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To make sure that we are talking about the same, I would like to post the relevant definitions I know first.

Definitions:

A pair $(G, +)$ where $G$ is a set and

$+: G \times G \rightarrow G$

is called a commutative group if it has all of the following three characteristics:

  • (associativity): $\forall x,y,z \in G: (x+y)+z=x+(y+z)$
  • (identity element): $\exists e \in G \forall x \in G: e + x = x = x +e$
  • (inverse elements): $\forall x \in G \exists x^{-1} \in G: x^{-1} + x = e = x + x^{-1}$
  • (commutativity): $\forall x,y \in G: x+y=y+x$.

A triple $(\mathbb{K}, +, \cdot)$, where $\mathbb{K}$ is a set and

$+: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$

$\cdot: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$

is called a field if it has all of the following three characteristics:

  • $(\mathbb{K}, +)$ is a commutative group
  • $(\mathbb{K} \setminus \{0\}, \cdot)$ is a commutative group
  • distributive properties: $\forall x,y,z \in \mathbb{K}: x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ and $\forall x,y,z \in \mathbb{K}: (y+z) \cdot x = (y \cdot x) + (z \cdot x)$

$\mathbb{N^+} := 1, 2, 3, 4, 5 ...$ (all positive integers)

$\mathbb{N}_0 := \{0\} \cup \mathbb{N}$ (the natural numbers with zero)

$\mathbb{N}$ can be both: $\mathbb{N^+}$ or $\mathbb{N}_0$.


My question: Does a field exist that has $\mathbb{N}$ as its set?

I know that $(\mathbb{Q}, +, \cdot)$ is a field and I know that $(\mathbb{N}, + )$ is only a commutative semigroup.

But maybe it's possible to define two mappings $\circ, *$ for $\mathbb{N}$ in such a way that $(\mathbb{N}, \circ, *)$ is a field.


Related:

The smallest field containing the integers is the field of rational numbers.

Source: en-Wikipedia: Integer

I don't know if this is true. If somebody can prove (or at least scribble the proof of) the quote it would be a good answer, I guess.

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1  
Brian and Gerry beat me to the punch, but to supplement their answers, the Wikipedia page quote probably means "the smallest field containing the integers as a subring (in the usual sense) is $\mathbb Q$"; that's because one way to define $\mathbb Q$ is as the "field of fractions" of $\mathbb Z$, so this is true essentially by definition. –  user1306 Sep 3 '12 at 3:56
    
Your definition of "commutative group" (more often called an "abelian group") omits the commutativity condition: $(\forall x,y\in G)(x+y = y+x)$. –  jwodder Sep 3 '12 at 16:31

3 Answers 3

up vote 18 down vote accepted

$\Bbb Q$ is a countably infinite set, so there is a bijection $\varphi:\Bbb N\to\Bbb Q$. For $m,n\in\Bbb N$ define

$$m\circ n=\varphi^{-1}\Big(\varphi(m)+\varphi(n)\Big)$$

and

$$m\ast n=\varphi^{-1}\Big(\varphi(m)\cdot\varphi(n)\Big)\;;$$

then $\langle\Bbb N,\circ,\ast\rangle$ is a field isomorphic to $\langle\Bbb Q,+,\cdot\rangle$.

Added: For your last question, the proof is to note that every field $F$ containing a copy of $\Bbb N$ with the usual operations must also contain a copy of $\Bbb Q$ with the usual operations. In order for $F$ to be an additive group, it must contain an element $-n$ for each $n\in\Bbb N$, and in order for it to be a field, it must contain an element $n^{-1}$ for each $n\in\Bbb Z^+$. Once you have all of the integers and all of the reciprocals of positive integers, it’s easy to see that you must have all of the rationals just to get closure under multiplication.

Note that I’m slurring over a lot of details about isomorphic copies in order to give the fundamental idea.

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Great answer (+1 and accept). Can you explicitly note those mappings $\circ$ and $*$? I know that $g: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$, defined as $g(n,m) := n+\frac{1}{2}(n+m-1)(n+m-2)$ is a bijective mapping, but I don't know a bijective function $\varphi: \mathbb{N} \rightarrow \mathbb{Q}$ and $\varphi^{-1}$. I know, it's not necessary, so I accepted your answer. But I simply would like to know how such a function $\varphi$ would look like. –  moose Sep 3 '12 at 4:39
1  
@moose: Bijections between $\Bbb N$ and $\Bbb Q$ are a bit messy. Given injections $\Bbb N\to\Bbb Q$ and $\Bbb Q\to\Bbb N$, this proof of the Cantor-Schröder-Bernstein theorem can be used to construct an explicit bijection. This pairing function can be used to construct an explicit injection $\Bbb Q\to\Bbb N$. –  Brian M. Scott Sep 3 '12 at 4:51

Sure, one can transport the structure of an algebraic structure to any set of the same cardinality, given any bijection between the two underlying sets. When the target set is $\Bbb N$ we can think of the naturals as indices (or computer memory addresses) of the field elements. To perform a field operation on the integer indices, dereference them to obtain the field elements, perform the field operation, then return the integer index of the result, e.g. $\: n + m = i\,(i^{-1}(n) + i^{-1}(m))\:$ where $\:i^{-1}(n)\:$ is the field element with index $\,n\,$ and $\:i(a)\:$ is the index of the field element $a,\:$ and analogously for all other operations.

Note that this is simply a procedural reinterpretation of $\:j(n+m) = j(n) + j(m),\,\ j = i^{-1}.\:$ Since this holds true for all operations, the map $\,j\,$ is not only a bijection of sets, but, further, a ring/field isomorphism - as desired.

Remark $\ $ This is intimately connected to the notion of isomorphism for algebraic structures. From an algebraic perspective it is crucial to forget about any internal structure possessed by the elements of the structure. Such internal structure is an artefact of the particular construction employed. Such representational information is not an essential algebraic property. It matters not whether ring elements are represented by (co)sets, or by sequences, matrices, functions, differential or difference operators, etc. Instead, all that matters algebraically are how the elements are related to one another under the operations of the structure. Thus the isomorphism type of a ring depends only upon its additive and multiplicative structure. Rings with the same addition and multiplication tables are isomorphic, independent of whatever 'names', or internal structure the elements might possess (possibly set-theoretic, or data-structures in programming languages, etc). We are free to "rename" the elements by transporting the algebraic structure along any bijection to any set of the same cardinality. This is the essence of the abstraction encapsulated in the notion of an algebraic structure.

For some further examples of transporting structure, see this simple example where $\rm\:n\ mod\ 7\:$ is relabeled by $\rm\:n - 3,\:$ and see this post which views normal forms from this perspective, e.g. in $\,\Bbb Z/7\Bbb Z,\,$ the integers mod $7,\,$ one frequently labels the coset $\rm\: n + 7\,\Bbb Z\:$ by the integer $\rm\:n\ mod\ 7,\:$ which effectively transports the quotient ring structure structure to the set $\rm\{0,1,2,3,4,5,6\}$ of least positive representatives of the congruence classes. See also here, 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.

Note: Some authors reserve the term "transport of structure" for less trivial examples, e.g. transporting the class group structure of quadratic fields from ideals to primitive binary quadratic forms - which greatly simplifies Gauss's presentation of composition of binary quadratic forms.

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The smallest field containing the integers with the usual operations on the integers is the rationals, since any field containing the integers must contain their quotients (other than dividing by zero), thus, must contain the rationals.

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