I'm reading Algebraic Number Theory by Milne, and in a footnote of Chapter 7 he proves that
$\mathbb{R},\mathbb{C}$ with usual absolute values are the only archimedean locally compact fields, up to isomorphism (of fields with absolute values, that is equivalent up to continuous isomorphism of fields).
He gave an argument that is fine to me, but before reading his one, it came this to me and i was wondering if it is wrong(it seems to me pretty easier than his one, so i suspect of a mistake):
Since the field K is complete and you know that on $\mathbb{Q}$ the absolute value is archimedean you know that is(up to equivalence) the usual one, so you can assume that $\mathbb{R} \subseteq K$ with the inclusion respecting the absolute values. Now you can look at K as a normed space on $\mathbb{R}$ that is locally compact. But this implies that K is finite dimensional(By Riesz Lemma this can be seen directly, constructing a sequence, or indirectly using with the sequence the existence of infinitely many bounded open ball of the same radius, and pairwise disjoint, that implies immediately no Haar measure, and so no local compactness, by Haar theorem), so K is a finite dimensional extension of $\mathbb{R}$ and so it has to be $\mathbb{R}$ or $\mathbb{C}$ by the fundamental theorem of algebra.
Thanks for any comment or correction.