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Eisenstein's Irreducibility Criterion

  1. Oct 9, 2010 #1
    I have a question about the application of Eisenstein’s Criterion. I want to show that

    g(x)=x4+4x3+7x+5 is irreducible over [tex]\mathbb{Q}[/tex].

    That means I need to find a prime number p such that

    [tex]p \nmid 1[/tex] , [tex]p | 4[/tex] , [tex]p | 7[/tex], [tex]p | 5[/tex] and [tex]p^2 \nmid 5[/tex].

    But unfortunately I can't see any prime number which would satisfies this! :confused:

    I think there is a theorem that says if g(x+1) is irreducible then g(x) is irreducible. So in this case

    g(x+1) = x4 + 8x3 + 18 x2 + 16 x + 17

    But again I cannot find a p to satisfiy Eisenstein’s irreducibility criterion... So why does the method fail? Then what other method can one use to establish g(x)'s irreducibility?
  2. jcsd
  3. Oct 9, 2010 #2
    You could use:
    [tex] (g\mod p) \text{ is irreducible in } (\mathbb{Z}/p\mathbb{Z})[X] \Rightarrow
    g \text{ is irreducible in } \mathbb{Z}[X] \Rightarrow
    g \text{ is irreducible in } \mathbb{Q}[X]. [/tex]

    where p is prime
    Last edited: Oct 9, 2010
  4. Oct 10, 2010 #3
    g(x+2) works for Eisenstein.
  5. Oct 11, 2010 #4
    Are you sure?

    [tex]g(x+2)= x^4 + 12 x^3 + 48 x^2 + 87 x + 67[/tex]

    2 does not divide 87 and 3 does not divide 67. So there are no primes to use for Eisenstein...

    I haven't seen this before but there s a similar test in my book called "Mod p irreducibility test", where p is a prime, here we must reduce the polynomial in Zp and see if the reduce polynomial is irreducible over Zp, if it is, it imples irreducibility over Q for the original polynomial. So here's what I've done:

    Let g(x)=x4+4x3+7x+5. Then over [tex]\mathbb{Z}_2[/tex], we have [tex]\bar{g}(x)=x^4+x+1[/tex]. If we show that [tex]\bar{g}(x)[/tex] is irreducible over [tex]\mathbb{Z}_2[/tex] that implies that g(x) is irreducible over [tex]\mathbb{Q}[/tex].

    But how do we show this?? We surely can't use Eisenstein here... :confused:
    Last edited: Oct 12, 2010
  6. Oct 12, 2010 #5
    In [tex]\mathbb{Z}_2[/tex] all coefficients are either 1 or 0, so ...
  7. Oct 12, 2010 #6
    What do you mean?? I know that [tex]\bar{g}(0)=1[/tex] and [tex]\bar{g}(1)=1[/tex] in [tex]\mathbb{Z}_2[/tex], so there are no zeros. But we can't use this to conclude that it is irreducible because this test is only for degrees 2 and 3, so I can't use it when the polynomial has degree 4...

    And for the Eisenstein we need a prime that divides 1 and doesn not divide 1 at the same time which is impossible!
  8. Oct 12, 2010 #7
    You have [tex]\bar{g}(x)=x^4+x+1[/tex]. You cannot divide out a factor x, so if there were some factor it would have to be of the form (x^n+q) for some polynomial q and some integer n. Any coefficient in these polynomials is 1, use this to argue that you must have [tex](x^n+q_1)(x^n+q_2) \neq x^4+x+1[/tex] for all [tex]q_1,q_2\in \mathbb{Z}_2[X][/tex].
  9. Oct 12, 2010 #8
    I think it's better to use xn and xm because they can be different as long as n+m=4.

    [tex](x^n+q_1)(x^m+q_2) = x^{n+m} + q_2 x^n + q_1 x^m +q_1q_2[/tex]

    In order for all coefficients to be 1 or 0, q1=q2=1 OR 0. Then we have:

    [tex](x^n+1)(x^m+1) = x^{n+m} + x^n + x^m + 1[/tex] .....(1)


    [tex](x^n+0)(x^m+0) = x^{n+m}[/tex] .....(2)

    The polynomial (2) is not equal to [tex]\bar{g}(x)=x^4+x+1[/tex], and in polynomial (2) regardless of our choice for m and n, we will never end up with [tex]\bar{g}(x)[/tex]:

    n,m = 4,0 [tex]\implies x^4+x^4+1[/tex]
    n,m = 3,1 [tex]\implies x^4 + x^3 + x^2 +1[/tex]
    n,m = 2,2 [tex]\implies x^4 + x^2 + x^2 +1[/tex]

    Is this correct now?
  10. Oct 13, 2010 #9
    Sorry, I must have made some terrible mistake :blushing:
  11. Oct 13, 2010 #10
    Yes, I should have done that

    No, q1 and q2 could be any polynomial just all the coefficients in those polynomials are either 0 or 1, i.e. all the ai in [tex]q_1 = \sum a_i X^i [/tex] are either 0 or 1 and all the bi in [tex]q_2 = \sum b_i X^i [/tex] are either 0 or 1.

    You can use your idea like this:

    if n,m = 4,0 then we have [tex] (x^4 + q_1) (1+q_2) =...[/tex] which cannot be equal to [tex]\bar{g}[/tex] because...
    if n,m = 3,1 then we have [tex] (x^3 + q_1) (x+q_2) =... [/tex] which cannot be equal to [tex]\bar{g}[/tex] because...
    if n,m = 2,2 then we have [tex] (x^2 + q_1) (x^2+q_2) =...[/tex] which cannot be equal to [tex]\bar{g}[/tex] because...

    (better not use "\implies" here, whenever you use "\implies" you need to make really sure that whenever the thing on the left of the arrow is true that then also the thing on the right is true, and it is really easy to forget something on the left)
    Last edited: Oct 13, 2010
  12. Oct 14, 2010 #11
    Thank you so much;

    We have [tex](x^n+q_1)(x^m+q_2) = x^{n+m} + q_2x^n + q_1x^m + q_1q_2[/tex]

    if n,m = 4,0 then we have [tex] (x^4 + q_1) (1+q_2) = x^4+q_2x^4+q_1+q_1q_2[/tex] which cannot be equal to [tex]\bar{g}[/tex] because regardless of what value (1 or 0) we choose for the coefficients we will never get [tex]\bar{g}[/tex].

    if n,m = 3,1 then we have [tex] (x^3 + q_1) (x+q_2) =x^4+q_2x^3+q_1x+q_1q_2[/tex] which cannot be equal to [tex]\bar{g}[/tex] because no possible choice of coefficients can make it equal to [tex]\bar{g}[/tex].

    if n,m = 2,2 then we have [tex] (x^2 + q_1) (x^2+q_2) =x^4+q_2x^2+q_1x^2+q_1q_2[/tex] which cannot be equal to [tex]\bar{g}[/tex] because whatver value we pick for [tex]q_1[/tex] and [tex]q_2[/tex] we will never end up with [tex]\bar{g}[/tex].

    This is okay now I guess?
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