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What are Isomorphisms

  1. Feb 12, 2004 #1
    Hello,

    I am currently taking course on an introduction to ring theory and I am having problems understanding what a isomorphism is. I can't follow the definition in Schuam's Outlines for isomorphism and the definition that was given in my lecture was brief and vague.

    If anyone can tell me:

    First what is an isomorphism between two systems?

    (An algerbraic system being defined as a set S together with any relations and operations defined on S).

    Secondly, how can I identify if two systems are isomorphic?

    Any help is appreciated. Thankyou.
     
  2. jcsd
  3. Feb 12, 2004 #2

    matt grime

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    Firstly, let S and T be two sets. Just sets. A BIJECTION is...?


    Now, if you've two sets with more structure an isomorphism is a bijection on the sets that preserves this structure.

    so in a ring, you have an additive group, and a multiplication. that's the structure we care about.

    actually, first let us know if you know about bijections. we'll go from there
     
  4. Feb 12, 2004 #3
    Hello matt,

    There hasn't been any mention of bijections yet. In fact I haven't heard of that term to date.

    As well, the definition of isomorphism in Schaums occurs long before the definitions of rings or fields. Is it possible to explain isomorphism without reference to rings, fields or bijections?

    Thanks for taking an interest by the way.

    Perhaps I will elaborate a little more.

    Schaums has a precursor to the definition of iosmorphism between to sets. I didn't follow it very well after several readings. But here is what I have been given in class and what schaums states:

    From my lecture:

    My prof goes on from here to use some notation which I don't follow and makes me question whether I copied it down correctly or not.

    To show that there is an isomorphism between to sets, my prof states that we should demonstrate the the mapping is one to one, onto, and something else I do not follow.
     
    Last edited by a moderator: Feb 12, 2004
  5. Feb 12, 2004 #4

    matt grime

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    isomorphism is contextual, a ring isomorphism preserves the structure of a ring, a field isomorphism preserves the field structure and so on. So, you see, one cannot define an iso of rings without rings.

    The first thing to do is to define a homomorphism (homo same, iso equal)

    a map, f, from a ring S to a ring T is a homomorphism if

    f(x+y)= f(x) + f(y)
    f(-x) =-f(x)
    f(xy)=f(x)f(y)

    so it sends sums to sums, additive inverses to additive inverses and products to products, that is what one means by preserving structure.

    It is an ISOmorphism if it has an inverse. Better perhaps if you know about injections (f(x)=f(y) iff x=y) and surjections (for all z in T there is an x in S with f(x)=z). An isomorphism needs to be an injection and a surjection.

    simple examples.

    let S be the integers mod 4, that is the set 0,1,2,3 with additon and multiplication taken modulo 4 and T by the integers mod 2). Then there is a homomorphism from S to T just by sending 0 to 0, 1 to 1, 2 to 0 3 to 1. check it by hand.


    this is not an isomorphism cos 0 and 2 get sent to the same thing.

    have you any explicit examples you want to discuss?
     
    Last edited: Feb 12, 2004
  6. Feb 12, 2004 #5

    matt grime

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    from your edited post....

    a map f from sets X to T is a bijection if for every t in T there is an x in X with f(x)=t (this can be thought of as covering) AND this x is unique, that is no two distinct elements of X get sent to the same thing.


    a map which covers is a surjection (french, sur) an injection is map where distinct elements are sent to distinct elements (not especially french)
     
  7. Feb 12, 2004 #6
    Yes. Perhaps I will give you a question that we did but I didn't follow:

    The prof goes on from here to say that we must prove it is one to one (which I can do). And we must prove that it is onto. (which I can do). And something else which I don't follow. Here is the prof's answer:

    Let gamma be a mapping of (N,+) -> (M,+) with

    (n)gamma = 2n which is an element of M

    = {2x | x is an element of N} is an isomorphism.

    The prof. then goes on to show one to one and onto. Which I get. Then he goes on with the third part which I don't follow:

    (n + n')gamma = (n+n')2 = n2 + n'2 = ngamma + n'gamma

    where

    (n + n')gamma is addition in N

    and

    ngamma + n'gamma is addition in M.

    So what is my prof doing here? Is he proving that the binary op which is valid in the set N is also valid in the set M?

    If that is the case, then to show isomorphism between two sets do I have to show that the mapping is one to one, onto, and that the binary op in the set N is valid in the set M to which N is mapped to?
     
  8. Feb 12, 2004 #7

    matt grime

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    OK so you've got a bijection on the sets, it now suffices to show that the map preserves the structure of the operation

    so that would mean, for x and y in N that f(x+y)=f(x)+f(y) (we assume the addition operation to be understood to be in the relevant ring)

    ie 2(x+y) = 2x+2y

    on the left addition inside N, on the right addition inside M

    [it might be confusing because the operation in M is that inherited from N considering M to be inside N.]

    here's a harder example to see how you are going, because the operations will be different

    show that 1,2 with mult. defined mod 3 is 'the same as' the set 0,1 with addition mod 2
     
  9. Feb 12, 2004 #8

    HallsofIvy

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    Pretty much but you have to understand that "valid" means
    f(a.b)= f(a)*f(b) where . is the operation in one set and * is the operation in the other set.


    Here's a simple example. The set of all real numbers is a group with the operation of addition. The set of all positive real numbers is a group with the operation of multiplication.

    for every real number, x, define f(x)= ex. That is an isomorphism from one group to the other. It is onto: if y is any positive real number, then take x= ln(y): y= ex. It is one-to-one: if ex= ez then, taking ln of both sides x= z. Finally f(x+y)= ex+y= ex.ey= f(x).f(y).
    Although that is enough to prove this is an isomorphism, it then follows that f(0)= 1 (an isomorphism always take the identity of one structure into the identity of the other) and the f(-x)= 1/f(x) (an isomorphism always take inverses into inverses).
     
  10. Feb 13, 2004 #9
    for rings of finite size (or infinite size) they are isomorphic if they have the same addition and multiplication tables just with different symbols. like if i have one addition table with a, b, c, d and convert that to the same table with differnent letters, a', b', c', and d', and perform the same conversion for the multiplication table, then the two rings are isomorphic. my prof. used to say "if you paint all the elements a different color ten they're isomorphic." the specific function that does this is like the paint brush.
     
  11. Feb 13, 2004 #10

    NateTG

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    A group isomorphism is a bijection (one-to-one and onto) that sends products to products. Two groups are said to be isomorphic if there is an isomorphism between them. For example, S3[/tex] is isomorphic to D3[/tex].

    A ring isomorphism is a bijection that sends sums to sums and products to products.
     
  12. Feb 15, 2004 #11

    HallsofIvy

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    That's really a very good definition of "isomorphism" in general.
    Two algebraic structures (group, field, vector space, ring, etc.) are isomorphic if and only if the are exactly the same except that they use different "names" for the elements (and possibly different symbols for the operations). The "isomorphism function" just associates the "same" object with a different name. As phoenixthoth's prof would say: they are exactly the same except that ones painted red and the other is painted blue!
     
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