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Is this irreducible?

  1. Mar 16, 2005 #1
    Hi,
    I need to figure out whether or not the polynomial
    [tex]x^6-2x^3-1[/tex]
    is irreducible (over Q).
    I don't think Eisenstein works in this case, and performing modulo 2 on this i get [tex]x^6-1[/tex]
    which is reducible over F2.
    Any ideas? Incidently, if i let y=x^3, then i get
    [tex]y^2-2y -1[/tex]
    which is irreducible over Q.....but i'm not sure if that means anything about the original polynomial.
     
  2. jcsd
  3. Mar 16, 2005 #2
    Sure enough, Eisenstein doesn't work as there is no prime [tex] p[/tex] such that [tex] p\mid -1[/tex]. However if you reduce modulo 3, you get (i hope)
    [tex] x^6-[2]x^3-[1] = x^6+[-2]x^3+[-1] = x^6+x^3+[2][/tex] and this polynomial in [tex] Z_2[x][/tex] seems to be irreducible (at a first try I couldn't factor it in 2 polynomials of smaller degree, but you can try yourself). So if that polynomial is irreducible in [tex] Z_2[x][/tex] then [tex]x^6-2x^3-1[/tex] is irreducible in [tex] Q[x][/tex].
     
  4. Mar 17, 2005 #3
    Hmm..okay. I was hoping I wouldn't have to resort to brute force hehe
    OK, so I have shown that the polynomial [tex] x^6+x^3+2[/tex] has no linear or quadratic factors over F3, but how did you show it can't factor into cubic terms? I don't really know the irreducible cubic polynomials over F3, and it seems like there might be quite a few of them?
     
  5. Mar 17, 2005 #4

    shmoe

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    It won't be horrid, there are only 18 candidates for monic irreducibles of degree 3 (constant term must be non-zero), 10 of them will have roots. Trial division by the remaining 8 will be tedious though. Actually you can cut this down quite alot, if you check that none of the monic irreducibles with constant term 2 divide your polynomial, then you know none with constant term 1 do (do you see why?).

    Sorry, I don't have anything quicker to offer.
     
  6. Mar 17, 2005 #5
    Yes, great, thanks for the tip. After a little tedious work, it turns out that it is irreducible over F3, so my original polynomial was irreducible over Q. Yay, thanks a lot :biggrin:
     
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