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Thread: Unit

  1. #1
    Aug 2012


    I can not solve this exercise.I need help

    Let A be any commutative ring with identity 1 $\displaystyle \neq{}$ 0 .Let R be the set of all group homomorphism of the additive group A to itself with addition defined as pointwise addition of functions and multiplication defined as function composition.Prove that these operations make R into a ring with identity .Prove that the units of R are the group automorphism of A
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Unit

    ok, we have

    R = {f:A→A : f(a+a') = f(a) + f(a'), for all a,a' in A}.

    we define:

    (f+g)(a) = f(a) + g(a) (the + on the LHS is in R, the + on the RHS is in A)

    (fg)(a) = f(g(a)).

    first: we need to check closure, that is, we need to show that f+g is a homomorphism of the group (A,+).

    now (f+g)(a+a') = f(a+a') + g(a+a') (by definition)

    = f(a) + f(a') + g(a) + g(a') (since f and g are additive homomorphisms)

    = f(a) + g(a) + f(a') + g(a') (since (A,+) is abelian)

    = (f+g)(a) + (f+g)(a') (by definition of f+g). bim, bam boom!

    now we know the map z(a) = 0, for all a in A, is a homomorphism (the trivial, or 0-map). i claim z is an additive identity for R.

    (note that + in R is commutative since f+g = g+f , by commutativity of + in A. the same goes for associativity of +).

    (f+z)(a) = f(a) + z(a) = f(a) + 0 = f(a), for all a in A, hence f+z = f.

    if we define -f by:

    (-f)(a) = -(f(a)), i claim -f is an additive inverse for f:

    (f + -f)(a) = f(a) + (-f)(a) = f(a) + -(f(a)) = 0 = z(a), hence f+ -f = z.

    hence (R,+) is an abelian group.

    now we need to show closure of * in R (associativity is trivial, because composition of functions is associative), so we need to show that fg is an additive homomorphism of A.

    (fg)(a+a') = f(g(a+a')) = f(g(a) + g(a')) (since g is an additive homomorphism)

    = f(g(a)) + f(g(a')) (since f is an additive homomorphism)

    = (fg)(a) + (fg)(a').

    define i(a) = a, for all a in A. clearly this is an additive homomorphism, since i(a+a') = a + a' = i(a) + i(a'). i claim i is a multiplicative identity for R:

    (fi)(a) = f(i(a)) = f(a)
    (if)(a) = i(f(a)) = f(a), so fi = if = f.

    now..."the big one" distributivity. note we have to prove BOTH distributive laws, since we have no idea if R is commutative (in fact, it usually isn't).

    we must show for all f,g, and h in R, that:

    f(g+h) = fg + fh
    (f+g)h = fh + gh

    f(g+h)(a) = f((g+h)(a)) = f(g(a) + h(a)) = f(g(a)) + f(h(a)) = (fg)(a) + (fh)(a) = (fg + fh)(a), so f(g+h) = fg + fh.
    (f+g)h(a) = (f+g)(h(a)) = f(h(a)) + g(h(a)) = (fh)(a) + (gh)(a) = (fh + gh)(a), so (f+g)h = fh + gh.

    so (R,+,*) is a ring with unity.

    to prove U(R) = Aut(A), we do 2 things.

    suppose u in R, is a unit. then we show u is in Aut(A).
    suppose f is in Aut(A). then we show f is a unit of R.

    if u is a unit in R, there exists an additive homomorphism r of A to A with ur = ru = i. what does this mean?

    it means u(r(a)) = a, for all a in A, and r(u(a)) = a, for all a in A.

    claim 1: u is injective. suppose u(a) = u(a'). then r(u(a)) = r(u(a')), so a = a'.

    claim 2: u is surjective. let a be any element of A. then u(r(a)) = a, so r(a) is a pre-image of a under u.

    thus u is a bijective homomorphism of (A,+) to (A,+), so is an isomorphism of (A,+) with (A,+), so is an automorphism of A.

    now suppose that f is in Aut(A). since f is bijective, it possesses an inverse automorphism, f-1 (we are appealing to the fact that Aut(A) is a group).

    hence (ff-1)(a) = a, so ff-1 = i, and

    (f-1f)(a) = a, so f-1f = i, thus f is a unit.
    Thanks from mathrld
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  3. #3
    Aug 2012

    Re: Unit

    Hi Deveno !

    thanks for your answer! I understand the exercise
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