The intuition behind the idea of 'embedding' in rings Apologies in advance if this was asked before.
Say we have $X$ is embedded into $Y$, my understanding is that there exists an injective ring homomorphism from $X$ to $Y$. I have the following questions about the idea of 'embedding': (as I want to get more familiar with this terminology)
$1)$ Is the idea of embedding, loosely speaking, saying that there is a subring in $Y$ that is isomorphic to $X$?
$2)$ I saw this question earlier today, where the answer says 'properly, you get an embedding $\mathbb N\to K$'. However though, can the naturals be embedded to an arbitrary field $K$ if $K$ is finite?
$3)$ Lastly, maybe just a check of my understanding, is an inclusion map always an embedding map?
 A: 1) Yes, exactly. That is why you consider an injective morphism of rings (gives you the isomorphic subring as the image of your homomorphism by the first isomorphism theorem).
2) No, embedding is a bad word there (I guess Martin had the hidden assumption that the characteristic of the field is $0$). It is better to say that there is a natural map from the natural numbers to every field (but I would not call it an embedding). Also possible embedding of what? Monoids?
3) Yes, inclusions are always embeddings. Whenever you have a notion of subobjects you certainly also want the inclusion to be an embedding. For example, the subspace topology is defined to be the coarsest topology such that the inclusion is continuous. So here we even define subobjects (in that category) with that in mind.
A: 1) Yes, not even loosely : if you have an embedding $X\to Y$ then the image of that embedding is a subring of $Y$ isomorphic to $X$.
More generally, if you have an isomorphism from $X$ to a subring of $Y$ you can compose it with the inclusion of the subring in $Y$ and get an embedding.
Moray, embedded rings are just subrings.
2) The answer is no for an arbitrary field, I'm assuming they meant an arbittary field of characteristic $0$. Then by definition the inclusion $\mathbb Z \to K$ is injective.
Note, however, that $\mathbb N $ is not a ring, so the embedding would have to be an embedding of something else than rings (semirings ?)
3) Yes : it is injective, and a ring map, so it's an embedding. In fact, they're the prototypical embeddings, any embedding is isomorphic to an inclusion map (see 1) for a more precise statement)
