David Cox's Primes of the form $x^2+ny^2$ defines infinite primes as
[Infinite primes] are determined by the embeddings of $K$ into $\mathbb{C}$. A real infinite prime is an embedding $\sigma: K \to \mathbb{R},$ while a complex infinite prime is a pair of complex conjugate embeddings $\sigma, \bar{\sigma} : K \to \mathbb{C}, \sigma \neq \bar{\sigma}$. Given an extension $K \subset L,$ an infinite prime $\sigma$ of $K$ ramifies in $L$ provided that $\sigma$ is real but it has an extension to $L$ which is complex. For example the infinite primes of $\mathbb{Q}$ is unramified in $\mathbb{Q}(\sqrt{2})$ but ramified in $\mathbb{Q}(\sqrt{-2})$.
i) I looked at this question and the comments mention calling primes as absolute places. I do know that places are classes of absolute values, but I don't see how they correspond to the real and conjugate embeddings of $K$ into $\mathbb{C}$? Like, for non-archimedean spaces, we work in $\mathbb{Q}_p$ and not $\mathbb{Q}$ so how does this work? If someone could give me an example I would be grateful.
ii) I do not follow the ramification example either? The minimal polynomials are $x^2-2$ and $x^2+2$ over $\mathbb{Q}$ for $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$ respectively, but I do not understand what they mean by the embeddings/primes ramifying/not ramifying here.
Thank you so much in advance!
For 1): Not all places correspond to embeddings of $K$ into $\mathbb{C}$!
Given an Archimedian absolute value $v=|\cdot|$ on $K$, the completion $K_v$ is a field of finite dimension over $\mathbb{R}$ (which is the completion of $\mathbb{Q}$ at its only Archimedian place), so $K_v$ is $\mathbb{R}$ or $\mathbb{C}$.
Thus, the canonical embedding $K \rightarrow K_v$ produces an embedding $K \rightarrow \mathbb{R}$ (or $K \rightarrow \mathbb{C}$ up to conjugacy, since the isomorphism $K_v \cong \mathbb{C}$ can be realized in two ways).
Note that $v$ is easily recovered from the embedding: it is (a power of) the restriction of the usual absolute value on $\mathbb{R}$ or $\mathbb{C}$.
For 2): consider a number field extension $L/K$. Let $v$ be an Archimedian place of $K$ and $w$ be an Archimedian place of $L$ above $v$. We say that $L/K$ is ramified at $w$ if the extension of Archimedian local fields $L_w/K_v$ is nontrivial (ie $L_w \cong \mathbb{C}$ and $K_v \cong \mathbb{R}$).
If $v$ is an Archimedian place of $K$, we say that $L/K$ is ramified at $v$ if there is some place $w$ of $L$ above $v$ such that $L/K$ is ramified at $w$.
For example, $\mathbb{Q}(\sqrt{-2})$ does not embed into $\mathbb{R}$: for any Archimedian place $v$, $\mathbb{Q}(\sqrt{-2})_v\cong \mathbb{C}$, which is not the same as the completion of $\mathbb{Q}$ at its Archimedian place. Therefore, $\mathbb{Q}(\sqrt{-2})/\mathbb{Q}$ is ramified at the Archimedian place of $\mathbb{Q}$.
On the other hand, the image of every embedding $\mathbb{Q}(\sqrt{2}) \subset \mathbb{C}$ lands in $\mathbb{R}$: so for any Archimedian place $v$ of $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{2})_v \cong \mathbb{R}$.
It follows that $\mathbb{Q}(\sqrt{2}) /\mathbb{Q}$ is not ramified at the Archimedian place of $\mathbb{Q}$.