Given a field $k$, show that an injection $\mathbb{Z} \rightarrow k$ implies an injection $\mathbb{Q} \rightarrow k$ I am trying to prove that the existence of an injection $\mathbb{Z} \rightarrow k$ implies the existence of an injection $\mathbb{Q} \rightarrow k$, where $k$ is a field.
I started by defining an equivalence relation $\sim$ on $\mathbb{Z} \times \mathbb{Z}_+$ such that $(a, b) \sim (c, d) \iff ad = bc$. Then I have a bijection from the set of equivalence classes of $\sim$ to $\mathbb{Q}$, i.e $f : [\mathbb{Z}\times\mathbb{Z}_+]_\sim{} \rightarrow \mathbb{Q}$.
This is where I get stuck: if $g : \mathbb{Z} \rightarrow \mathbb{Q}$ is the injection, I tried to construct injection $h : [\mathbb{Z}\times\mathbb{Z}_+]_\sim{} \rightarrow k$ from $g$. The only thing I could think of was $h([(a, b)]) = g(a)g(b)^{-1}$, but I can't prove that $g(b) \ne 0$ or that $h$ is an injection without assuming $g$ is a ring homomorphism. If $h$ is an injection, then I have the injection $h \circ f^{-1} : \mathbb{Q} \rightarrow k$.
Am I approaching this the right way? Is this provable at all?
EDIT: For context, this is question was asked in an abstract algebra class, but the question did not explicitly state that the function $\mathbb{Z} \rightarrow k$ is a ring homomorphism. To answer it as posed, I would have to construct a bijection from $\mathbb{Q} \rightarrow \mathbb{Z}$. This question has been answered here.
 A: I wonder if the question is correctly stated and correctly tagged.
The tags abstract-algebra, ring-theory and field-theory seem to imply that you are talking about injective ring homomorphisms rather than just injections. In this case, your approach is correct (but perhaps a bit overcomplicated).
However you explicitly stated "without assuming $g$ is a ring homomorphism". If this means that you are talking about set-theoretic injections, then the answer is to note that there is a bijection between $\Bbb Q$ and $\Bbb Z$. This would be more suitable as an exercise in set theory.
A: There are two different interpretations of injections here.

*

*$\mathbb{Z}\to k$ is an injective ring homomorphism.  Then your $\mathbb{Q}\to k$ is an injective ring (or field) homomorphism.


*$\mathbb{Z}\to k$ is an injection of underlying sets.  Then you get $\mathbb{Q}\to k$ an injection of underlying sets by composing with an injection $\mathbb{Q}\to\mathbb{N}\hookrightarrow\mathbb{Z}$ (since $\mathbb{Q}$ is countable).
