Are p-adic fields isomorphic for different primes?

[caji]

Hey,

If $$p \neq q$$ are two primes, then the p-adic fields $$\mathbb{Q}_p$$ and $$\mathbb{Q}_q$$ are non isomorphic, right?

Actually I've read this in my book and I'm not sure, if that's obvious (which means its just me who doesn't recognize it) or a statement which has to be proven.

The p-adic fields as I know them are defined as:
Let $$\mathcal{C}_p$$ be the set of all rational Cauchy-Sequences, and $$\mathcal{N}_p$$ be the ideal of $$\mathcal{C}_p$$ of all sequences converging to zero. Then $$\mathbb{Q}_p := \mathcal{C}_p / \mathcal{N}_p$$

caji

 

[fresh_42]

This is the definition of $\mathbb{R}$ in general. It depends on the used metric, because the metric defines Cauchy sequences. So the $p-$adicness is hidden in the metric with this definition. If we use the usual Euclidean (Archimedean) metric, we get the usual reals. We need to use the p-adic metric to get $\mathbb{Q}_p\,.$

$\mathbb{Q}_p \ncong \mathbb{Q}_q \Longleftrightarrow p\neq q$.

Looking at the number of roots of unity in your field suffices to distinguish all $\mathbb{Q}_p$ for odd values of $p$, because the number of roots of $1$ there is precisely $p−1\,.$ It's different for the $2-$adic numbers, since they have two roots of unity, same as the $3$-adics. But the $2-$adics have a square root of $−7$ and the $3-$adics don't, whereas the $3-$adics have a square root of $10$ and the $2-$adics don't.

https://math.stackexchange.com/questions/93633/is-mathbb-q-r-algebraically-isomorphic-to-mathbb-q-s-while-r-and-s-denote/95128#95128