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First, if I may, I would like to ask for help in getting an intuitive understanding regarding embeddings.

Wikipedia gives examples such as $\mathbb N$ in $\mathbb Z$. My first question is: were they not there already? And my second question is what advantage (or advantageous constraint) do you get by embedding.

This general question is motivated by a Theorem in Appendix 2 in Marcus's "Number Fields" on p. 259.

He states: We are interested in the embeddings of L in $\mathbb C$ which fix K pointwise. (L and K are subfields of $\mathbb C$ and K $\subset$ L.) Clearly such an embedding sends $\alpha \in$ L to one of its conjugates over K.

As a second question: How does an "embedding" of something that is already there $\alpha \in$ L $\subset \mathbb C$ get sent to a conjugate? Rather than call it a injective map or permutation?

Lastly in the proof of the theorem, he says: let $\sigma$ be an embedding of K in $\mathbb C$ ... and f be the monic irreducible polynomial for $\alpha$ over K. He then goes on to apply $\sigma$ to the coefficients of f to get g. But I thought (as above) that the embeddings fix K, so why would he say then say that g is irreducible (no problem) over $\sigma$K? Shouldn't that just be K?

As always thanks for your help and patience with what is probably a pretty obvious question.

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up vote 2 down vote accepted

On your first question: One way to define $\mathbb Z$ is as the set of equivalence classes of pairs of natural numbers under the equivalence relation $(k,l)\sim(m,n)\Leftrightarrow k+n=l+m$. The natural numbers are "not there already" – they have to be embedded in this structure via $f:\mathbb N\hookrightarrow\mathbb Z:n\to[(n,0)]$.

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@jorikiThanks that's an eye-opener. I guess the same goes for $\mathbb Q$ and $\mathbb R$. So, if I may, does everything originate from $\mathbb N$? Or is the yet a precursor and a prime unmoved source? And in the context of verbiage, why is the word "embed" used since that has the implication of putting something into another thing that was already there? Thanks – TheBirdistheWord Aug 5 '12 at 21:11
Yes, everthing originates from $\mathbb N$. $\mathbb Q$ is a set of equivalence classes of pairs in $\mathbb Z$, $\mathbb R$ is a set of equivalence classes of sequences in $\mathbb Q$, and $\mathbb C$ is a set of pairs in $\mathbb R$; though all these structures can also be defined in other ways. On your second question, I guess you're right about the connotations of "embed"; but I can't off the top of my head think of a word that would have fit the concept better. – joriki Aug 5 '12 at 21:20
Was $\mathbb N$ (or $L$) already there? There is certainly a sub-monoid of $\mathbb Z$ (or sub-field of $\mathbb C$) that looks like $\mathbb N$ (resp. $L$) since this is after all the groupification of the monoid, but how do we know there are not different copies? And if so, which one of these would have priority? I interpret "embed" as a choice of such a copy. – Andrew Aug 5 '12 at 21:23
@Andrew: You may find this link interesting too:… – Andrew Aug 5 '12 at 21:24
@Andrew: I'm not sure I understand the question. What's an extension in this context? How would you extend the natural numbers to pairs of natural numbers? – joriki Aug 7 '12 at 11:01

Fair warning: my understanding of algebra is a bit shaky at points. But as I understand it...

joriki's answer can essentially be extended to answer all the questions. We tend to think about certain objects as though they were other objects, and when we're not being too rigorous there's nothing wrong with this. But for technical reasons relating to how things are defined, we sometimes need to deal with embeddings instead of the objects themselves.

I'm not entirely sure what your objection is in the second question. But you can call it an injective map if you prefer, because it is that. However, it would be awkward to call it a permutation, because the mapping is from L to $\mathbb{C}$, which is generally not the behavior one wants in a permutation. There is of course a natural extension from $\mathbb{C}$ to $\mathbb{C}$ (with non-L elements unchanged) or a natural "restriction" from L to L. But these change the triple (domain, range, relation), so they're not technically the same function.

Your third question makes me thing that either he is using a different definition of subfield that I am, or that he did not actually say that L and K are subfields of $\mathbb{C}$. When I think of a subfield, it is a subset of the original field which is also a field. However, it would be perfectly reasonable to define it as a field which is field-isomorphic to a subset of the original.

I'll assume the second definition. $g$ cannot be irreducible over K because it's coefficients are members of $\sigma$K, and in general there is no way to define a multiplication on two groups. Remember, even though $\sigma$K is in some sense "the same" as K, it is not technically the same object since the elements of $\sigma$K are elements of $\mathbb{C}$. No such restriction exists on the elements of K.

Now, if you take the first definition, and he explicitly states that K is a subfield of $\mathbb{C}$, then all of this is redundant, because the elements of K are also elements of $\mathbb{C}$, and there is the obvious multiplication on any two subsets of a field.

(I used the phrase "a multiplication on X and Y" a bit poorly, but I hope the meaning was clear.)

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To provide a different example, it is not technically true that $D_3$ is a subgroup of $D_6$, if we mean that for $X$ to be a subgroup of $Y$ we must have $X\subseteq Y$. We must first create an embedding from $D_3$ into $D_6$; then the image of that embedding can be a subgroup. However, it is true if we mean that $X$ must be group-isomorphic to a subset of $Y$. (which is one of the advantages to using that definition of a subgroup) – Eric Stucky Aug 5 '12 at 21:32
@EricStuckyThanks. Regarding the $\sigma$K issue, what happened to the introductory remark about embeddings that fix K pointwise? So if $a$ is a coefficient of f, wouldn't $\sigma$$a$ = $a$ as the analagous coeffieient of g? – TheBirdistheWord Aug 5 '12 at 21:36
@EricStuckyI was typing my comment-question as you wrote you comment. So that's probably what you are addressing? – TheBirdistheWord Aug 5 '12 at 21:40
Ooh, I did forget that. So the "redundant" interpretation seems to be most valid here. Aesthetically, since $g$ was defined to be a member of $\sigma$K[$x$], it makes sense that we might want the $x$'s to take on values in $\sigma$K. But you're probably correct in that it is not technically necessary. – Eric Stucky Aug 5 '12 at 21:50
@Eric Stucky When $K$ is a field and one speaks of a $K$-embedding or $K$-monomorphism $\sigma:L\rightarrow L^\prime$, it is implicit that $L$ and $L^\prime$ are extensions of $K$, meaning that $K$ has been identified with a subfield of each, and these identifications are fixed. One then simply regards $K$ as a subset of $L$ and $L^\prime$. So if $f\in K[X]$ is a polynomial, $\sigma(f)=f$, because $\sigma$ fixes $K$ pointwise. Precisely what this means is that, if $\alpha\in K$, $i:K\rightarrow L$ and $i^\prime :K\rightarrow L^\prime$ are the (implicit, fixed) embeddings, then – Keenan Kidwell Aug 5 '12 at 23:34

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