Automorphisms, Galois Groups & Splitting Fields

[mathjam0990]

Let f(x)=x⁴-2x²+9
Find the splitting field and Galois group for f(x) over ℚ

Here is what I have written out so far. If I have found the splitting field E correctly, have I proceeded with the Gal(E/F) group correctly?

Also, how would I go about finding the roots of this equation by hand without a calculator?

If anyone could help me out with the answer and a detailed explanation to follow that would be very helpful. Thank you.

View attachment 5498

 

[Deveno]

OK, you've found the roots, that's good.

To know the degree of the splitting field over $\Bbb Q$, we need to know if $h(x)$ is irreducible over $\Bbb Q$.

We know that $h(x)$ has no linear factors in $\Bbb Q[x]$ (since none of the roots are rational), and thus also no rational cubic factors. Does it have any quadratic factors in $\Bbb Q[x]$?

If $h(x) = (x^2 + ax + b)(x^2 + cx + d)$ we have:

$a+c = 0$
$ac + b + d = -2$
$ad + bc = 0$
$bd = 9$.

So $c = -a$, and the third equation becomes $a(d - b) = 0$. Thus either $a = 0$, or $b = d$.

If $a = c = 0$, then the second equation is $b+d = -2$. Since $d = \dfrac{9}{b}$, we get:

$b + \dfrac{9}{b} = -2$, that is:

$b^2 - 2b + 9 = 0$. Checking the discriminant: $4 - 36 = -32 < 0$, we see this has no real, thus no rational, solutions.

Thus $b = \pm 3$. Substituting back in equation 2, either:

$-a^2 - 6 = -2 \implies a^2 = -4$ (for $b = -3$) , this is impossible;

$-a^2 + 6 = -2 \implies a^2 = 8$ (for $b = 3$), which has no rational solution, as $8$ is not a perfect square.

So we conclude that $h(x)$ is indeed irreducible, as its roots are all distinct it is separable, so the splitting field has degree $4 = \text{deg }h$ over $\Bbb Q$.

As this splitting field (let's call it $E$) is Galois (being a splitting field), we see that $|\text{Gal}(E/\Bbb Q)| = 4$.

So we are looking for a group of automorphisms of $E$, that fix $\Bbb Q$, of order $4$.

Now it should be clear that $E \subseteq \Bbb Q(\sqrt{2},i)$, since the roots of $h$ are contained in the latter field.

Furthermore, by factoring $h$ over the reals, we see that:

$h(x) = (x^2 + 2\sqrt{2}x + 3)(x^2 - 2\sqrt{2}x + 3)$.

Since $i \not\in \Bbb Q(\sqrt{2})$, and $i$ is a root of $x^2 + 1 \in \Bbb Q(\sqrt{2})[x]$, we see that:

$[\Bbb Q(\sqrt{2},i):\Bbb Q(\sqrt{2})] = 2$, and clearly $[\Bbb Q(\sqrt{2}:\Bbb Q] = 2$, thus:

$[\Bbb Q(\sqrt{2},i):\Bbb Q] = [\Bbb Q(\sqrt{2},i):\Bbb Q(\sqrt{2})][\Bbb Q(\sqrt{2}:\Bbb Q] = 4$.

Since $[E:\Bbb Q] = 4$, as well, we see that $\Bbb Q(\sqrt{2},i)$ *is* $E$.

Complex-conjugation clearly gives us an element of $\text{Gal}(E/\Bbb Q)$ of order $2$ which fixes the subfield $\Bbb Q(\sqrt{2})$.

The map: $\sqrt{2} \mapsto -\sqrt{2}, i \mapsto i$ also gives us an automorphism of order 2, so we conclude the Galois group is the Klein 4-group.

As far as "finding the roots by hand", since $h(x)$ is a bi-quadratic (a quadratic in $x^2$), it's really pretty easy:

Solve $y^2 - 2x + 9$ for $y$ by using the quadratic formula.

Sub $x^2$ in for $y$, and then solve the resulting two quadratics (you'll need to know how to take the square root of a complex number),

OR: solve the two (real) quadratics I found up above.

 

[mathjam0990]

Very helpful information, thank you!...if I could ask one more question, say we have the polynomial x⁴+x³+x²+x+1 and we wanted to find a splitting field over ℚ. Could we approach this by recognizing that this is the minimum polynomial of Φ₅(x) thus root ϑ = e^2πi/5. So the minimum polynomial is (x-ϑ)(x-ϑ²)(x-ϑ³)(x-ϑ⁴) and because ϑ↦ϑ^k for k=1,2,3,4 then E = ℚ[e^2πi/5] ? Thank you.

 

[Deveno]

The phrase "minimum polynomial of $\Phi_5(x)$" makes no sense. It *is* the minimal polynomial of $e^{2\pi i/5}$, and the reason is that $\Phi_n(x)$ is irreducible for every $n \in \Bbb Z^+$. It's a non-trivial proof to show this for arbitrary $n$, but you can use a clever substitution to show it for $n = p$, a prime, so that you can apply the Eisenstein Criterion.

And yes, the splitting field is indeed $\Bbb Q(\zeta_5)$, where $\zeta_5 = e^{2\pi i/5}$ (the other roots are all powers of $\zeta_5$-in fact, the powers of $\zeta_5$ (including $1$) are a cyclic group of order 5, and the Galois group of $x^4 + x^3 + x^2 + x + 1$ is a cyclic group of order $4$, generated by the automorphism:

$\sigma:\zeta_5 \mapsto (\zeta_5)^2$

($\zeta_5 \mapsto (\zeta_5)^3$ also works).

Interestingly enough, since $\text{Gal}(\Bbb Q(\zeta_5)/\Bbb Q) \cong \Bbb Z_4$, it has a unique subgroup of order 2, and thus we have a unique subfield of $\Bbb Q(\zeta_5)$ of degree (dimension) 2 over the rationals.

Can you guess what this field might be? (Here is a helpful hint: $\sigma^2$ is complex-conjugation restricted to $\Bbb Q(\zeta_5)$, so how would you find real quadratic factors of $x^4 + x^3 + x^2 + x + 1$?).