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Is it possible to define addition and/or multiplication on the set of

a) natural numbers (including $0$: $0,1,2,3,...$)

b) integers $(..., -2, -1, 0, 1, 2, ...)$

in such way that they will become fields?

Thanks in advance.

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3 Answers 3

up vote 32 down vote accepted

Let $X$ be any countably infinite set (such as $\mathbb{Z}$ or $\mathbb{N}$). Choose any bijection $f:X\to\mathbb{Q}$. Define operations $\oplus,\otimes$ on $X$ by $$a\oplus b=f^{-1}(f(a)+f(b)),\qquad a\otimes b=f^{-1}(f(a)\cdot f(b))$$ where $+$ and $\cdot$ are the operations on $\mathbb{Q}$. Then these operations make $X$ into a field, indeed one that is isomorphic to $\mathbb{Q}$.

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7  
For an even greater generality, you can replace $\Bbb Q$ with any other countable field, for example $\bar{\Bbb F_2}$. –  Asaf Karagila May 22 '13 at 23:00
    
Unfortunately i don't lnow what $\bar{\Bbb F_2}$ is. –  Igor May 23 '13 at 8:07
    
@Igor: The ring of integers modulo $2$ is a field (see the Wikipedia article on finite fields). It is usually denoted $\mathbb{Z}/2\mathbb{Z}$, but also as $\mathbb{F}_2$ depending on the context. $\overline{\mathbb{F}_2}$ is the algebraic closure of the field $\mathbb{F}_2$. –  Zev Chonoles May 23 '13 at 8:09
    
It's clear now. Thank you. –  Igor May 23 '13 at 9:39
    
@AsafKaragila since in derivation there is inverse wouldn't you need the bijections (i.e. infinite countable field)? Otherwise I don't the identity can be reconstructed even bending the $f^{-1}$ to mean one on numbers mapped to this value - take $a, b \neq 0$. As $a^{-1}$ exists the $f(a) = 1$. Therefore $a \otimes b = g(1)$ for any such $a, b$. Therefore there is no identity. –  Maciej Piechotka May 23 '13 at 9:52

Apart from choosing arbitrary bijections to countable fields, there is actually a quite natural field structure on the set $\mathbb{N}$ which is motivated by game theory, namely nim addition and nim multiplication. See the wikipedia article on nimbers, or better directly consult Conway's On numbers and games. This field is the union (or rather colimit) of the finite fields $\mathbb{F}_{2^{2^n}}$

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Hint: index a countably infinite field $\,F\,$ by any bijection $\,f\,:\,\Bbb N\to F.\,$ To perform a field operation on the indices, dereference them to obtain the corresponding field elements, then perform the field operation, then index the result, e.g. $\: n + m = f^{-1}\,(f(n) + f(m)),\:$ and similarly for other operations. This "pulled back" field structure on $\,\Bbb N\,$ is isomorphic to the field $\,F\,$ since, by design, the map $\,f\,$ is a ring hom: $\,f(n+m) = f(n)+f(m), \,$ and similarly for all other operations. This is how many algebraic structures are represented in computers, where the indices are addresses of the elements in memory or an array.

The same idea works to pull back (along a bijection) the operations of any algebraic structure, inducing an isomorphic algebraic structure on any set of the same cardinality. This is a trivial special case of transport of structure.

Here are some more interesting examples. This answer contains an instructive easily-grasped finite example, where $\, n\ {\rm mod}\ 7\:$ is labeled/indexed by $\, n - 3.$

See also this answer, where the reason behind a trick boils down to the fact that an operation on rationals is associative and commutative, simply because it is addition or multiplication transported to rationals labeled by inverses or increments.

A beautiful and important example is transporting the class group structure of quadratic fields from ideals to primitive binary quadratic forms - which greatly simplifies Gauss's presentation of composition of forms. For example, Gauss's proof in Disq. Arith. of the associativity of composition of forms involved many pages of abstruse calculations. But nowadays we deduce it immediately since it is simply transported ideal multiplication, which is clearly associative.

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"This is how many algebraic structures are represented in computers, where the indices are addresses of the elements in memory or an array." Very interesting, I didn't know that! –  Martin Brandenburg May 22 '13 at 23:59

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