What is the intuition behind defining this isomorphism?

Solution 1:

Here's a fairly intuitive explanation.

First, let's consider that for any ring $R$, there is a unique ring homomorphism $\mathbb{Z} \to R$. This is because a ring homomorphism $\mathbb{Z} \to R$ is always the unique group homomorphism $\mathbb{Z} \to R$ sending $1$ to $1$, and conversely this group homomorphism is always a ring homomorphism.

Consider the smallest subring of $F$. This will always be the image of the unique ring homomorphism $f : \mathbb{Z} \to F$. For consider any subring $R \subseteq F$. Then let $f_R : \mathbb{Z} \to R$ be the unique ring homomorphism, and let $i : R \to F$ be the inclusion homomorphism. Then $f = i \circ f_R$, since both these maps are ring homomorphisms. Then $f$ factors through $i$; therefore, the image of $f$ is a subset of $R$.

Now the image of $f$ is isomorphic to $\mathbb{Z} / \ker(f)$. Using the fact that $\mathbb{Z}$ is a PID, take some $p$ be such that $(p) = \ker(f)$; without loss of generality, take $p \in \mathbb{N}$. Note that since $im(f)$ is a subring of $F$, it must be an integral domain; therefore, $\mathbb{Z} / \ker(f)$ must be an integral domain; therefore, $(p)$ must be a prime ideal.

There are two cases:

Case 1: $p$ is a prime number

In this case, $\mathbb{Z} / \ker(f) = \mathbb{Z} / (p)$ is already a field. So the smallest subring of $F$ is in fact already a field; hence, it's the smallest subfield.

Case 2: $p$ is 0

In this case, $f : \mathbb{Z} \to F$ has the property that for all $x \neq 0$, $f(x)$ is a unit. Then $f$ extends to a unique function $h : \mathbb{Q} \to F$ such that $h|_\mathbb{Z} = f$ by the definition of $\mathbb{Q}$ as the field of fractions of $\mathbb{Z}$. The image of $h$ forms a subfield of $F$.

Consider any subfield $G \subseteq F$ with inclusion map $i : G \to F$. Then consider the unique ring homomorphism $f_G : \mathbb{Z} \to G$. Then consider the unique extension of this map $h_G : \mathbb{Q} \to G$. Then we see that $i \circ f_G = f$. Therefore, $h|_\mathbb{Z} = (i \circ h_G)|_{\mathbb{Z}} = i \circ (h_G|_\mathbb{Z})$. Then $h$ factors through $i$. Therefore, the image of $h$ is a subfield of $G$.

So we see that the image of $h$ is the smallest subfield.