# There is, up to isomorphism, a unique smallest field $\mathbb{Q}$, which contains $\mathbb{Z}$ as a subring

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

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.