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Is $ \sqrt{-1}$ of $ \Bbb{Q}_2( \sqrt{-1})$ and $ \Bbb{Q}( \sqrt{-1})$ the same number?

The former $ \sqrt{-1}$ is a root of $x^2＋1＝0$ in $ \overline{ \Bbb{Q}_2}$, and the latter $ \sqrt{-1}$ is a root of $x^2＋1＝0$ in $ \overline{ \Bbb{Q}}$.

Until now, I thought these $ \sqrt{-1}$ are completely different number, but if we regard $ \Bbb{Q}_2( \sqrt{-1})$ as completion of $ \Bbb{Q}( \sqrt{-1})$ at $(2)$, $ \Bbb{Q}( \sqrt{-1})$ must contain $ \sqrt{-1}$ in $ \overline{ \Bbb{Q}}$.

Poitou-Tate
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    I feel like this question doesn't really make sense. Ultimately it depends on how you view these structures, for example is $1$ the same in both fields. – Fishbane Aug 24 '22 at 18:08
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    There are two homomorphisms $\mathbb{Q}(i) \to \mathbb{Q}_2(i)$ which differ by $i \mapsto -i$ so there's an unavoidable sign ambiguity. Up to that ambiguity, this homomorphism allows you to directly relate the two. – Qiaochu Yuan Aug 24 '22 at 18:21
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    I wrote a lot about this and related issues in https://math.stackexchange.com/a/4007515/96384. It is good that you are careful with these symbols! As Fishbane and Qiaochu say, the best way to think of elements of different fields is not to wonder if they are "the same" (what's the criterion for that?, and for that matter, maybe "my" $i \in \mathbb C$ is not "the same" as "your" $i \in \mathbb C$), but to clarify for oneself what possible field embeddings exist between those different fields, and what these embeddings do to the elements in question. – Torsten Schoeneberg Aug 24 '22 at 19:02

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