1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prep for Algebra Comprehensive Exam #3

  1. Jul 24, 2007 #1
    3. Show that the polynomial [tex]p(x) = x^4 + 5x^2 + 3x + 2[/tex] is irreducible in [tex] \bbmath{Q} [x] [/tex].

    This on has me totally stumped. I've seen many other problems on polynomial reducibility, but they are all solvable by Eisenstein, or by checking for irreducibility in a quotient ring by a proper ideal. Neither applies here.

    I know that if it does reduce, then there are only two cases to check.

    Case 1: [tex]p(x) = a(x)b(x)[/tex] where [tex]deg(a(x)) = 3, deg(b(x)) = 1[/tex].

    This case is easy to eliminate, since it implies that [tex]p(x)[/tex] must have a root in [tex]\mathbb{Q}[/tex]. Since the only possible choices are [tex]\pm 1, \pm 2[/tex], it is easy to check that none of these work.

    So, Case 1 is impossible.

    Therefore, if [tex]p(x)[/tex] is reducible, then it must reduce into two monic polynomials of degree 2.

    Case 2: [tex]p(x) = a(x)b(x)[/tex] where [tex]deg(a(x)) = deg(b(x)) = 2[/tex].

    Here is where I hit a dead end. I've tried my entire bag of tricks and I have nothing to show for it. Can someone get me started?
  2. jcsd
  3. Jul 24, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    Suppose it factors into (x^2 + ax + b)(x^2 + cx + d) over [itex]\mathbb{Z}[/itex]. Multiply out, equate coefficients, and derive a contradiction.
  4. Jul 24, 2007 #3
    I did that - but I did it over [tex] \mathbb{Q} [/tex] which made it impossible to find a contradiction (or at least difficult enough that I gave up). Why am I allowed to restrict my coefficients to [tex] \mathbb{Z} [/tex]?

    Thanks for your quick response!
  5. Jul 24, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    Gauss's lemma.
  6. Jul 24, 2007 #5
    So, if I'm reading this correctly, Gauss's Lemma says that if R is a UFD (Z in our case) and F is its field of fractions (Q in our case), then a polynomial p(x) in R[x] (which ours is) that is reducible in F[x] is also reducible in R[x].

    So I'm taking the contrapositive stance, right? If p(x) is irreducible in Z[x], then it is irreducible in Q[x].

    I see here a corollary which I think I can use: Same R and F - if the g.c.d. of the coefficients of p(x) is 1 then p(x) is irreducible in R[x] if and only if it is irreducible in F[x].

    Thanks for pointing me in the right direction!
  7. Jul 24, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    No problem.
  8. Jul 24, 2007 #7
    We can state Gauss' Lemma a little stronger than that.

    Theorem: Let [tex]D[/tex] be a UFD and [tex]F[/tex] be the field of quotients of [tex]D[/tex]. Given a polynomial [tex]f(x) \in D[x][/tex]. The polynomial [tex]f(x)[/tex] factors into degrees [tex]n\mbox{ and }m[/tex] in [tex]D[x][/tex] if and only if [tex]f(x)[/tex] factors into polynomials of degrees [tex]n \mbox{ and }m[/tex] in [tex]F[x][/tex].

    Since [tex]\mathbb{Q}[/tex] is the field of quotients of [tex]\mathbb{Z}[/tex] Gauss' Lemma applies. We can restrict our attention to the integers because they are easier to work with.
  9. Jul 24, 2007 #8
    Okay, I think I've got it now. Equating [tex](x^2 + ax + b)(x^2 + cx + d)[/tex] to [tex]p(x)[/tex], we arrive at the following system of equations:

    [tex]a + c = 0[/tex]

    [tex]b + ac + d = 5[/tex]

    [tex]ad + bc = 3[/tex]

    [tex]bd = 2[/tex]

    From the first equation, I know that c = -a. Substituting into the second equation and moving a few terms gives me [tex] a^2 = (b + d) - 5 [/tex]. (*)

    Now, from the bottom equation, we know [tex]b, d \in \{ \pm 1, \pm 2 \} [/tex]. But, once we know b, we automatically know d. Looking back at (*) the only possible values for [tex] b + d [/tex] are -3 or 3. Both produce a contradiction, however, since [tex] a^2 = -2 [/tex] and [tex] a^2 = -8 [/tex] have no solutions in [tex] \mathbb{Z} [/tex].
  10. Jul 30, 2007 #9


    User Avatar
    Science Advisor
    Homework Helper

    mod 3 this reduces to an irreducible polynomial of degree 4. done.
  11. Aug 9, 2007 #10
    I think I understand what you're saying. Since I can check for irreducibility in [tex] \mathbb{Z} [/tex] I can use the proper ideal [tex] 3 \mathbb{Z} [/tex] to show irreducibilty by Eisenstein.

    There is a second part to this question that I did not post earlier. I'll post it now (I've been working on papers for the past 2 weeks...):

    Show that the same polynomial does not have any roots in [tex] \mathbb{Q} (\sqrt[5]{2}).[/tex]

    Is this just a matter of comparing orders of extensions? If [tex] k [/tex] is a root of the polynomial, then [tex] [ \mathbb{Q} (k) : \mathbb{Q} ] = 4[/tex] which does not divide [tex] [ \mathbb{Q} (\sqrt[5]{2}) : \mathbb{Q} ] = 5[/tex].

    It seems like the answer to this question is simple, but I'm having a hard time justifying it in the right language.
  12. Aug 9, 2007 #11
    Theorem: Let E be an extension field over F if [tex]\alpha[/tex] is algebraic over F and [tex]\beta \in F(\alpha)[/tex] then [tex]\deg \left<\beta, F\right> \mbox{ divides }\deg \left< \alpha , F \right>[/tex].
  13. Aug 9, 2007 #12
    You define the extension E, but then never use it in your statement. I'm confused... Is my explanation correct?
  14. Aug 9, 2007 #13
    The degree of k is 4 since the polynomial is irreducible over F. Now if k is in Q(2^{1/5}) the by theorem deg(k,Q) divides deg (2^{1/5},Q). But deg (2^{1/5},Q)=5 and so we have that 4 divides 5 a contradiction.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Prep for Algebra Comprehensive Exam #3
  1. Lie 3-Algebra (Replies: 3)