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Math Help - Proving a ring is a field

  1. #1
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    Proving a ring is a field

    I was given the following question:

    A is the set Q of the rational numbers, and the operations are \oplus and \otimes defined as follows:

    a \oplusb = a + b + 1
    a \otimesb = ab + a + b

    I have proved that it is a ring. So all I have to do is..

    Prove that A is commutative

    a \otimesb = ab + a + b
    b \otimesa = ba + b + a

    ab + a + b = ba + b + a
    ab + a + (b - b) - a = ba + b + (a - a) - b
    ab + a + 0b - a = ba + b + a0 - b
    ab + a + 0 - a = ba + b + 0 - b
    ab + a - a = ba + b - b
    ab + (a - a) = ba + (b - b)
    ab + a0 = ba + b0
    ab + 0 = ba + 0
    ab = ba

    I think I am off to the wrong start if someone can correct me that would be great

    Prove that A has unity
    Prove that every non zero element of A is invertible
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  2. #2
    MHF Contributor

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    Re: Proving a ring is a field

    a⊗b = ab + a + b = ab + b + a (because ORDINARY addition is commutative)

    = ba + b + a (because ORDINARY multiplication is commutative)

    = b⊗a. you're over-thinking it.

    if A has a unity, we should be able to FIND it, ya?

    such a unity has the property that

    a⊗x = x⊗a = a for all a in A. so let's write this in terms of operations we understand better:

    ax + a + x = a
    ax + x = 0
    x(a+1) = 0

    now if a is not -1, this means x= 0. so the multiplicative identity is 0:

    a⊗0 = a0 + 0 + a = a.

    now to prove any non-zero element of A is invertible, we need to know what "zero" (the additive identity) means.

    it should be clear that the additive identity is -1:

    a⊕(-1) = a + -1 + 1 = a.

    so in A, "non-zero" means A-{-1}. so we have to show that if a is not -1, there is some b in A with a⊗b = 0 (the multiplicative identity).

    again, let's try to FIND b, by solving for b in terms of the ordinary operations:

    a⊗b = 0
    ab + a + b = 0
    ab + b = -a
    b(a+1) = -a since a is not -1, a+1 is not 0, so we can divide by it:

    b = -a/(a+1).

    verifying:

    a⊗(-a/(a+1)) = -a2/(a+1) + a - a/(a+1) = (-a[SUP]2 - a)/(a+1) + a(a+1)/(a+1) = (-a2 - a + a2 + a)/(a+1) = 0/(a+1) = 0.

    ***************

    a faster way, given that A is already a ring:

    define φ:Q-->A by φ(q) = q-1.

    a) φ is injective:

    φ(q') = φ(q) => q'-1 = q-1 => q' = q

    b) φ is surjective:

    let a be any element of A, then for q = a+1, φ(q) = (a+1) -1 = a.

    c) φ is a ring homomorphism:

    φ(q+q') = q+q'-1 = (q-1) + (q'-1) + 1 = (q-1)⊕(q'-1) = φ(q)⊕φ(q')

    φ(qq') = qq' - 1 = qq' - q - q' + 1 + q - 1 + q' - 1 = (q-1)(q'-1) + (q-1) + (q'-1) = (q-1)⊗(q'-1) = φ(q)⊗φ(q')

    together, (a) through (c) say φ is a ring isomorphism, thus φ(Q) = A is a field.
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