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How many maximal ideals of Z_3[x]

  1. Apr 24, 2017 #1
    1. The problem statement, all variables and given/known data
    How many maximal ideals generated by quadratic polynomials does ##\mathbb{Z}_3 [x]## contain?

    2. Relevant equations


    3. The attempt at a solution
    Here is my reasoning. There are a total of ##2 \cdot 3 \cdot 3 = 18## quadratic polynomials in ##\mathbb{Z}_3 [x]##. If we write ##(x+a)(x+b)##, then there are ##\binom{3}{2} = 3## ways to choose ##a## and ##b##, up to the order of the factors. Since these are the only ways to factor a polynomial, there are 18 - 3 = 15 irreducible quadratic polynomials.

    Is this correct?
     
  2. jcsd
  3. Apr 24, 2017 #2

    andrewkirk

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    There a few glitches:

    1. Combinations is not a correct way to count possibilities for a and b, since Combinations are for choices without replacement, and the choice here is with replacement - ie a can equal b.

    2. Your answer is described as the number of (distinct) irreducible quadratic polynomials. What you are required to count is the number of distinct maximal ideals generated by such. Are you sure there's a 1-1 correspondence between the two? If so, can you demonstrate that, or is it a result you've already been given? Every irreducible polynomial generates a maximal ideal, but does that mean every irreducible polynomial generates a different ideal?
     
  4. Apr 24, 2017 #3
    So taking into account 1., there are 3 + 3 = 6 reducible quadratics, so there are 18 - 6 irreducible quadratics.

    Now, I need to find the number of distinct maximal ideals generated by these irreducible quadratics, but I am not sure how to see when two polynomials generate the same ideal.
     
  5. Apr 24, 2017 #4

    andrewkirk

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    Interesting. I get the same answer, but I don't see how 3+3 comes into it. For me it's 3+2+1.
    Are they principal ideals? What rules do you know about things one can do to the generator of a principal ideal without changing the ideal?
     
  6. Apr 24, 2017 #5
    You can take the additive inverse of the generator, right? Like that for ##\mathbb{Z}##, ##\langle 1 \rangle = \langle -1 \rangle##
     
  7. Apr 24, 2017 #6

    andrewkirk

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    A more general way to approach that uses the concept of units, together with Principal Idealness and commutativity of the ring. In this case I expect it'll give the same answer, but it's good to be aware of the most general approach possible.
     
  8. Apr 24, 2017 #7
    Don't units have to do with multiplicative inverses though? Do two quadratics generating the same principal ideal have to do with multiplicative inverses or additive inverses?
     
  9. Apr 25, 2017 #8

    andrewkirk

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    'Have to do with' is too vague a concept.

    What are the units of ##\mathbb Z_3[x]##? What happens if you multiply a non-trivial polynomial in ##\mathbb Z_3[x]## by one of them?
     
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