# Fundamental subfield of a field $F$

I must prove that every fundamental subfield of a field $$F$$ is a $$\textit{prime field}$$.

Definition. A field $$F$$ is said $$\textit{prime}$$ if does not posses any proper subfields, that is if $$L\subset F$$ is a subfield, then $$F=L$$.



Proposition 1. Let $$\{F_i\}$$ a family of subfields of a field $$F$$, then $$\bigcap_i F_i$$ is also a subfield of $$F$$.

proof. We observe that $$(F_i^{\times},\cdot)$$ is a subgroups of $$(F^{\times},\cdot)$$ for all $$i$$, then $$1_{F_i}=1_F$$ for all $$i$$, the $$1_F\in\bigcap_i F_i$$. Therefore $$\bigcap_i F_i\ne\emptyset$$, moreover if $$a,b\in\bigcap_i F_i$$, then $$a-b\in F_i$$ for all $$i$$, in conseguence of this $$a-b\in\bigcap_i F_i$$. In the same way we can prove that if $$b\ne 0$$, then $$ab^{-1}\in \bigcap_i F_i$$.



Definition. Let $$F$$ a field and let $$S\subseteq F$$ a subset, we define $$(S):=\bigcap\bigg\{F_i\;|\;F_i\;\text{subfield of F with}\;S\subseteq F_i\bigg\}.$$

For the last proposition $$(S)$$ is a subfield of $$F$$ and is called $$\textit{subfield generated by}\;S$$.

Proposition 2. The subfield $$(S)$$ is the smallest subfield of $$F$$ which contains $$S$$.

proof. We place $$\mathcal{F}:=\bigg\{F_i\subseteq F\;|\;F_i\;\text{subfield of}\;F,S\subseteq F_i\bigg\},$$ and $$(S):=\bigcap_{F_i\in\mathcal{F}} F_i.$$ The family $$\mathcal{F}$$ is not empty, as $$F\in\mathcal{F}$$. Naturally $$S\subseteq (S)$$, moreover if $$\tilde{F}$$ is an arbitrary subfield of $$F$$ such that $$S\subseteq \tilde{F}$$, then $$\tilde{F}\in\mathcal{F}$$ therefore $$(S)\subseteq\tilde{F}$$.

If $$S=\{1_F\}$$, we denote with $$F_f=(1_F)$$ the fundamental subfield of $$F$$. As every subfield $$F_i$$ of $$F$$ contains $$1_F$$, $$F_f$$ must be contained in each subfield of $$F$$, therefore $$F_f$$ is the intersection of all the subfield of $$F$$.

We prove that $$F_f$$ is a prime field.

If $$K$$ is a subfield proper of $$F_f$$, then $$K\subset F_i\subseteq F$$ for all $$i$$, therefore $$K$$ is a subfield of $$F$$, then $$1_k=1_F\in K$$; in consequence of this we have $$F_f\subseteq K$$, then $$F_f=K$$.

Could someone tell me if I did something wrong during the reasoning? Thanks!

Hint: Every subfield of $$F$$ contains the prime field of $$F$$.
• @WuestenfuxThanks for your answer. Now, if $K\subset F_f$ is a subfield, then $K\subset\bigcap_{F_i\subseteq F}\{F_i\}$, however $\bigcap\{F_i\}$ is a subfield of $F$, then $K$ is a subfield of $F$, but $F_f$ by definition is the intesection of all subfield of $F$, then $F_f\subseteq K$. Right? – Jack J. Nov 25 '18 at 12:02