Here is the question I am trying to answer:
Let $F$ be a field. Prove that $F$ contains a unique smallest subfield $F_0$ and that $F_0$ is isomorphic to either $\mathbb Q$ or $\mathbb Z/ p \mathbb Z$ for some prime $p$ ($F_0$ is called the prime subfield of $F$).[See Exercise $26,$ Section 3.]
Here is a solution I found online:
Proof:
Let $F$ be a field and let $S$ be a set of subfields of $F.$ Then, $\bigcap S$ is a subfield of $F.$ Since, $\bigcup S$ is a subring of $F$ and contains $1$, since $1\in A,$ for every $A \in S.$
Let $b \in F.$ Then $b \in A$ for each $A \in S,$ and so, $b^{-1} \in A.$ Then, $b^{-1} \in F,$ and hence, $F$ is a field.
Let $F$ be a field and let $f$ denote the set of all subfields of $F.$ By using assumption, $F_0 = \cap f$ is a subfield of $F$ and is contained in every other subfield.
Now consider the ring homomorphism $\varphi: \mathbb Z \rightarrow F_0$ with $\varphi(1) = 1.$
But I am not sure what is the intuition of defining the homomorphism like this? why $n$ has to be either a prime or zero? why we are sure that the induced map $\psi$ exists? where is the proof that $F_0$ is unique? why we need $\Phi$ to be an isomorphism?
Best Answer
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.