Ytukyg

I must prove that $mathbb{Q}$ is a prime field, that is $mathbb{Q}$ does not posses any proper subfield.

We suppose that $Ksubsetmathbb{Q}$ is a proper subfield of $mathbb{Q}$ and we consider the morphism $fcolonmathbb{Z}to K$ defined as $nmapsto ncdot 1$. The $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(m)$, where $minmathbb{Z}$. If $m=0$, we have that $mathbb{Z}cong f(mathbb{Z})subseteq Ksubseteqmathbb{Q}$, in particular $mathbb{Z}subseteq K$, then $text{Frac}(mathbb{Z})subseteq K$, but $text{Frac}(mathbb{Z})=mathbb{Q}$, then $mathbb{Q}subseteq K$.

Question. In the current situation it could happen that $mne0$?

My attempt If $mne 0$, then $mathbb{Z}_m cong f(mathbb{Z})subseteq K$, then since $K$ is a field $f(mathbb{Z})$ must be a integral domain, then $m$ must be a prime $p$, therefore $mathbb{Z}_p cong f(mathbb{Z})subseteq Ksubsetmathbb{Q}$. Now, can I conclude?

Thanks

You already have that ${Bbb Z}_p$ is isomorphic to a subfield of $Bbb Q$. The unit element 1 is inherited. Then $pcdot 1= 0$ in ${Bbb Z}_p$ but not in $Bbb Q$.