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Math Help - Factor Rings

  1. #1
    Super Member Bernhard's Avatar
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    Factor Rings

    Can someone please help with the following problem taken from Beachy and Blair: Abstract Algebra ch 5 Commutative Rings

    Give the multiplication table for the ring  Z_3 [x] / <x^2 - 1>
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  2. #2
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    Re: Factor Rings

    note that the elements of Z3[x]/<x2 - 1> will be cosets that we can represent as polynomials of degree < 2, to save on notation instead of writing p(x) + I, i will simply write p(u), and thus u = x + I, and i will write a instead of a + I

    (taking advantage of the ring isomorphism Z3-->Z3[x]/<x2 - 1> given by: a--> a + <x2+1>).

    there are exactly NINE of these cosets:

    0,1,2,u,u+1,u+2,2u,2u+1,2u+2. since this is a commutative ring (since Z3 is commutative) i will only write "half" the products (a little more than half, actually).

    0*0 = 0
    0*1 = 0
    0*2 = 0
    0*u = 0
    0*(u+1) = 0
    0*(u+2) = 0
    0*(2u) = 0
    0*(2u+1) = 0
    0*(2u+2) = 0

    1*1 = 1
    1*2 = 2
    1*u = u
    1*(u+1) = u+1
    1*(u+2) = u+2
    1*(2u) = 2u
    1*(2u+1) = 2u+1
    1*(2u+2) = 2u+2 so far, these are all trivial

    2*2 = 1 (4 = 1 (mod 3))
    2*u = 2u
    2*(u+1) = 2u+2 (remember distributivity still works)
    2*(u+2) = 2u+1
    2*(2u) = u (associativity still applies, as well)
    2*(2u+1) = u+2
    2*(2u+2) = u+1

    u*u = u2 = (u2 - 1) + 1 = 1 (since u = x + I, we are essentially saying x2 - 1 + I = I, so x2 + I = 1 + I, that is u2 = 1. u is a "third square root of 1". note this can't happen in a field)
    u*(u+1) = u2 + u = 1 + u = u+1
    u*(u+2) = 2u+1 (what did i do here?)
    u*(2u) = 2
    u*(2u+1) = u+2
    u*(2u+2) = 2u+2

    (u+1)*(u+1) = 2u+2
    (u+1)*(u+2) = 0 <---a zero divisor. interesting....
    (u+1)*(2u) = 2u+2 <---note that cancellation does NOT hold (u+1)*(u+1) = (u+1)*(2u), but u+1 does not equal 2u.
    (u+1)*(2u+1) = 0 <---more zero divisors.
    (u+1)*(2u+2) = u+1...but u+1 is NOT the multiplicative identity. very bizarre, hmm?

    (u+2)*(u+2) = u+2 <--more strangeness
    (u+2)*(2u) = u+2
    (u+2)*(2u+1) = 2u+1 <---u+2 behaves VERY strangely. note that 1 is a root of x2-1, and so is u, so u+2 is "sort of like" 1+2 = 0.
    (u+2)*(2u+2) = 0

    (2u)*(2u) = u
    (2u)*(2u+1) = 2u+1
    (2u)*(2u+2) = u+1

    (2u+1)*(2u+1) = u+2
    (2u+1)*(2u+2) = 0

    (2u+2)*(2u+2) = 2u+2

    all in all, a very strange ring of order 9.
    Thanks from Bernhard
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Factor Rings

    Will work through this now

    Peter
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Factor Rings

    I assume you know the Chinese Remainder Theorem, so that \mathbb{Z}_3[x]/(x^2-1)\cong \mathbb{Z}_3[x]/(x-1)\times\mathbb{Z}_3[x]/(x+1)\cong\mathbb{Z}_3^2.
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  5. #5
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    Re: Factor Rings

    that's true, alex, but it's not the "usual" (field) multiplication on Z3xZ3.

    but let's use this:

    say we want to multiply (u+1)(u+2).

    first, which element of Z3xZ3 does u correspond to? a little thought shows it must be (1,2) (a root of x-1 in the first factor, and a root of x+1 in the second factor).

    since we want to preserve the identity, we have to send 1-->(1,1), the identity of Z3xZ3.

    so u+1 = (1,2) + (1,1) = (2,0) <---this explains why u+1 is a 0-divisor.

    similarly, u+2 = (1,2) + (2,2) = (0,1) <--explains why u+2 is a 0-divisor.

    so (u+1)(u+2) = (2,0)(0,1) = (0,0) = 0

    continuing, we can now create a complete correspondence between the 2 versions:

    0<-->(0,0) (duh!)
    1<-->(1,1)
    2<-->(2,2)
    u<-->(2,1)
    u+1<-->(0,2)
    u+2<-->(1,0)
    2u<-->(1,2)
    2u+1<-->(2,0) wait...was 2u+1 a 0-divisor? yep, sure was.
    2u+2 <-->(0,1)

    proving this IS an isomorphism is tedious (additively it's clear, but multiplication is a chore) perhaps the distributive law might help....
    Thanks from Bernhard
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Re: Factor Rings

    No, it's the multiplication given by the product ring structure, which is not that bad.
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  7. #7
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    Re: Factor Rings

    i think this is a great example of how the additive structure of a ring can have very little to do with the multiplicative structure. and that different multiplicative structures on the very same abelian group can produce VERY different results. put another way: the cardinality of a finite ring tells us a lot less than we might hope.
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