If $K$ is a number field, it is well known that the non-archimedean places of $K$ are in bijection with the prime ideals $\mathfrak p$ of $O_K$. Whereas the archimedean places are in bijection with the field embeddings $K\to\mathbb C$ (up to conjugation).
Now, if $K=\bar{\mathbb{Q}}$ (algebraic closure of $\mathbb Q$), what are the places in this case? I have problems in visualising them, especially the non-archimedean ones. Do the places satisfy a product formula like in the number field case?
Let $\Bbb Q = K_0 \subset K_1 \subset K_2 \subset K_3 \subset \cdots$ be finite extensions of $\Bbb Q$ such that $\bigcup_n K_n = \overline{\Bbb Q}$ (this is possible because $\overline{\Bbb Q}$ is countable).
Then a place of $\overline{\Bbb Q}$ restricts to a sequence of places $v_n$ of $K_n$ such that $v_m$ restricts to $v_n$ for any $m \ge n$.
Conversely, a sequence of compatible places extends to a place of $\overline{\Bbb Q}$.
The terminology for this is "inverse limit": let $\operatorname{Spv}(K)$ be the places of $K$, then: $$\operatorname{Spv}(\overline{\Bbb Q}) = \varprojlim \operatorname{Spv}(K_n)$$
More generally, for any algebraic field extension $K \subset L$: $$\operatorname{Spv}(L) = \varprojlim_{\substack{K \subset K' \subset L \\ [K':K] < \infty}} \operatorname{Spv}(K')$$