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Q&A Are there multiple (small class) extensions of the naturals?

posted 6mo ago by clemens‭  ·  edited 6mo ago by clemens‭

Answer
#4: Post edited by user avatar clemens‭ · 2026-03-08T02:02:42Z (6 months ago)
  • No, at least in some sense. Any structure that plays the "role" of the integers will *be* either the integers or a homomorphic image thereof (i.e. $\mathbb{Z}/n\mathbb{Z}$ for some $n$).
  • This is because the integers are the [free](https://ncatlab.org/nlab/show/free+functor#Examples) group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (More generally, there is a natural correspondence between set homomorphisms $X → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a [free/forgetful adjunction](https://bartoszmilewski.com/2016/06/15/freeforgetful-adjunctions/).)
  • More concretely: _any group containing the naturals will contain the integers_. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every *morphism*, whether injective or not, has to contain the integers!<sup>2</sup>
  • <sup>1</sup>We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a *universal property* of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)
  • <sup>2</sup> That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.
  • No, at least in some sense. Any structure that plays the "role" of the integers will *be* either the integers or a homomorphic image thereof (i.e. $\mathbb{Z}/n\mathbb{Z}$ for some $n$).
  • This is because the integers are the [free](https://ncatlab.org/nlab/show/free+functor#Examples) group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (Much as there is a natural correspondence between set homomorphisms $X → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a [free/forgetful adjunction](https://bartoszmilewski.com/2016/06/15/freeforgetful-adjunctions/).)
  • More concretely: _any group containing the naturals will contain the integers_. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every *morphism*, whether injective or not, has to contain the integers!<sup>2</sup>
  • <sup>1</sup>We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a *universal property* of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)
  • <sup>2</sup> That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.
#3: Post edited by user avatar clemens‭ · 2026-03-07T19:44:53Z (6 months ago)
  • No, at least in some sense.
  • This is because the integers are the [free](https://ncatlab.org/nlab/show/free+functor#Examples) group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (More generally, there is a natural correspondence between set homomorphisms $X → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a [free/forgetful adjunction](https://bartoszmilewski.com/2016/06/15/freeforgetful-adjunctions/).)
  • More concretely: _any group containing the naturals will contain the integers_. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every *morphism*, whether injective or not, has to contain the integers!<sup>2</sup>
  • <sup>1</sup>We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a *universal property* of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)
  • <sup>2</sup> That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.
  • No, at least in some sense. Any structure that plays the "role" of the integers will *be* either the integers or a homomorphic image thereof (i.e. $\mathbb{Z}/n\mathbb{Z}$ for some $n$).
  • This is because the integers are the [free](https://ncatlab.org/nlab/show/free+functor#Examples) group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (More generally, there is a natural correspondence between set homomorphisms $X → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a [free/forgetful adjunction](https://bartoszmilewski.com/2016/06/15/freeforgetful-adjunctions/).)
  • More concretely: _any group containing the naturals will contain the integers_. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every *morphism*, whether injective or not, has to contain the integers!<sup>2</sup>
  • <sup>1</sup>We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a *universal property* of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)
  • <sup>2</sup> That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.
#2: Post edited by user avatar clemens‭ · 2026-03-07T19:42:30Z (6 months ago)
  • No, at least in some sense.
  • This is because the integers are the [free](https://ncatlab.org/nlab/show/free+functor#Examples) group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (More generally, there is a natural correspondence between set homomorphisms $X → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a [free/forgetful adjunction](https://bartoszmilewski.com/2016/06/15/freeforgetful-adjunctions/)/)
  • More concretely: _any group containing the naturals will contain the integers_. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every *morphism*, whether injective or not, has to contain the integers!<sup>2</sup>
  • <sup>1</sup>We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a *universal property* of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)
  • <sup>2</sup> That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.
  • No, at least in some sense.
  • This is because the integers are the [free](https://ncatlab.org/nlab/show/free+functor#Examples) group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (More generally, there is a natural correspondence between set homomorphisms $X → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a [free/forgetful adjunction](https://bartoszmilewski.com/2016/06/15/freeforgetful-adjunctions/).)
  • More concretely: _any group containing the naturals will contain the integers_. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every *morphism*, whether injective or not, has to contain the integers!<sup>2</sup>
  • <sup>1</sup>We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a *universal property* of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)
  • <sup>2</sup> That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.
#1: Initial revision by user avatar clemens‭ · 2026-03-07T19:42:13Z (6 months ago) 
No, at least in some sense.

This is because the integers are the [free](https://ncatlab.org/nlab/show/free+functor#Examples) group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (More generally, there is a natural correspondence between set homomorphisms $X  → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a [free/forgetful adjunction](https://bartoszmilewski.com/2016/06/15/freeforgetful-adjunctions/)/)

More concretely: _any group containing the naturals will contain the integers_. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every *morphism*, whether injective or not, has to contain the integers!<sup>2</sup>


<sup>1</sup>We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a *universal property* of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)

<sup>2</sup> That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.