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Math Help - Abstract Alegbra - Ring Homomorphism

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    Abstract Alegbra - Ring Homomorphism

    Consider the ring R = {0,2,4,6,8,10,12} with the operations being addition and multiplication mod 14.


    (a) Show that R is a ring with identity by finding the multiplicative identity (and prove that it’s the identity by computation).


    (b) Find the multiplicative inverse of 10 in R.

    Would I be doing this correctly:

    For part (a):

    8 would be the multiplicative identity since every element in R multiplied by 8 equals itself (mod 14)?
    Last edited by AwesomeHedgehog; April 16th 2014 at 03:27 PM.
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  2. #2
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    Re: Abstract Alegbra - Ring Homomorphism

    Correct. Is that your only question?
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    Re: Abstract Alegbra - Ring Homomorphism

    Then would for part (b) the inverse of 10 just be 8?
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    Re: Abstract Alegbra - Ring Homomorphism

    Quote Originally Posted by AwesomeHedgehog View Post
    Then would for part (b) the inverse of 10 just be 8?
    No. You are looking for $10x \equiv e \pmod{14}$ so that $x \equiv 10^{-1} \pmod{14}$. Since $e = 8$, you want $10x \equiv 8 \pmod{14}$. So, in your multiplication table, what times 10 gives 8?
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    Re: Abstract Alegbra - Ring Homomorphism

    Oh, so the multiplicative inverse of 10 would be 12.

    I did a whole multiplication table on my paper, so that's how I figured that out.
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    Re: Abstract Alegbra - Ring Homomorphism

    Quote Originally Posted by AwesomeHedgehog View Post
    Oh, so the multiplicative inverse of 10 would be 12.

    I did a whole multiplication table on my paper, so that's how I figured that out.
    Correct
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    Re: Abstract Alegbra - Ring Homomorphism

    I hate multiplication tables. Honestly, I do.

    What we want to do here is try to find a ring homomorphism between $R = \{0,2,4,6,8,10,12\}$ and $\Bbb Z_7$. The trick here is to find an IDEMPOTENT element of $R$, because identity elements are idempotent (they satisfy: $e^2 = e$).

    We can eliminate 2 right off the bat, and:

    4*4 = 2
    6*6 = 8
    10*10 = (-4)*(-4) = 4*4 = 2
    12*12 = (-2)*(-2) = 2*2 = 4

    so the only idempotent (besides 0) is 8: 8*8 = 8.

    Next we need to show that 8*a = a, for all of our a's. Note that we have a = 2b (b = 0,1,2,3,4,5,6). So, mod 14, we have:

    8a = 8(2b) = 16b = 2b = a. That establishes identity.

    Since any non-zero element is a generator (because $(R,+)$ is an additive cyclic subgroup of $(\Bbb Z_{14},+)$, and since any ring homomorphism maps identity to identity), this suggests the homomorphism:

    $\phi: \Bbb Z_7 \to R$ given by $\phi(k \text{ (mod }7)) = 8k \text{ (mod }14)$.

    Of course, we have to verify the multiplicative property:

    $\phi(km) = \phi(k)\phi(m)$. This means that:

    $\phi(km) = 8km = (8)(8)(km) = (8k)(8m) = \phi(k)\phi(m)$, which is true since 8 is idempotent.

    Now to find the inverse of 10 in $R$, we need to do 2 things:

    a) find the pre-image of 10 under $\phi$
    b) find the inverse of the pre-image in $\Bbb Z_7$

    To accomplish (a), let's calculate explicitly what $\phi$ does:

    $\phi(0) = 0$
    $\phi(1) = 8$
    $\phi(2) = 2$ (16 = 2 (mod 14))
    $\phi(3) = 10$ <--we can stop, now, the pre-image of 10 is 3.

    Now we need to find the inverse of 3 (mod 7). The easiest way to do this is to find integers r,s with 3r + 7s = 1. This means that 3r = 1 (mod 7), so taking r (mod 7) will give us the inverse. To find these integers, use the gcd division algorithm:

    7 = 2(3) + 1, so:

    1 = 7 - 2(3), that is: (-2)(3) + (1)(7) = 1, so r = -2, and s = 1. We don't need s, so forget about it. We DO need r, and we need to find -2 (mod 7), which is 5. Hence 5 is the multiplicative inverse of 3 (mod 7):

    (3)(5) = 1 (mod 7) (because 15 = 1 (mod 7)).

    So the inverse of 10 (mod 14) is going to be $\phi(5)$, which is 40 (mod 14), which reduces to 12 (40 = 12 + 2(14)).

    The advantage to doing it this way is avoiding explicitly creating a multiplication table with 49 entries (although if you're clever, you can reduce it to just 21 entries, which is still more calculation than I like to do).
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