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

We suppose that $K\subset\mathbb{Q}$ is a proper subfield of $\mathbb{Q}$ and we consider the morphism $f\colon\mathbb{Z}\to K$ defined as $n\mapsto n\cdot 1$. The $\ker f$ is an ideal of $\mathbb{Z}$, then $\ker f=(m)$, where $m\in\mathbb{Z}$. If $m=0$, we have that $\mathbb{Z}\cong f(\mathbb{Z})\subseteq K\subseteq\mathbb{Q}$, 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 $m\ne0$?

My attempt If $m\ne 0$, 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 $\mathbb{Z}_p \cong f(\mathbb{Z})\subseteq K\subset\mathbb{Q}$. Now, can I conclude?

Thanks

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  • $\begingroup$ @Andreas CarantiThanks for your answer, but I would like to know what happens in this case if $m\ne 0$. $\endgroup$ – Jack J. Nov 25 '18 at 16:34
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You already have that ${\Bbb Z}_p$ is isomorphic to a subfield of $\Bbb Q$. The unit element 1 is inherited. Then $p\cdot 1= 0$ in ${\Bbb Z}_p$ but not in $\Bbb Q$.

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