, finite fields, etc. But remember: don't fret about any details, don't worry if stuff is way beyond you. Just pretend you've snuck into Google HQ and poke around. Sep 11, 2014 at 15:12

Here's the plainest way I can think of:

A field is a place where you can do division and multiplication commutes.

• Is that the same as saying "A field is a place where addition and multiplication are commutative?"
– Hal
Sep 11, 2014 at 14:56
• No. While it is true that addition in a field is commutative, this is not its "defining" property. What makes a field special is that you can do division with every nonzero element of the field and ab=ba for all a,b. Sep 11, 2014 at 15:04
• I agree in essence, bit this a bit too short. You need to mention addition and subtraction (neither the unit circle nor the positive reals form a field), and division must be exact (polynomials do not form a field). Even if we take the ring axioms as implicit, I'd say "multiplication is commutative, and multiplication by any nonzero value is invertible". Sep 13, 2014 at 14:38
• The addresses “what a field is” simply, but does not seem to tackle “the significance of a field.” Sep 5, 2021 at 21:57