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Irreducible polynomial in Q[x] but redicuble in Z[x]

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Give an example of a polynomial irreducible in Q[x], but reducible in Z[x]


    2. Relevant equations



    3. The attempt at a solution

    I think there is no example of this. The coefficients of a reducible polynomial with integer coefficients will be rational, or am I mistaken?
     
  2. jcsd
  3. Aug 31, 2012 #2

    jbunniii

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    The question makes no sense. If a polynomial can be factored in Z[x], then it can be factored (with the same factors) in Q[x].

    Also, I'm not sure what you mean by this question:
    Since integers are rational numbers, it's (trivially) true that integer coefficients are rational coefficients. But what does this buy you?
     
  4. Aug 31, 2012 #3

    Hurkyl

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    There is an example.

    Recall the definition of irreducible:
    A non-unit x is irreducible if and only if it cannot be written as yz where y and z are non-units​

    Now, as is very often a useful, we simply translate the question according to the definition:

    We seek polynomials f,g,h with integer coefficients such that:

    • f=gh, and none of f,g,h are units in Z[x]
    • There do not exist non-units u,v in Q[x] such that f=uv

    A more symmetric phrasing might be useful, to spot what you're missing:

    We seek a polynomial f with integer coefficients such that

    • There exist non-units u,v of Z[x] such that f=uv
    • There do not exist non-units u,v of Q[x] such that f=uv
     
  5. Aug 31, 2012 #4
    The question is nonsense as written. If a polynomial f(x) in Q[x] is reducible (spelling!) in Z[x], then f(x) = g(x)h(x) were g(x) and h(x) are in Z[x]. We assume that the degrees of g(x) and h(x) are > 0. This means that g(x) and h(x) have integer coefficients. When multiplied, the product polynomial, f(x), must have integer coefficients in the first place.
    g(x) and h(x) are also in Q[x], so f(x) IS reducible in Q[x].

    The non-trivial result is that if a polynomial in Z[x] is irreducible, then it is also irreducible in Q[x]. This takes a bit of proof and is usually given in a first course in so-called abstract algebra.
     
  6. Aug 31, 2012 #5

    Hurkyl

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    Not only does the question make sense, the request can be satisfied.

    And thus, we conclude that any example must violate the assumption....
     
  7. Sep 1, 2012 #6
    I am still stumped. Do you have another hint to try and make me see what I am missing?
     
  8. Sep 1, 2012 #7

    HallsofIvy

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    If a polynomial is reducible in the rational numbers, it can be written as a fraction (1 over the least common multiple of the denominators of the coefficients) times a polynomial that is reducible over the integers. That does NOT mean the original poynomial was reducible over the integers.
     
  9. Sep 1, 2012 #8

    Hurkyl

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    Have you noticed that I've already said:
    • If f=uv is an example, then at least one of u and v are units of Q[x],
    • If f=uv is an example, then at least one of u and v have degree 0?
    ?

    If not then be aware that your problem is not algebra, but logic and its application to problem solving.
     
    Last edited: Sep 1, 2012
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