# Is $\mathbb{Q}$ the smallest ordered field up to isomorphism?

Pretty simple question. Does there exist a ordered field smaller than (i.e. is a strict subset of) $$\mathbb{Q}$$? It seems like we can't go any smaller than $$\mathbb{Q}$$. Is this true? Why?

• Well, you have to argue that no finite field can be ordered. More broadly, try to argue that no field of finite characteristic $p$ can be ordered. Beyond that, if $\mathbb F$ is a subfield of $\mathbb Q$ it must contain $1$ hence it contains $\mathbb Z$, hence it contains $\mathbb Q$. – lulu Nov 9 at 1:02