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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 $F=L$.

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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 $ab^{-1}\in \bigcap_i F_i$.

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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!

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Hint: Every subfield of $F$ contains the prime field of $F$.

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  • $\begingroup$ @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? $\endgroup$ – Jack J. Nov 25 '18 at 12:02

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