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Irreducibility Question (Rationals, Eisenstein!)

  1. Dec 1, 2005 #1
    Hello. I have this question from the book:

    Prove that the polynomial [tex]x^4 + 2x^2 + 2[/tex] is irreducible in Q[x]. (Q being the rational fancy Q)

    So I used Eisenstein Criterion (because we are dealing with the rationals) and said that the coefficents are: 1 (for x^4), 2(for x^2), 2 with the last 2 coefficients being the important ones. So I then said that lets use 2 as the prime, and then 2 | 2 (the middle coefficient), and 4 (2^2) does not divide 2 (the last coefficient), so [tex]x^4 + 2x^2 + 2[/tex] is not reducible in Q[x].

    That seems quite acceptable to me, but the book has no examples, did I do the problem correctly? Danke!
    Last edited: Dec 1, 2005
  2. jcsd
  3. Dec 1, 2005 #2


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    All the coefficients are important, 2 divides the x^3 and x coefficients trivially. More importantly, there's a condition about the lead coefficient that you haven't mentioned...
  4. Dec 1, 2005 #3
    Thanks, I did not think about those. However, I am not sure what you are talking about with the leading coefficient (I am guessing you are talking about the 1 in front of x^4).

    In my book the definition of The Eisenstein Criterion is:

    Let [tex]f(x) = x^n + a_{1}x^{n-1} + .. + a_{n}[/tex] be a polynomial with integer coefficients. Suppose that there is some prime p such that [tex]p|a_{1}, p|a_{2}, ..., p|a_{n}[/tex], but p^2 does not divide [tex]a_{n}[/tex] Then f(x) is irreducible in Q[x].

    Hmm... I just looked in another book, and on wikipedia, at the definition of The Eisenstein Criterion and they both mention a property about the first coefficient (1 for x^4); specifically: p cannot divide the first coefficient(1 for x^4). Which looks true too. Not sure why one of my books says that and the other does not. Thanks
    Last edited: Dec 1, 2005
  5. Dec 2, 2005 #4


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    Your book is assuming a "monic" polynomial- the "leading coefficient" (the coefficient of the highest power) is 1 and no prime divides that so it isn't necessary to mention it. Since factoring out a constant doesn affect irreducibility, the two forms are equivalent.
  6. Dec 2, 2005 #5
    You are absolutely correct, I missed that part! Thanks.
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