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I've read the axioms of a field. To understand the generality of the axioms, could you give me an example of a field which is not (isomorphic to) a subset of complex number (with or without modulus operations).

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Something worth noting is that $\mathbb{Q}$ (the rationals) or $\mathbb{Z}_p$ (integers mod a prime) are always lurking around somewhere. Fields of characteristic $0$ must contain an isomorphic copy of $\mathbb{Q}$ and those of characteristic $p$ must contain a copy of $\mathbb{Z}_p$. That might help you understand why all of the examples given seem to be built out of "regular old numbers" or "stuff mod $p$". – Bill Cook Mar 21 '12 at 20:16
Pretty interested. That exactly addresses my question. I guess I have to read up more of the basics :) – Gerenuk Mar 21 '12 at 22:24
up vote 13 down vote accepted
  1. Any field of positive characteristic. For example, $\mathbb{F}_2 = \{0,1\}$, or more generally $\mathbb{F}_p$ for prime $p$ (integers modulo $p$ with modular addition and multiplication); $\mathbb{F}_{p^n}$, the Galois Field with $p^n$ elements (the splitting field over $\mathbb{F}_p$ of the polynomial $x^{p^n}-x$). Since they have finite characteristic, they cannot be isomorphic to any subfield of $\mathbb{C}$. There are also infinite fields of positive characteristic, such as $\mathbb{F}_p(x)$, the field of rational functions with coefficients in $\mathbb{F}_p$.

  2. Let $X$ be a set of indeterminates of cardinality greater than $\mathfrak{c}$, the cardinality of the continuum. Then $\mathbb{C}(X)$, the field of rational functions on $X$ with coefficients in $\mathbb{C}$ are a field of cardinality $\mathfrak{c}|X| =|X|\gt\mathfrak{c}=|\mathbb{C}|$, so it cannot be isomorphic to a subfield of $\mathbb{C}$.

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Thanks for the examples! Maybe I didn't put it clear enough. I do also exclude modulus operation (giving "simple" finite fields). Is your example 1. just some kind of modulus operation? Sorry, if this question is silly :) I don't know much about abstract algebra yet, and I suppose I will have to read up all of the answers given :D – Gerenuk Mar 21 '12 at 19:32
@Gerenuk: I don't know what you mean by "modulus operation", sorry; to me, if you are talking about complex numbers and you say "modulus", I assume you mean $z\mapsto |z|$. – Arturo Magidin Mar 21 '12 at 20:14
Sorry for being imprecise. I meant that modulus arithmetics with integer numbers "doesn't surprise me" either. So if examples are isomorphic to this, "it's not new to me". Btw, surreal numbers form a field? That's different? – Gerenuk Mar 21 '12 at 20:32
@Gerenuk: The fields with a prime number of elements are indeed obtained that way; on the other hand, the field with, for example, $4$ elements is not. Yes, the surreal numbers form a field, as do the nonstandard real numbers (which include infinitesimals and so cannot be a subfield of the complex numbers). – Arturo Magidin Mar 21 '12 at 20:37

Let $I$ be a set of cardinality greater than the cardinality of the reals (or equivalently the complex numbers). For every $i$ in $I$, invent a symbol $x_i$. Let $F$ consist of all ratios of polynomials in the $x_i$. By polynomial we mean say polynomial with real coefficients. Any polynomial will "mention" only finitely many of the $x_i$. Add and multiply in the natural way.

Remark: The "with or without modulus operations" seems to allow the possibility of taking a subring of the complex numbers, defining an equivalence relation on it ("modulus") and forming the quotient structure. It may be very difficult to describe what sorts of fields can arise in this way. That's why I took the safe route and made my $F$ so big that it could never be produced by this kind of process!

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The field of p-adic numbers is not canonically isomorphic to a subset of $\mathbb C$.

In fact, the claim that the field of p-adic numbers is isomorphic to a subset of $\mathbb C$ is equivalent to some form of choice, therefore not provable in ZF, so p-adic numbers and their completions might also qualify as an example. See e.g.

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$\mathbb Z/p\mathbb Z$ is a (finite) field for every prime $p$. It is not isomorphic to a subfield of $\mathbb C$ since it has characteristic $p$, i.e. in $\mathbb Z/p\mathbb Z$ we have $1+\ldots+1 = 0$. In $\mathbb C$, however, such an equation is not true.

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