(adsbygoogle = window.adsbygoogle || []).push({}); Determine the minimal polynomial [tex] f(x) [/tex] over [tex] \mathbb{Q} [/tex] of the element [tex] \sqrt3 + \sqrt7 [/tex]. Determine the Galois group of [tex] f(x) [/tex] and all the subfields of the splitting field of [tex] f(x) [/tex].

SOLUTION

This one seems pretty straightforward, I just need checks on my reasoning. I apologize in advance for any redundancies. Your comments are welcome.

The splitting field must include [tex] \sqrt3 + \sqrt7 [/tex] so specifically, it must contain [tex] \sqrt3 [/tex] and [tex] \sqrt7 [/tex]. The smallest field extension of [tex] \mathbb{Q} [/tex] that contains [tex] \sqrt3 [/tex] is [tex] \mathbb{Q}(\sqrt3) [/tex] and likewise, the smallest field extension of [tex] \mathbb{Q} [/tex] that contains [tex] \sqrt7 [/tex] is [tex] \mathbb{Q}(\sqrt7) [/tex]. But, since neither root is contained in the other's extension, neither of these can be the splitting field.

The smallest field extension of [tex] \mathbb{Q} [/tex] that contains [tex] \sqrt3 [/tex] and [tex] \sqrt7 [/tex] is [tex] \mathbb{Q}(\sqrt3, \sqrt7) [/tex]. It follows from the above argument that the degree of this extension is 4.

(Again, I realize this may be superfluous... I'm just contrasting this type of problem to, say, a similar one with the element [tex] \sqrt3 + \sqrt[6]3 [/tex] which would have degree 6, and the above argument would be different)

So, we're looking for a minimal polynomial of degree 4. Since the polynomial is in [tex] \mathbb{Q} [/tex] the other 3 roots must be the distinct conjugates of [tex] \sqrt3 + \sqrt7 [/tex], which are [tex] \pm \sqrt3 \pm \sqrt7 [/tex]. Thus, the minimal polynomial is:

[tex] (x - (\sqrt3 + \sqrt7))(x - (\sqrt3 - \sqrt7))(x - (- \sqrt3 + \sqrt7))(x - (- \sqrt3 - \sqrt7)) = x^4 - 20x^2 + 16 [/tex]

Do I need to show irreducibility here? If so, by Gauss's Lemma, taking mod 7, and using Eisenstein (p = 2), we have it to be so.

From this, we know that the order of the Galois group is also 4, which is obvious, since any automorphism is completely determined by its action on [tex] \sqrt3 [/tex] and [tex] \sqrt7 [/tex], each of which can either be left alone (identity automorphism) or sent to their additive inverse (conjugate automorphism).

Let [tex] \sigma [/tex] be the conjugate automorphism on [tex] \sqrt3 [/tex] and let [tex] \tau [/tex] be the conjugate automorphism on [tex] \sqrt7 [/tex]. Then, [tex] \{ 1, \sigma, \tau, \sigma\tau \} [/tex] is the Galois group. Since each element has order 2, this group is isomorphic to [tex] \mathbb{Z} _2 \times \mathbb{Z} _2 [/tex] which has 3 subgroups of order 2.

Therefore, the splitting field [tex] \mathbb{Q}(\sqrt3, \sqrt7) [/tex] must have three subfields of index 2 (yes?) which are [tex] \mathbb{Q}(\sqrt3) [/tex] , [tex] \mathbb{Q}(\sqrt7) [/tex] and [tex] \mathbb{Q}(\sqrt21) [/tex] which all contain the subfield [tex] \mathbb{Q} [/tex].

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