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I am reading Section 9. The Rational Numbers from textbook Analysis I by Amann/Escher, where there is a theorem:

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I would like to confirm if my understanding about the proof (which leaves details to readers) is correct or not. Thank you for your help!


My thought:

  1. The injective homomorphism $\mathbb Z \to \mathbf{Q}$ is unique.

Let $\mathbf{Q}$ be a field containing two copies of $\mathbb Z$. Then there are two injective homomorphisms $f_1 , f_2$ from $\mathbb Z$ to $\mathbf{Q}$. For $\mathbb Z \ni n > 0$, we have $$\begin{aligned} f_1(n) &= f_1(\underbrace{1+\cdots+1}_{n \text{ times}}) \\ &= \underbrace{f_1(1)+\cdots + f_1(1)}_{n\text{ times}} \\ &= \underbrace{1'+\cdots+1'}_{n \text{ times}} \\ &= \underbrace{f_2(1)+\cdots+f_2(1)}_{n\text{ times}} \\ &= f_2(\underbrace{1+\cdots+1}_{n \text{ times}}) \\ &= f_2(n) \end{aligned}$$

Here $1,1'$ are multiplicative identities of $\mathbb Z, \mathbf{Q}$ respectively.

Similarly, $f_1(n) = f_2(n)$ for $\mathbb Z \ni n < 0$. As a result, there is a unique injective homomorphism $f$ from $\mathbb Z$ to $\mathbf{Q}$.

  1. $\mathbb Q$ is one of the smallest fields that contain $\mathbb Z$ as a subring.

We define a mapping $h: \mathbb Q \to \mathbf{Q}$ by $$h([(m,n)]) = \begin{cases} \dfrac{f(m)}{f(n)} & \text{if } n \neq 0\\ f(0) & \text{otherwise}\end{cases}, \quad m,n \in \mathbb Z$$

It is easy to verify that $h$ is an injective homomorphism.

  1. $\mathbb Q$ is unique up to unique isomorphism.

Let $\overline{\mathbf{Q}}$ be another smallest field that contains $\mathbb Z$ as a subring and $f$ a unique injective homomorphism from $\mathbb Z$ to $\overline{\mathbf{Q}}$. Then there are two injective homomorphisms $\psi: \mathbb Q \to \overline{\mathbf{Q}}$ and $\varphi: \overline{\mathbf{Q}} \to \mathbb Q$. Clearly, $\psi \restriction \mathbb Z = f$ and $\varphi \restriction f[\mathbb Z] = f^{-1}$. If $x \in \overline{\mathbf{Q}}$ then $\varphi (x) = \dfrac{p}{q} \in \mathbb Q$ for some $p,q \in \mathbb Z$.

We have $\psi (\varphi (x)) = \psi \left(\dfrac{p}{q}\right) = \dfrac{\psi(p)}{\psi(q)} = \dfrac{f(p)}{f(q)} = \dfrac{\varphi^{-1} (p)}{\varphi^{-1} (q)} = \varphi^{-1} \left(\dfrac{p}{q}\right) = \varphi^{-1} (\varphi (x)) = x$. It follows that $\psi = \varphi^{-1}$ and thus $\psi, \varphi$ are isomorphisms. Moreover, the isomorphism between $\mathbb Q$ and $\overline{\mathbf{Q}}$ is unique.

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