# $\mathbb{Q}$ is a prime field

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

• @Andreas CarantiThanks for your answer, but I would like to know what happens in this case if $m\ne 0$. – Jack J. Nov 25 '18 at 16:34

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 $$\Bbb Q$$.