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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


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    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


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    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


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    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


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    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.
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