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Math Help - Something about isomorphism

  1. #1
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    Exclamation Something about isomorphism

    The question is:

    Let f : R ! S be a homomorphism. Let
    T = { f(r) : r 2 R}.
    T is called the image of f. Show that T is a subring of S.
    Hint: In this problem, you can use the following facts:
    f(0R) = 0S,f(−x) = −f(x) for x 2 R, and f(x − y) = f(x) − f(y) for x, y 2 R. These facts were stated in class for isomorphisms only but the same properties (with the same proofs) hold
    for homomorphisms.


    Thanks very much.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by suedenation
    The question is:

    Let f : R ! S be a homomorphism. Let
    T = { f(r) : r 2 R}.
    T is called the image of f. Show that T is a subring of S.
    Hint: In this problem, you can use the following facts:
    f(0R) = 0S,f(−x) = −f(x) for x 2 R, and f(x − y) = f(x) − f(y) for x, y 2 R. These facts were stated in class for isomorphisms only but the same properties (with the same proofs) hold
    for homomorphisms.


    Thanks very much.
    You forgot to mention it, but R and S must be rings.

    T is, of course, a subset of S.

    By definition of the homomorphism, f(a+b)=f(a)+f(b) and f(a*b)=f(a)*f(b).

    If a set T is a ring, it must have the following properties:
    1. a+b=b+a
    2. (a+b)+c=a+(b+c)
    3. An identity (0) for addition exists.
    4. An additive inverse exists for all elements of T.
    5. (a*b)*c=a*(b*c)
    6. An identity (1) for multiplication exists.
    7. * distributes over +.

    You can use the homomorphism to assist you. For example, the first property:
    f(a+b)=f(a)+f(b)
    f(b+a)=f(b)+f(a)
    Since f(a+b)=f(b+a) we have that f(a)+f(b)=f(b)+f(a). So addition in T is commutative.

    The second property follows the same pattern.
    The third property: f(a+0)=f(a)+f(0)=f(a), so f(0)=0T exists.
    The fourth property: f(a+-a)=f(a)+f(-a)=f(0)=0T, so f(-a)=-f(a) exists.
    The fifth property:
    f((a*b)*c)=(f(a)*f(b))*f(c)
    f(a*(b*c))=f(a)*(f(b)*f(c))
    and the two of these are equal.

    The sixth property f(a*1)=f(a)*f(1)=f(a) so f(1)=1T exists.
    The seventh property: f(a*(b+c))=f(a)*(f(b)+f(c)) etc.

    And, of course, you should put in a statement that 0T and 1T correspond to 0S and 1S respectively, which is easy to show.

    -Dan
    Last edited by topsquark; March 27th 2006 at 01:07 PM. Reason: Addition
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