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Galois Theory (degree 4 polynomial)

  1. Jul 30, 2007 #1
    I have 4 questions for the same problem. I worked on this for days but after while, I didn't get very far.

    Q1. To show that [itex]f(x)=x^4-2x^2-1[/itex] is irreducible, which irreducibility test is the most efficient? Reduction mod p doesn't really work (when p=2) and neither does Eisenstein's Irreducibility Criterion.

    I supposed that f(x) can be written as a linear factor and a cubic polynomial and came up with a contradiction. I then supposed that f(x) can be written as a product of two quadratic polynomials and came with a contradiction. These processes work but they're just too long.

    Any efficient irreducibility suggestions/tests for this particular polynomial: [itex]f(x)=x^4-2x^2-1[/itex]?

    Q2. Let [itex]a= (1+\sqrt{2})^{1/2} [/itex] and [itex]b= (1-\sqrt{2})^{1/2} [/itex]. Does [itex]\mathbb{Q}(b)= \mathbb{Q}(a,i)[/itex]?

    Q3. Let [itex]E=\mathbb{Q}(a)[/itex]. To find the index [itex] [E:\mathbb{Q}] [/itex], is this 4 or 8? I thought it would equal the degree of the irreducible polynomial for [itex]a[/itex], which is 4, but [itex]b[/itex] is also a root of the above polynomial and [itex]b[/itex] is a complex number, not real. So since we need to adjoin [itex]i[/itex], I thought the index should be 8.

    Q4. What is the Galois group of E over [itex]\mathbb{Q}[/itex]? I found a similar problem from Dummit and Foote and its Galois group is [itex]\mathscr{D}_4[/itex] but someone mentioned that it should be [itex]\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2[/itex]. So which one is it and how does one know?

    Thanks for your time!
    Last edited: Jul 30, 2007
  2. jcsd
  3. Jul 30, 2007 #2


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    Q1. what about mod 3? dont give up so soon.
  4. Jul 30, 2007 #3


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    Q4. first of all convince yourself that E is a galois, i.e. normal extension of Q. if so, then as you say the order of G is the degree of E over Q.

    but in any case, galois or not, the order of G divides the degree of E over Q. what does that tell you?
  5. Jul 31, 2007 #4
    Hi Mathwonk, thanks for your reply!! As for Q1., I misunderstood the original statement of the theorem (Reduction mod p). I thought f is irreducible if f is irreducible in [itex]\mathbb{Z}_p[x][/itex] for every prime p, but this is wrong! If we can find one such prime p such that f(x) is irreducible in [itex]\mathbb{Z}_p[x][/itex], then we're done!

    Thanks for the other replies. Although I read them briefly, I didn't get a chance to think about them and apply to what I have learned.

    But I do have one question:

    QUESTION: let f be a polynomial of degree 5 over [itex]\mathbb{Q}[/itex] and let [itex]Gal_\mathbb{Q}(f)[/itex] be isomorphic to the alternating group [itex]A_5[/itex]. Then how do we know that f must be irreducible over [itex]\mathbb{Q}[/itex]?

    MY ATTEMPT: Well since the alternating group cannot be solved by radicals, the corresponding polynomial cannot be solved by radicals. So f has a degree 5 irreducible polynomial as its factor, thus it must be itself.

    I think there is something more to the reasoning. That is, since the Galois group is isomorphic to [itex]A_5[/itex], the galois group has a 5-cycle (it permutes 5 roots). But perhaps two roots are conjugated together while the other three roots are permuted together. I feel like I'm almost there, yet I'm missing something. Please help....
  6. Jul 31, 2007 #5


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    i ike your argumnt. it seems enough to me.

    the point is i gues, that if the polynomil were a product, then the splitting field of one factor would be a normal subfield, and correspond to a normal subgroup, of which A(5) has only two.

    so the spliting field of one factor would be the full splitting field of f. but then this field would have degree dividing 24 = 4! i think. but with group A(5) the field has degree 30.

    this is basically your argument, to look at the interplay between subgroups of A(5) and subfields of the field of f.
  7. Jul 31, 2007 #6
    This problem is a little bit computational.

    Note, [tex][\mathbb{Q}(\sqrt{\sqrt{2}+1},i):\mathbb{Q}]=8[/tex].
    So they are not equal.
  8. Jul 31, 2007 #7


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    It might be easier to notice that ab = i...

    BTW, you can use \left[ ... \right] to make large brackets. (And these work with many different kinds of delimiters)
  9. Aug 2, 2007 #8
    Thanks Mathwonk, Kummer, and Hurkyl for your replies!!

    So according to Hurkyl ab = i implies a = i/b where [itex]a= \sqrt{\left[1+\sqrt{2}\right]}[/itex] and [itex]b= \sqrt{\left[1-\sqrt{2}\right]} [/itex]. So since [itex]b\not\in E=\mathbb{Q}(a)[/itex] (because i is not in E), E over [itex]\mathbb{Q}[/itex] is not Galois (because all the roots of the polynomial [itex]f(x)=x^4-2x^2-1[/itex] are not in the field extension E). So [itex]|Gal(E/\mathbb{Q})| \leq [E:\mathbb{Q}]=\deg_\mathbb{Q} (f(x)) = 4 [/itex].

    I don't remember that the order of the Galois group must divide the degree of E over Q, but since the Galois group of E over Q has only two roots of the polynomial f, namely, a and -a, [itex]Gal(E/\mathbb{Q})\cong \mathbb{Z}_2[/itex]. That is, the Galois group has two automorphisms: one takes a to itself and another automorphism takes a to -a.

    So my answer to Q3. is:
    [itex] [E:\mathbb{Q}] = \deg_\mathbb{Q} (f) = 4 [/itex]
    and if we were to adjoin b to E, then
    [itex][E(b):\mathbb{Q}]=[\mathbb{Q}(a,b):\mathbb{Q}]= [\mathbb{Q}(a,i):\mathbb{Q}]=8 [/itex].

    As for Q4., [itex]Gal(E/\mathbb{Q})\cong \mathbb{Z}_2[/itex] and [itex]Gal(\mathbb{Q}(a,i)/\mathbb{Q})\cong \mathbb{Z}_2\times \mathbb{Z}_2[/itex].

    And thanks again for the LaTeX tips Hurkyl!

    Last edited: Aug 2, 2007
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