Fields, Characteristics, and Prime Numbers

Jan 16 2008 Published by under Group Theory

When we start looking at fields, there are a collection
of properties that are interesting. The simplest one - and
the one which explains the property of the nimbers that
makes them so strange - is called the
characteristic of the field. (In fact, the
characteristic isn't just defined for fields - it's defined
for rings as well.)

Given a field F, where 0F is the additive
identity, and 1F is the multiplicative identity,
the characteristic of the field is 0 if and only if no
sequence of adding 1F to itself will ever result
in 0F; otherwise, the characteristic is the
number of 1Fs you need to add together to get
0F.

That sounds confusing - but it really isn't. It's just
hard to write in natural language. A couple of examples will
make it clear.

The field of nimbers has characteristic two - because
1+1=0 in the nimbers - so adding to 1s together gives you
0.

The field of numbers modulo-7 is a nice simple field,
with characteristic 7: in mod-7, you need to add 1+1=2;
2+1=3; 3+1=4; 4+1=5; 5+1=6; 6+1=0. So 1+1+1+1+1+1+1=0.

The real numbers have characteristic 0, because adding
one repeatedly will never give you 0. In fact, the
characteristic is 0 for any ordered field O - that
is, a field with a properly defined "&lt" operator (one
where ∀x∈O, ∃y∈O : x<y ∧
∃z∈0 : z<y, and < is total, transitive,
antisymmetric, and antireflexive).

See? Not hard, right?

What makes nimbers so strange is that they're an
infinite set with characteristic two. Our intuitions about
numbers all rely on the structure of an infinite set with
characteristic 0. All of the strange properties of nimbers
can be traced back to the fact that their basic structure is
strange; and the root of that strangeness always traces back
to the field characteristic.

There's an interesting notion also associated with the
characteristic of a field. Like groups could have
sub-groups, fields can have sub-fields: a sub-field of a
field (F,+,×) is a field (G,+,×) where G⊆F.
For every field, there is a minimal subfield - a field where
removing any elements would require removing 1 to maintain
closure. The minimal subfield of a field, F, is called the
prime subfield of F.

Here's where we get something fascinating. There are
numerous limits to the kinds of structures that can fulfill
the field axioms to become proper fields. One of the amazing
ones (at least to me) is that every field has either
characteristic 0 (in which case it's isomorphic to a
sub-field of the complex numbers), or it has a
prime characteristic, and it's prime subfield is
isomorphic to a finite field whose size equals it's
characteristic.

What does this tell us? Well, looking at it
philosophically, it means that the prime numbers are very
deeply embedded in the structure of algebra. Even if we
throw away our standard number system and go into the realm
of the abstract, the numbers that are prime in the basic set
of natural numbers remain fundamental. Both the concept and
the specific values of the prime numbers are deeply embedded
all the way down in the foundations of what we know as
mathematics. Even when we start from scratch, with abstract
sets, and build upwards into mathematical structures, when
we get to the point where we can build things that behave
even just a little bit like numbers, the prime numbers are
inevitable - even though they aren't necessarily prime in
every field!

After all, think about the nimbers! The number 11 for example, is clearly prime in the natural numbers. But in nimbers, it's 8×4: and yet, even if we play with the nimbers, we'll wind up finding out that 11 is prime.

Incidentally, I'm still trying to hack together an implementation of the nimbers. As one of the commenters (Xanthir?) mentioned, it's hard to get them right, and I haven't had much free time to try it. I've got a bunch of Scheme code that's close, but it's still got one recursive case that never terminates. Nimbers are truly annoying.

No responses yet

  • Coin says:

    every field has either characteristic 0 (in which case it's isomorphic to a sub-field of the complex numbers)
    Hmm...
    I'm trying to figure out how to reconcile this statement with the idea of a field whose set cardinality is greater than that of the reals-- such a field clearly couldn't be isomorphic to a subfield of the complexes.
    Is the idea that no such field exists? Or that if such a field existed its cardinality could not be zero?

  • Mark C. Chu-Carroll says:

    Coin:
    A field is a *set* of values. If something has cardinality greater than the reals, it's not a set - it's a proper class.
    It's a bit of a weasel in some ways, but a lot of field theory won't work for proper classes - in the basic logic of the proofs, it relies on ZFC.

  • Antendren says:

    If something has cardinality greater than the reals, it's not a set - it's a proper class.
    This is wrong. Recall Cantor's proof that P(X) has cardinality strictly greater than that of X. Further, ZFC says that if X is a set, then P(X) is a set. R (the reals) is a set. So P(R) is a set with cardinality strictly greater than that of R.
    There are fields of all infinite cardinalities. The correct statement about fields being subfields of C is "every field of characteristic 0 and cardinality no greater than that of the reals is isomorphic to a sub-field of the complex numbers."

  • Joshua Zucker says:

    If you're trying to implement Nimbers in scheme, you might find
    http://home.att.net/~numericana/answer/fields.htm
    useful -- search for the mention of nimbers -- just below the multiplication table, there's a nice suggestion of how to implement the multiplication recursively to run in fairly reasonable time.

  • Joshua Zelinsky says:

    Antendren is correct; there are fields which are of small cardinality that are characteristic zero and are not isomorphic to a subset of C. Consider for example the ring S) that is formed by C adjoint |P(R)| distinct formal variables (where P(R) is the powerset of the real numbers). Take the Field of fractions of S (we can do this because S is Dedekind). This resulting field is small enough that we can represent it as a set but it has cardinality |P(R)| which is greater than the cardinality of C, so this field cannot be isomorphic to a subfield of C.

  • Anonymous says:

    If something has cardinality greater than the reals, it's not a set - it's a proper class.
    It's a bit of a weasel in some ways, but a lot of field theory won't work for proper classes - in the basic logic of the proofs, it relies on ZFC.
    Not only is the first statement wrong (as Antendren pointed out), but the second is as well. I wouldn't swear that nothing ever breaks down, but in general field theory works fine for proper classes. This is usually an irrelevant fact, since it's very rare to find a "naturally occurring" example of a field that's a proper class. The only example I've ever run across is Conway's field of surreal numbers, and even there you can do a version using sets if you'd like (that's just not how Conway did it, because for example he wanted to include all ordinals).
    Strictly speaking, the word "field" doesn't apply to proper classes, since it is almost universally defined in terms of sets, but I don't know of a compelling reason why it couldn't.

  • Eric says:

    every field has either characteristic 0 (in which case it's isomorphic to a sub-field of the complex numbers), or it has a prime characteristic, and it's prime subfield is isomorphic to a finite field whose size equals it's characteristic.

    Wouldn't the numbers modulo-6 make just as good a field as numbers modulo-7, with characteristic 6 (ie neither 0 nor prime)?

  • JC says:

    I would like if it were possible a proof, proving that
    if finite field does not have characteristic zero than i must have charateristic prime.

  • Anonymous says:

    Wouldn't the numbers modulo-6 make just as good a field as numbers modulo-7, with characteristic 6 (ie neither 0 nor prime)?
    Modulo 6, the numbers 2 and 3 are not zero but don't have multiplicative inverses, so you don't get a field.
    I would like if it were possible a proof, proving that
    if finite field does not have characteristic zero than it must have charateristic prime.

    In any field, when the product of two numbers is zero, one of the numbers must be zero itself. Now suppose you have characteristic n. In other words, the sum 1+1+...+1 (with n 1's) vanishes. If n factors, then this sum factors as well. For example, because 2*3=6, (1+1)*(1+1+1)=1+1+1+1+1+1. If n factors, then one of these factors must vanish. However, by definition the characteristic is the smallest possible number of 1's that sum up to zero. That means it must be prime, since any nontrivial factor would be even smaller.
    Notice that this breaks down modulo 6. Mod 6, we have 2*3=0, but neither 2 nor 3 vanishes.

  • Darin Brown says:

    Mark, I find it amazing that you continue to spew patent nonsense, confusing explanations, and constipated thought week after week, month after month, year after year, like
    "For every field, there is a minimal subfield - a field where removing any elements would require removing 1 to maintain closure."
    COMMENT: Why don't you just say "it's the smallest subfield containing 1"?
    "What makes nimbers so strange is that they're an infinite set with characteristic two."
    COMMENT: The nimbers are strange for many reasons, but that isn't really one of them. There are much less exotic fields, e.g. (Z/2Z)(x), the rational function field over Z/2Z, which are infinite and have characteristic 2.
    "Even when we start from scratch, with abstract sets, and build upwards into mathematical structures, when we get to the point where we can build things that behave even just a little bit like numbers, the prime numbers are inevitable - even though they aren't necessarily prime in every field!... After all, think about the nimbers! The number 11 for example, is clearly prime in the natural numbers. But in nimbers, it's 8×4: and yet, even if we play with the nimbers, we'll wind up finding out that 11 is prime."
    COMMENT: I have no idea what this is supposed to mean. Fields don't have prime elements, a prime element is a non-zero non-unit, and every non-zero element of a field is a unit.
    "In fact, the characteristic is 0 for any ordered field O - that is, a field with a properly defined 'LT' operator ["less than", couldn't get the symbol to process] (one where ∀x∈O, ∃y∈O : x LT y ∧ ∃z∈0 : z LT y, and LT is total, transitive, antisymmetric, and antireflexive)."
    COMMENT: I can't even make sense of this so-called "definition". Assuming your "O"'s [letter O] and "0"'s [numeral 0] are the same, first of all, "LT" is not an "operator", it's called a "relation" (you seem to be constantly confused about even the most elementary mathematical terms). Also, I hate to break it to ya, but every total relation is trivially necessarily reflexive, not "antireflexive", whatever the hell that means (I think you mean "irreflexive"??) And what the hell does "∀x∈O, ∃y∈O : x LT y ∧ ∃z∈0 : z LT y" mean?? "For all x in O, there is a y in O such that (x LT y and there is a z in O such that z LT y)"?? Doesn't the "x" in "x LT y" itself satisfy "there is a z in O such that z LT y"? And is anyone else bothered by the fact that Mark's "definition" of an ordered field makes no reference to axioms combining the order and field operations??
    "every field has either characteristic 0 (in which case it's isomorphic to a sub-field of the complex numbers)"
    COMMENT: Simply false. I think you're missing a hypothesis somewhere.
    "If something has cardinality greater than the reals, it's not a set - it's a proper class."
    COMMENT: I can understand someone goofing on the previous statement, but to make this statement demonstrates profound mathematical illiteracy.
    "Arrow composition has to meet two properties, which are going to look pretty familiar from group theory: associativity, and identity. Associativity says that aº(bºc)=(aºb)ºc; and identity says that that for any arrow f:a→b, there are arrows 1a and 1b such that 1bºf = f = fº1a."
    COMMENT: I think you really mean to say that for every object a, there is an identity arrow 1a: a→a such that for every arrow f:a→b, 1bºf = f = fº1a. But as usual, you got a bit tongue-tied.
    "For an example that's important for looking at group theory related topics, we say than an arrow is iso (i.e., it is a iso-morphism) if it's reversable - which in category theoretic terms means that if f is an iso-arrow, then there is an arrow f-1 such that fºf-1=1."
    COMMENT: Again, I think you mean, f:a→b is iso if there is f-1:b→a such that f-1ºf=1a and fºf-1=1b. You must have been tongue-tied again.
    "That's a common error - I can't count how many CS papers I've seen that show a diagram without proving that it actually commutes!"
    COMMENT: Or maybe you just don't understand what you're reading. Not every diagram commutes. That's why we call the ones that DO commute commutative diagrams. Go figure.
    "For an example of a diagram, there's a type of arrow between two objects A and B called a principle morphism. The idea of a principle morphism is that a morphism M from A to B is principle if and only if every self-arrow of A (or endomorphism, an arrow from A to A), composed with an arrow from A to B yields M. As a very visual thinker, I find that paragraph hard to follow. But in category theory, we can say that m is principle if and only if for every arrow x from A to A, and for every arrow y from A to B, the diagram to the right commutes."
    COMMENT: First of all, it's "principal" morphism not "principle" morphism. Second of all, you fucked up the picture whose arrows unfortunately go the wrong direction from your definition.
    "I was just perusing my stats on sitemeter - and to my amazement, I discovered that Good Math/Bad Math had its 2 millionth view this weekend. 2 million pageviews! I never dreamt that this little blog would ever see a number like that."
    COMMENT: Mark, I'm surprised you can count that high.
    What really gets me is that someone who is so OBVIOUSLY mathematically ILLITERATE (as is obvious to anyone with a real math degree) is able to attract such a cult-like following and praise on the internet. Friggin unbelievable.
    darin

  • Mark C. Chu-Carroll says:

    For those who don't remember him, Darin Brown is an AIDS denialist, who's pissed off at pretty much everyone at ScienceBlogs for debunking his incredibly idiotic anti-HIV arguments. If you want to see both why Darin dislikes me so much, and the brilliance of his allegedly qualified mathematical opinions, take a look here.
    While criticizing me for things like mis-spelling "principal" (which I admit was a stupid mistake...), Darin makes some truly astonishingly innumerate claims. One of his major arguments muddles the difference between aggregate
    correlations and individual correlations.
    Specifically, he wants to dispute HIV as the cause of AIDS.
    Now, if we look at statistics, we can see that in the population of HIV+ individuals as a whole, increased viral load correlates amazingly well with onset of symptoms. But taken on an individual level, the viral load isn't a terrible good predictor of the onset of symptoms for an individual. According to Darin, that indicates that the correlation between viral load in aggregate must be false.
    To anyone with the least clue about statistics, this is obvious nonsense. The fact that he's willing to stand by that argument indicates one of two things. Either he doesn't understand anything about statistics, or else he's a deliberate liar. In either case, he's not someone I take seriously.

  • Yuri_KYAD says:

    I have never studied this subject before, but I find it interesting. Would it be at all possible for you to show a quick proof of why the char(F) is either 0 or a prime number?

  • Sketch of proof.
    Consider a field F with multiplicative identity 1. Now consider the numbers 2=1+1, 3=1+1+1, 4=1+1+1+1, and so on.
    Either these numbers are all different, in which case F is said to have characteristic 0, or two of them will be equal.
    In the latter case, it is straightforward to show that, for some number p, we have 1+1+1...+1 (p times) = 0. If p is chosen to be as small as possible, then p will always turn out to be a prime, and we say that F has characteristic p.
    For p a prime, the finite field GF(p^n) has characteristic p.