In what mathematically rigorous sense does $ \mathbb{Q}$ extend $\mathbb{Z}$?

Question

I was trying to understand rigorously what the word "extends" means in this context, pin it down formally with the correct mathematical language.

First, let me explain some of my thoughts and the context of what I expect for my answer. I was learning that $ \mathbb{Q} = \{ \frac{p}{q} : q \neq 0 \}$ and where $ \frac{p}{q}$ is the equivalence class $ [(p,q)] = \frac{p}{q} = \{ (a, b) \mid a, b, p, q \in \mathbb{Z}, aq = bp, b \neq 0, q \neq 0\}$. This is more or less the context of my question.

Anyway, it makes intuitive sense to me that the set of equivalence classes that should behave the same way as the integers is:

$$\{ \frac{n}{1} \mid n \in \mathbb{Z} \}$$

but its not exactly the same as $\mathbb{Z}$, since isn't a set of ordered tuples nor equivalence classes. Therefore, I was wondering, is the reason $\mathbb{Q}$ is considered an extension of the integers $\mathbb{Z}$ is because there exists an isomorphism between a subset of $\mathbb{Q}$ and the integers (that retains the algebraic structure of $\mathbb{Z}$)?

i.e. the isomorphism being the map (from equivalence classes of fraction to integers):

$ f(\frac{n}{1}) = n$

it seems intuitive, to me but how do I show that the algebraic structure is actually retained?

Answer 1

You are right about what it means that $\Bbb Q$ extends $\Bbb Z$. $\Bbb Q$ contains an isomorphic copy of $\Bbb Z$, namely the one you point out. To show that the algebraic structure, you only need to show that your map is indeed an isomorphism.

When we say $\Bbb Q$ extends $\Bbb Z$ it is an abuse of language used for convenience.

Answer 2

Your intuition is correct about the isomorphism between $\Bbb Z$ and a subset of $\Bbb Q$. To show the algebraic structure is maintained, you need to show that addition and multiplication work the same way. Let $\times$ and $+$ represent the operations in $\Bbb Z$ and $\oplus$ and $\otimes$ represent the operations in $\Bbb Q$. Then you need to show that $f(\frac n1 \oplus \frac m1)=f(n)+f(m)$ and $f(\frac n1 \otimes \frac m1)=f(n)\times f(m)$ Note that the operations have the proper sort of operands. In fact $\oplus$ and $\otimes$ are defined to make this true. Presumably you have already shown that $\oplus$ and $\otimes$ are well defined, so whichever elements of an equivalence class you choose for the operands, the result is in the same class.

Answer 3

The extension of $\mathbb Z$ to $\mathbb Q$ is the paradigmatic example of the extension of an integral domain to its field of fractions, and a motivating example for extending a ring by adding the multiplicative inverses of some of its elements.

In these more general cases there is a construction involving ordered pairs, and an injection $z\to (z,1)$.

You are right to say these things are not identical, but in the cases we care most about they are isomorphic. Isomorphism often means "we don't care so much about the details of the elements provided they behave the same way" i.e. it doesn't matter in the end of we treat things as the same (even though they strictly aren't).

Answer 4

If you really want to boggle your mind, $\Bbb Z$ itself is an equivalence relation on ordered pairs of natural numbers, where we define:

$(m,n) \sim (m',n')$ if $m + n' = m' + n$.

We define addition, like so:

$[(m,n) + [(m',n')] = [(m+m',n+n')]$.

This gives a monoid isomophism $(\Bbb N,+) \to (\Bbb Z,+)$ given by:

$m \mapsto [(m,0)]$.

We think of $[(m,n)]$ as what we get by subtraction: $m - n$, so $m$ is the "positive part" and $n$ is the "negative part".

It is instructive to verify that this indeed yields an abelian group.