If $A$ is a $\mathbb{Q}_p$- and $\mathbb{Q}_q$-algebra then $p=q$?
问题内容
Let $A$ be a (associative with unit) $\mathbb{Q}_p$- and $\mathbb{Q}_q$-algebra where $p,q \in \mathbb{P}\cup \{\infty\}$. Does it follow that $p=q$?
If $A$ would be a Hausdorff locally compact skew-field and topological $\mathbb{Q}_p$- and $\mathbb{Q}_q$-algebra then it would be necessarily $p=q$.
But what if there is no such topology on $A$? Is there a counter-example for the non-topological case? What if $A$ has finite dimension over $\mathbb{Q}_p$?
回答 (1)
Consider algebraic closures $\overline{\mathbb{Q}_p}$ and $\overline{\mathbb{Q}_q}$. It's clear that these are both algebras over $\mathbb{Q}_p$ and $\mathbb{Q}_q$ (respectively), but moreover they are both algebraically closed fields of characteristic zero and cardinality continuum. Thus they're isomorphic as fields (and in fact they're both isomorphic to $\mathbb{C}$). Crucially this field isomorphism is highly noncontinuous, since it is constructed using the axiom of choice.
I hope this helps ^_^