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Proving a polynomial over Q[x] is irreducible

  1. Mar 24, 2009 #1
    Show that the polynomial x^2+1 is irreducible in Q[x].

    Hint: If not, it must factor as (ax+b)(cx+d) with a,b,c,d in Q. Show that this is impossible.

    So I got this far:

    ac =1

    ad+bc=0

    bd=1

    I'm not sure how to go further than this.
     
  2. jcsd
  3. Mar 24, 2009 #2
    Hmm, I don't understand why they restrict to just Q, as it's fairly easy to show that it's irreducible over R as well.

    Hint: Square.
     
  4. Mar 24, 2009 #3
    Well the question restricts it to Q[x].

    I asked my prof if just saying it factors to (x-i)(x+i)...however he said it needs to be general. Hence why I have the ac, ac+db, and bd. I'm not sure how to 'generalize' the proof.
     
  5. Mar 24, 2009 #4
    Looking at the equation, the core issue is that squares can't be negative in Q, right? So, you have a set of mixed products. Try to mix them together in a way that generates an absurdity.
     
  6. Mar 24, 2009 #5
    So does this work:

    ad=-bc
    a=-bc/d

    sub into the first equation

    -bc^2=1
    c^2=-1/b


    ^ This is impossible in Q
     
  7. Mar 24, 2009 #6
    No, b could be negative. Try substituting in the other direction.
     
  8. Mar 24, 2009 #7
    well i'm saying that c^2 can't equal a negative number in Q, it would imply an i somewhere
     
  9. Mar 25, 2009 #8
    Yes, but if b is negative, then -1/b will be positive, which is not an issue.
     
  10. Mar 25, 2009 #9
    hmm...so I'm not sure what you mean then.
     
  11. Mar 25, 2009 #10
    By substitute in the other direction, did you mean:

    a=1/c and b=1/d

    d/c+c/d=0

    This is impossible
     
  12. Mar 25, 2009 #11
    That works, though you did it differently from what I had in mind. I'd like to see you complete the reasoning though.

    Here's what I was hinting at: (ad+bc)^2=0

    (ad)^2 + 2abcd + (bc)^2 = 0

    (ad)^2 + (bc)^2 = -2
     
  13. Mar 25, 2009 #12
    Well to play with my idea more:

    d^2+c^2/cd=0

    d^2+c^2=0

    This is not possible...to be honest I'm not sure how to follow the logic of yours.
     
  14. Mar 25, 2009 #13
    What if c and d are 0?

    In moving from the second to the third line, I replaced acbd with 1*1.
     
  15. Mar 25, 2009 #14
    But c and d can't be zero because ac=1 and bd=1
     
  16. Mar 25, 2009 #15
    Well then, it seems our work here is done.
     
  17. Mar 25, 2009 #16
  18. Mar 25, 2009 #17

    Office_Shredder

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    Gold Member

    You can basically show how it's just split into x-i and x+i.


    If x2 + 1 =a(x-b)(x-c) then b, c are roots of the equation, so in particular b2 = -1. This is impossible, hence it can't factor
     
  19. Mar 25, 2009 #18
    You can use Eisenstein and the fact f(x) is irreducible if and only if f(x+1) is irreducible.
     
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