Why $\mathbb{R}$ is not isomorphic to $\mathbb{Q}_p$ as fields?

Question

I know Ostrowsky's theorem that says that the only valuations of $\mathbb{Q}$ are the real one and every p-adic, so metrically, $\mathbb{R}$ is not the same that $\mathbb{Q}_p$. However, I don't know the argument to say that, as fields, $\mathbb{R}$ is not isomorphic to $\mathbb{Q}_p$.

Any help would be appreciated.

Answer 1

The fields $\mathbb{Q}_p$ can't be ordered.

How to show that $\mathbb{Q}_p$ cannot be ordered?

Answer 2

I find reuns's comments the easiest: $X^2=p$ has solutions in $\mathbb R$ but not in $\mathbb Q_p$.

Or, $X^2 = 1-p^n$ (where $n \ge 3$; for $p \neq 2$, any $n \ge 1$ works) has solutions in $\mathbb Q_p$ but not in $\mathbb R$.