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Math Help - Embeddings

  1. #1
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    Embeddings

    Hi all,

    Could somebody help me try and understand the idea of an "embedding" please?

    I can understand field embeddings, since the properties (addition and multiplication) between two fields will be the same and it's easy to see it from examples like \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}

    But I don't quite understand embedding 'rings into fields' and 'groups into fields'..

    How are the properties of a field "lost" in those embeddings?..
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  2. #2
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    Quote Originally Posted by superbaros View Post
    Hi all,

    Could somebody help me try and understand the idea of an "embedding" please?

    I can understand field embeddings, since the properties (addition and multiplication) between two fields will be the same and it's easy to see it from examples like \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}

    But I don't quite understand embedding 'rings into fields' and 'groups into fields'..

    How are the properties of a field "lost" in those embeddings?..
    embedding is nothing but a one-to-one homomorphism. so, for exampple, embedding a group (ring) A in a field F, means that you may assume that A is a subgroup (subring) of F.

    for exxample, by embedding a group (G,*) in a field F we mean a one-to-one map f: (G,*) \longrightarrow (F,+) such that, for all a,b \in G: \ f(a*b)=f(a)+f(b). an immediate result is

    that (G,*) must be abelian because (F,+) is abelian. similarly, embedding a ring R in a field F means that there exists a map g: R \longrightarrow F such that, for all a,b \in R we have:

    g(a+b)=g(a)+g(b) and g(ab)=g(a)g(b). also if R has 1, then g(1_R)=1_F. clearly, such R has to be a (commutative) integral domian.
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  3. #3
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    Hiya, thanks for your reply!

    A definition from wolfram is

    One space is embedded in another space when the properties of restricted to are the same as the properties of .

    Let me see if I get this...
    If we had a group (G,\ast) embedded in a field (F,+,\cdot) then when we restrict the properties of F to (G,\ast), we must have that \ast is a commutative operation since + is a commutative operation. So like you said, G must be an abelian group.

    If I am correct, how does the same idea work for a ring embedded into a field?... If I use the same reasoning, then I will deduce that the ring must be a field?...=S
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  4. #4
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    For example, the set of all integers, with the usual multiplication and addition, is a ring. The set of all rational numbers, with the usual multiplication and addition, is a field. Obviously, the set of all integers is a subset of the set of all rational numbers but is NOT a field because the multiplicative inverse of 2 is 1/2, a rational number but not an integer.

    Any ring that is isomorphic to the integers can be "embedded" in the field of rational numbers with that isomorphism- but still doesn't have all the properities of a field.
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  5. #5
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    I think for that example its easier to see.

    I'm trying to think about it from the real basics now, so going back to the fact that an embedding is an injective homomorphism from a structure X to Y, which properties does the homomorphism preserve?... The properties of X, or the properties of Y?

    For group and ring homomorphisms this is pretty clear since both X and Y are groups or rings, but if they are 'different', does a homomorphism preserve the structure of X or Y?

    I think I'm confusing myself too!...
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  6. #6
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    Quote Originally Posted by superbaros View Post
    I think for that example its easier to see.

    I'm trying to think about it from the real basics now, so going back to the fact that an embedding is an injective homomorphism from a structure X to Y, which properties does the homomorphism preserve?... The properties of X, or the properties of Y?

    For group and ring homomorphisms this is pretty clear since both X and Y are groups or rings, but if they are 'different', does a homomorphism preserve the structure of X or Y?

    I think I'm confusing myself too!...
    well, when you embed X in Y must have a common structure. they might have other structures as well. that's why we always say that X is embdded in Y "as" a group, ring, module, etc.

    for example a ring X is also an additive group. a field Y is also a ring and an additive group. so it might be possible that X is embdded in Y as a group or ring. but if Y is just a vector space,

    then X,Y have only a group structure in common and thus we can only talk about embedding X in Y "as" as a group. it's exactly like the concept of isomorphism. when we say X and Y are

    isomorphic, we must also say "as" what if X, Y have more than one structure in common. for example \mathbb{R}^2 and \mathbb{C} are isomorphic "as" \mathbb{R} vector spaces but they are not isomorphic "as" rings

    because \mathbb{C} is a field but \mathbb{R}^2 is not even an integral domain.
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  7. #7
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    Thanks, I think that helps a bit more.

    However, before, you said

    by embedding a group in a field we mean a one-to-one map such that, for all an immediate result is

    that must be abelian because is abelian
    But then, is it not possible to embed a non-abelian group into a field?.. Since a field can preserve all it's operations...
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