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When constructing $\mathbb{Q}$ as equivalence classes containing pairs of integers, a natural inclusion of $\mathbb{Z}$ in $\mathbb{Q}$ arises: $$f: \mathbb{Z} \to \mathbb{Q}, \; n \to \frac{n}{1}. $$ This mapping is not surjective, but it is injective, and the function is well-defined. My question is: is injectivity required for us to conclude that $\mathbb{Z} \subset \mathbb{Q}$?

It seems to me that this isn't the case. I could assign two distinct integers to the same rational, and $\mathbb{Z}$ would still be embedded in $\mathbb{Q}$. I could assign every $n \in \mathbb{Z}$ to the rational $\frac{0}{1}$ and still prove the result.

Is this correct?

As a side question, when we talk about an inclusion map, must I map $\mathbb{Z}$ to $\mathbb{Q}$ or could I instead map $\mathbb{Q}$ to $\mathbb{Z}$ by the rule $\frac{n}{1} \to n$? In this case, perhaps it wouldn't work because the function would not be well-defined for all $\mathbb{Q}$. I'm interested in particular on how we defined an inclusion map.

  • When you say "$\Bbb Z$ would still embedded in $\Bbb Q$" you are implicitly thinking to an injective map from $\Bbb Z$ to $\Bbb Q$. You can assign two different integers to the same rational, but this map is not the embedding you have in mind. – InsideOut Jan 24 '20 at 21:30
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    Assigning the rational $\frac01$ to every $n$ doesn' t make $\mathbf Z$ isomorphic to subring of $\mathbf Q$. Furthermore, in commutative ring theory, ring homomorphisms have to map the unit onto the unit. – Bernard Jan 24 '20 at 21:32

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You can do whatever you like. However an embedding is by definition an injective function (with some other properties, depending on the context). Assigning $n\mapsto \frac{0}{1}$ is not an embedding. It is just a constant function. Since you can't distinguish image of say $2$ and $3$ then how can you say that these represent integers? For something to be a set of integers it has to be infinite at least.

Also $\frac{n}{1}\mapsto n$ is not even a function on $\mathbb{Q}$. Not every rational is of the form $\frac{n}{1}$, e.g. $\frac{1}{2}$. So this is not well defined.

Therefore yes: being injective is crucial.

freakish
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