Automorphisms, Galois Groups & Splitting Fields

In summary, Let f(x)=x4-2x2+9 You have found the splitting field, E, and Galois group, G, for f(x) over ℚ.
  • #1
mathjam0990
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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.

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  • #2
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.
 
  • #3
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 = e2πi/5. So the minimum polynomial is (x-ϑ)(x-ϑ2)(x-ϑ3)(x-ϑ4) and because ϑ↦ϑk for k=1,2,3,4 then E = ℚ[e2πi/5] ? Thank you.
 
  • #4
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 $x^4 + x^3 + x^2 + x + 1$?).
 

FAQ: Automorphisms, Galois Groups & Splitting Fields

1. What is an automorphism?

An automorphism is a mathematical function that maps a mathematical structure onto itself while preserving its structure and properties. In the context of Galois theory, automorphisms refer to functions that preserve the algebraic structure of a field.

2. What is a Galois group?

A Galois group is a specific type of group that is associated with a field extension. It consists of all the automorphisms of the field that leave the base field fixed. The Galois group provides important information about the properties of the field extension, such as its degree and its solvability by radicals.

3. How do automorphisms relate to Galois groups?

Automorphisms are closely related to Galois groups, as they are used to define and characterize the group. In fact, the elements of a Galois group are precisely the automorphisms of the field extension, and the group operation is composition of functions.

4. What is a splitting field?

A splitting field is a field extension that contains all the roots of a given polynomial. It is the smallest field extension that is needed to split the polynomial into linear factors. In other words, it is the smallest field in which the polynomial can be completely factored.

5. Why are splitting fields important in Galois theory?

Splitting fields are important in Galois theory because they provide a way to study the algebraic properties of a field extension. The Galois group of a field extension is closely related to its splitting field, and the structure of the group can reveal important information about the field and its extensions.

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