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Computing the order in a polynomial quotient ring

  1. May 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider the quotient ring ##F = \mathbb{Z}_3 [x] / \langle x^2 + 1 \rangle##. Compute the order of the coset ##(x+1) + \langle x^2 + 1 \rangle## in the group of units ##F*##.

    2. Relevant equations


    3. The attempt at a solution
    I was thinking that I just continually compute powers of (x+1) until I reach a polynomial with ##x^2+1## as a factor, but this is cumbersome. I guess my question, after each computation, how do I reduce the expression to get a simpler description of the coset? Do I just subtract ##x^2 + 1## as many times as a please?
     
    Last edited: May 6, 2017
  2. jcsd
  3. May 5, 2017 #2

    andrewkirk

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    The question, as stated, presupposes that ##(x+1)+\langle x^2+1\rangle## is a unit of ##F##. Is that true? Why, or why not?
     
  4. May 5, 2017 #3
    Well ##F## is a field, since ##\langle x^2 + 1 \rangle## is a maximal ideal since ##x^2 + 1## is irreducible in ##\mathbb{Z}_3##, so ##(x+1)+\langle x^2+1\rangle## must be a unit of ##F## since all non-zero elements are units by definition of a field.
     
  5. May 6, 2017 #4

    andrewkirk

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    Good.

    But in this latest post you wrote ##\mathbb Z_3##, implying that the parent polynomial ring is ##\mathbb Z_3[x]##, whereas in the OP you wrote ##\mathbb Z[x]##. Which is it?
     
  6. May 6, 2017 #5
    Oops, it's ##\mathbb{Z}_3 [x]##. Sorry.
     
  7. May 6, 2017 #6

    andrewkirk

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    That makes the problem make more sense.

    You can shortcut all the polynomial multiplication by making use of the fact that ##\mathbb R[x]/\langle x^2+1\rangle\cong \mathbb C## via the field isomorphism that maps 1 to 1 and ##x## to ##i##. It would seem to be a natural guess then that the quotient field here is isomorphic to the Gaussian Integers modulo 3, ie ##\mathbb Z_3+i\mathbb Z_3##. What is the lowest positive integer to which we need to raise ##1+i## to get a real number?
     
  8. May 6, 2017 #7
    Since ##(1+i)^4 = -4##, the answer is 4.
     
  9. May 6, 2017 #8

    andrewkirk

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    Not so fast. Firstly, the question should have been 'what is the lowest positive integer ##m## to which we need to raise ##1+i## to get a number that is 1 mod 3.' My mistake, but we wouldn't want the hints on here to be too accurate would we, or where would be the satisfaction for the students?

    Secondly, having found ##m##, since we have only guessed that the isomorphism holds, we need to verify that ##(x+1)^m=1\mod (x^2+1)## in ##\mathbb Z_3[x]##. Taking the appropriate row of Pascal's Triangle will help us do this.
     
  10. May 6, 2017 #9
    Okay, so ##(1+i)^8 = 16 = 1 \pmod 3##. Also, rather than using pascal's in this example, is there a more general way? For example, what if we have an irreducible quadratic of the form ##ax^2 + bx + c##?
     
  11. May 6, 2017 #10

    andrewkirk

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    Provided the roots of the quadratic are in the Gaussian Rationals ##\mathbb Q+i\mathbb Q## we can proceed the same way. We find the smallest ##m## such that ##(f+id)^m=1\mod 3## where ##f+dx## represents the coset we are considering.

    If we have proven the isomorphism, it ends there. If not, we need to divide ##(f+dx)^m## by the irreducible polynomial, in ##\mathbb Z_3[x]## and show that the remainder is ##1\mod 3##.
     
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