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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.

Ajay Tatachar
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  • Given that it's an abstract algebra class, perhaps the intent of the question is that if $\mathbb{Z}\to k$ is an injective ring map, then $k$ contains an isomorphic copy of $\mathbb{Z}$, hence must contain the quotient field $\operatorname{Quot}(\mathbb{Z})\simeq\mathbb{Q}$ as a subfield, which is to say there is an injective field map $\mathbb{Q}\to k$. This actually makes use of the hypothesis that $k$ is a field. – Ben West Aug 08 '20 at 02:14
  • For what it's worth, this is an immediate result of the universal property of localization. – Ayman Hourieh Aug 08 '20 at 11:30

2 Answers2

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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.

WhatsUp
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    Apologies -- I am new to this forum. The question I was trying to answer did not explicitly state that the function $\mathbb{Z} \rightarrow k$ is a ring homomorphism, so I interpreted it as a set-theoretic injection. I will re-tag the question. – Ajay Tatachar Aug 08 '20 at 01:53
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    It then depends on the context. If the question comes from algebra context such as ring theory or field theory, then it probably wants to say that $g$ is a ring homomorphism, which makes it a quite standard exercise. Sometimes people simply say "injections" when they mean "injective homomorphisms". – WhatsUp Aug 08 '20 at 01:55
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    Thanks for the response, I just updated the question -- it probably meant 'ring homomorphism' – Ajay Tatachar Aug 08 '20 at 02:01
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There are two different interpretations of injections here.

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

  2. $\mathbb{Z}\to k$ 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).

user10354138
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