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The automorphisms of an extension that leave the base alone: why they permute roots, why the generators determine them, and why there are never more of them than the degree.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to define the Galois group of an extension, prove that an automorphism fixing the base permutes the roots of any polynomial over it, list the automorphisms of a small extension from the images of its generators, identify the resulting group, say why the order never exceeds the degree, and give the standard extension where it falls short.
Adjoining one root of a cubic singles that root out; adjoining all three treats them alike. The splitting field is therefore the place where the roots are interchangeable, and this lesson measures exactly how interchangeable they are — by collecting the symmetries of the field that leave the base untouched.
An automorphism of $L$ is a bijection $L \to L$ respecting addition and multiplication. It fixes $K$ pointwise when $\sigma(c) = c$ for every $c \in K$. The Galois group $\operatorname{Gal}(L/K)$ is the group of all such automorphisms under composition. For a subgroup $H$, the fixed field $L^{H}$ is the set of elements every element of $H$ leaves alone.
The definition. $\operatorname{Gal}(L/K)$ is the set of automorphisms of $L$ fixing every element of $K$. It is a group under composition: the identity qualifies, composites and inverses of such automorphisms are such automorphisms.
Roots go to roots. If $f$ has coefficients in $K$ and $f(\alpha) = 0$, then applying $\sigma$ to the equation and pushing it inside gives
$$0 = \sigma(f(\alpha)) = \sum_i \sigma(c_i)\sigma(\alpha)^{i} = f(\sigma(\alpha)),$$
because $\sigma$ fixes each coefficient. So $\sigma$ permutes the roots of $f$ lying in $L$.
Determined by the generators. If $L = K(\alpha_1, \ldots, \alpha_r)$ then every element of $L$ is a rational expression in the $\alpha_i$ over $K$, and $\sigma$ respects all the operations — so knowing $\sigma(\alpha_i)$ determines $\sigma$ entirely. Combined with the previous point, the images are confined to the roots of the minimal polynomials.
So the group is small, and computable. $\operatorname{Gal}(L/K)$ embeds in the symmetric group on the roots, and
$$|\operatorname{Gal}(L/K)| \le [L : K].$$
A search over an infinite field has become a search over permutations of a handful of numbers.
When equality fails. $\mathbb{Q}(\sqrt[3]{2})$ has degree $3$ and Galois group of order $1$: an automorphism must send $\sqrt[3]{2}$ to a root of $x^{3} - 2$ in the field, and the other two are not real. The group has lost sight of the extension, and no correspondence between subgroups and subfields can work there. Equality is what the word Galois will mean, and the next lesson gives the two conditions that guarantee it.
Three worked groups.
| Extension | Degree | Group |
|---|---|---|
| $\mathbb{Q}(\sqrt2)/\mathbb{Q}$ | $2$ | order $2$, swapping $\pm\sqrt2$ |
| $\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}$ | $4$ | Klein four-group, flipping each root |
| splitting field of $x^{3}-2$ | $6$ | $S_3$ on the three cube roots |
| $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ | $3$ | trivial |
Galois theory replaces an infinite search among fields with a finite list of subgroups. Every question about what lies between $K$ and $L$ — how many fields, which are normal, what their degrees are — is answered by reading a subgroup lattice upside down.
Another way: picture
Think of the base field as what is nailed down and the rest of the extension as free to move. An automorphism is a rearrangement the field cannot tell apart from doing nothing: every equation with coefficients in the base that held before still holds after. The roots of a polynomial are interchangeable exactly to the extent that such rearrangements exist, and the Galois group is the complete list of them.
Another way: steps
To compute a Galois group: 1. Find the field and its degree over the base — for a polynomial, take the splitting field. 2. Name generators and their minimal polynomials. 3. Each generator may go to any root of its own minimal polynomial that lies in the field. 4. Keep the assignments that are consistent, and count. 5. Check the count against the degree; a shortfall means the extension is not Galois. 6. Identify the group from the orders of its elements and whether they commute.
The embedding into $S_n$. Let $f$ have $n$ distinct roots and $L$ be its splitting field. Every $\sigma \in \operatorname{Gal}(L/K)$ permutes the roots, and since they generate $L$, different automorphisms give different permutations. So $\operatorname{Gal}(L/K)$ is a subgroup of $S_n$.
Which subgroup depends on how the roots are related. If they satisfy no relations over $K$ beyond the symmetric ones, the group is all of $S_n$; each extra relation cuts it down.
$x^{2} - 2$. Two roots, $\pm\sqrt2$. Both permutations occur, so the group is $S_2$, of order $2$ — matching the degree.
$x^{3} - 2$. Three roots $\sqrt[3]{2}, \sqrt[3]{2}\omega, \sqrt[3]{2}\omega^{2}$, splitting field of degree $6$. So the group has order $6$ and sits in $S_3$, which has order $6$: it is all of $S_3$.
$x^{4} - 1$. Four roots $\pm 1, \pm i$, but the splitting field is $\mathbb{Q}(i)$ of degree $2$. The group has order $2$ and sits in $S_4$ as a very small subgroup — because two of the roots are rational and cannot be moved at all.
$(x^{2}-2)(x^{2}-3)$. Four roots, splitting field of degree $4$. The group is the Klein four-group inside $S_4$: $\sqrt2$ and $\sqrt3$ can be flipped independently, but $\sqrt2$ can never be sent to $\sqrt3$, because $\sigma$ must respect $(\sqrt2)^{2} = 2$.
The moral. The group is not all permutations of the roots; it is those the base field cannot forbid. Every relation among the roots with coefficients in $K$ is a constraint the group must respect, and the group is exactly what survives them all.
Assuming the group has order equal to the degree. At most, and equality is a condition to be checked. $\mathbb{Q}(\sqrt[3]{2})$ is the standing counterexample.
Taking the group to be all of $S_n$. Only when the roots satisfy no unexpected relations. $x^{4} - 1$ gives a group of order $2$ inside $S_4$.
Sending a generator to a root that is not in the field. An automorphism of $L$ must land inside $L$. This is exactly why $\mathbb{Q}(\sqrt[3]{2})$ has no non-trivial automorphism.
Forgetting that the base is fixed pointwise, not just setwise. Fixing $K$ as a set would allow automorphisms permuting elements of $K$, which is a different and much weaker condition.
Expecting an automorphism to fix things outside $K$. It usually moves everything else; fixing more is what defines the smaller subgroups.
Treating the group as acting on the field's elements arbitrarily. It is determined by the generators, so it has far fewer elements than the field has.
Any automorphism fixes $\mathbb{Q}$, and must send $\sqrt2$ to a root of $x^{2} - 2$ lying in the field: either $\sqrt2$ or $-\sqrt2$.
Two candidates.
Both work: $a + b\sqrt2 \mapsto a + b\sqrt2$ and $a + b\sqrt2 \mapsto a - b\sqrt2$ both respect the operations.
Both are automorphisms.
So the group has order $2$, equal to the degree, and is cyclic. The fixed field of the non-trivial automorphism is exactly $\mathbb{Q}$.
Order two.
An automorphism must send $\sqrt[3]{2}$ to a root of $x^{3} - 2$.
Roots to roots.
The three roots are $\sqrt[3]{2}$, $\sqrt[3]{2}\omega$ and $\sqrt[3]{2}\omega^{2}$, and the last two are not real — while $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$.
Only one root is available.
So $\sqrt[3]{2}$ must stay put and the automorphism is the identity. The group has order $1$ against a degree of $3$, and the extension is not Galois.
A trivial group.
$\mathbb{F}_4 = \mathbb{F}_2(\alpha)$ with $\alpha^{2} = \alpha + 1$, so the degree is $2$ and there are at most two automorphisms.
At most two.
An automorphism must send $\alpha$ to a root of $x^{2} + x + 1$ in the field: either $\alpha$ or $\alpha + 1$.
Two candidates.
Both work, and the non-trivial one is the Frobenius map $x \mapsto x^{2}$ — indeed $\alpha^{2} = \alpha + 1$. So the group is cyclic of order $2$, equal to the degree, and the extension is Galois.
$\mathbb{Q}(\sqrt2, \sqrt3)$ has four automorphisms fixing $\mathbb{Q}$, each sending $\sqrt2$ to $\pm\sqrt2$ and $\sqrt3$ to $\pm\sqrt3$. Those two signs are given. Fill in the sign it puts on $\sqrt6$, and the order of the automorphism.
| Sign on $\sqrt2$ | Sign on $\sqrt3$ | Sign on $\sqrt6$ | Order of the automorphism | |
|---|---|---|---|---|
| The identity | 1 | 1 | ||
| The second automorphism | -1 | 1 | ||
| The third automorphism | 1 | -1 | ||
| The fourth automorphism | -1 | -1 |
How many automorphisms fixing the base does this extension have: $\mathbb{F}_{16}$ over $\mathbb{F}_2$?
Answer:
Select every statement that is true of an automorphism $\sigma$ of $L$ fixing a subfield $K$ pointwise.
This task has no paper form; do it on a device.
Match each extension of $\mathbb{Q}$ to the group of automorphisms fixing $\mathbb{Q}$.
| Cyclic of order $2$ | The Klein four-group | $S_3$, of order $6$ | The trivial group | Cyclic of order $3$ | |
|---|---|---|---|---|---|
| $\mathbb{Q}(\sqrt2)$ | |||||
| $\mathbb{Q}(\sqrt2, \sqrt3)$ | |||||
| The splitting field of $x^{3} - 2$ | |||||
| $\mathbb{Q}(\sqrt[3]{2})$ |
Put in order the steps of computing the Galois group of a polynomial over $\mathbb{Q}$.
Number the steps in order (write the number in the box):
Build the proof that an automorphism of $L$ fixing $K$ pointwise sends each root in $L$ of a polynomial over $K$ to another root of that polynomial.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$\mathbb{Q}(\sqrt2, \sqrt3)$ has four automorphisms fixing $\mathbb{Q}$, each sending $\sqrt2$ to $\pm\sqrt2$ and $\sqrt3$ to $\pm\sqrt3$. Those two signs are given. Fill in the sign it puts on $\sqrt6$, and the order of the automorphism.
| Sign on $\sqrt2$ | Sign on $\sqrt3$ | Sign on $\sqrt6$ | Order of the automorphism | |
|---|---|---|---|---|
| The identity | 1 | 1 | ||
| The second automorphism | -1 | 1 | ||
| The third automorphism | 1 | -1 | ||
| The fourth automorphism | -1 | -1 |
You can list the automorphisms of a small extension and identify the group they form. Say in your own words why an automorphism cannot send a generator anywhere but to a root of its own minimal polynomial. Next: the two conditions that make the count equal the degree.
9. Your turn: the Galois group of $\mathbb{F}_4$ over $\mathbb{F}_2$, step 3