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The two conditions that make the automorphism count equal the degree: normality keeping whole families of roots together, separability keeping them distinct, and why the second is free in characteristic zero.
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 state what makes an extension normal and what makes it separable, decide whether a named extension is Galois, compare an automorphism count with a degree and diagnose a shortfall, use the formal derivative to test for repeated roots, give an inseparable extension in characteristic $p$, and prove that irreducible polynomials in characteristic zero have distinct roots.
The last lesson found that $\mathbb{Q}(\sqrt[3]{2})$ has degree three and only one automorphism, and warned that no correspondence between subgroups and subfields could work there. The shortfall has a reason, and naming it is this lesson's job: the field holds one root of a family and has left the others outside.
$L/K$ is normal when every irreducible polynomial over $K$ with a root in $L$ splits in $L$ — equivalently, $L$ is a splitting field over $K$. A polynomial is separable when it has no repeated roots; an extension is separable when every element's minimal polynomial is. A field is perfect when every irreducible polynomial over it is separable — every field of characteristic zero and every finite field is. Galois means normal and separable.
Normal. $L/K$ is normal when every irreducible polynomial over $K$ with one root in $L$ has all of its roots in $L$. Equivalently, for a finite extension, $L$ is the splitting field over $K$ of some polynomial.
$\mathbb{Q}(\sqrt[3]{2})$ fails: $x^{3} - 2$ is irreducible over $\mathbb{Q}$ and has exactly one root there. $\mathbb{Q}(\sqrt2)$ succeeds; $\mathbb{Q}(\sqrt[4]{2})$ fails, since it holds $\pm\sqrt[4]{2}$ and not $\pm i\sqrt[4]{2}$.
Separable. An irreducible polynomial is separable when its roots are distinct; an extension is separable when every element's minimal polynomial is. The test is the formal derivative: $f$ has a repeated root exactly when $f$ and $f'$ share a non-constant factor.
In characteristic zero this never happens for an irreducible $f$: a shared factor would have to be $f$ itself, so $f \mid f'$, while $\deg f' = n - 1$ and $f' \ne 0$. The same argument works over a finite field. So over $\mathbb{Q}$, over $\mathbb{R}$, over $\mathbb{C}$ and over every $\mathbb{F}_q$, separability is free.
It can fail: over $\mathbb{F}_p(t)$, the polynomial $x^{p} - t$ is irreducible and equals $(x - t^{1/p})^{p}$ in a splitting field — one root, repeated $p$ times, and $f' = px^{p-1} = 0$.
Together they give the count. For a finite extension,
$$|\operatorname{Gal}(L/K)| = [L : K] \iff L/K \text{ is normal and separable}.$$
Separability makes every minimal polynomial offer its full quota of distinct roots; normality guarantees those roots are actually available in $L$ for an automorphism to use. Lose either and automorphisms go missing.
Another characterisation, often the handiest. $L/K$ is Galois exactly when the elements of $L$ fixed by every automorphism are precisely $K$:
$$L^{\operatorname{Gal}(L/K)} = K.$$
For $\mathbb{Q}(\sqrt[3]{2})$ the group is trivial, so its fixed field is the whole of $\mathbb{Q}(\sqrt[3]{2})$ and not $\mathbb{Q}$ — the failure again, seen from the other side.
Normality is not transitive. In the splitting field of $x^{3}-2$ over $\mathbb{Q}$, the subfield $\mathbb{Q}(\sqrt[3]{2})$ is not normal over $\mathbb{Q}$ even though the big field is. Which intermediate fields are normal is decided by which subgroups are normal, and that is the next lesson.
Another way: picture
Imagine the roots of an irreducible polynomial as siblings, indistinguishable from the base field's point of view. Normality is the demand that a field admitting one sibling admits them all; a field with one of three has secretly chosen a favourite, and no symmetry can survive that choice. Separability is the demand that the siblings are genuinely distinct people rather than one person counted several times.
Another way: steps
To decide whether $L/K$ is Galois: 1. If the characteristic is zero or $K$ is finite, separability holds; go to step 3. 2. Otherwise check that each generator's minimal polynomial shares no factor with its derivative. 3. Check normality: is $L$ the splitting field of some polynomial over $K$? 4. Equivalently, count the automorphisms and compare with the degree. 5. Or compute the fixed field of the whole group and check it is exactly $K$.
The derivative here is defined by the rule
$$\left(\sum_i c_i x^{i}\right)' = \sum_i i\,c_i x^{i-1},$$
with no limits anywhere. It is a purely algebraic operation and satisfies the product rule, which is all that is needed.
Why it detects repeated roots. If $f = (x - \alpha)^{2}g$ then $f' = 2(x-\alpha)g + (x-\alpha)^{2}g'$, which vanishes at $\alpha$. Conversely, if $f(\alpha) = f'(\alpha) = 0$ then writing $f = (x-\alpha)h$ gives $f' = h + (x-\alpha)h'$, so $h(\alpha) = 0$ and $(x-\alpha)^{2} \mid f$. So
$$f \text{ has a repeated root} \iff \gcd(f, f') \ne 1,$$
and the gcd is computed by the Euclidean algorithm over $K$ itself — no splitting field required, which is what makes this a usable test.
Characteristic zero. For irreducible $f$ of degree $n$, $\gcd(f, f')$ divides $f$, so it is $1$ or an associate of $f$. The second would force $f \mid f'$, impossible since $0 \le \deg f' = n-1 < n$ and $f' \ne 0$. So the gcd is $1$: no repeated roots, ever.
Characteristic $p$. Now $f'$ can be zero, and it is zero exactly when every exponent with a non-zero coefficient is a multiple of $p$ — that is, when $f(x) = g(x^{p})$. Over a perfect field every coefficient is a $p$th power, so $g(x^{p}) = (h(x))^{p}$ for some $h$ and $f$ was not irreducible after all. That is why finite fields, where Frobenius is surjective, are perfect.
The genuine failures need an imperfect field. $\mathbb{F}_p(t)$ is the standard one: $t$ has no $p$th root in it, so $x^{p} - t$ is irreducible and inseparable.
Why any of this is stated. Almost every course works in characteristic zero, where separability is invisible. It is stated because the theorems are true as stated and false without it, and because the arithmetic of function fields over finite fields — central to coding theory and to number theory — lives exactly where it fails.
*Thinking normal means has a normal subgroup.* The word is reused; here it is about roots staying together. The two meanings are linked by the fundamental theorem, and only there.
Thinking normality is transitive. It is not: $\mathbb{Q} \subset \mathbb{Q}(\sqrt2) \subset \mathbb{Q}(\sqrt[4]{2})$ has each step normal and the composite not.
Ignoring separability because it is usually automatic. It is automatic in characteristic zero and over finite fields, and nowhere else. The theorems need it.
Computing the derivative as a limit. It is defined formally, coefficient by coefficient, and works over any field including finite ones.
*Concluding not Galois from a small group without checking the degree.* The comparison is with the degree; a group of order $2$ for a degree-$2$ extension is perfectly Galois.
Expecting $x^{p} - t$ to be reducible because it has one root. It is irreducible over $\mathbb{F}_p(t)$; having one root of multiplicity $p$ is exactly what inseparability looks like.
$x^{4} - 2$ is irreducible over $\mathbb{Q}$ by Eisenstein at $2$, and its roots are $\pm\sqrt[4]{2}$ and $\pm i\sqrt[4]{2}$.
Four roots.
The field is real, so it contains the two real roots and neither imaginary one.
Two of four.
So an irreducible polynomial has a root there without splitting: not normal. Counting automorphisms agrees — there are $2$ against a degree of $4$.
Not Galois.
$\mathbb{Q} \subset \mathbb{Q}(\sqrt2)$ is normal: it is the splitting field of $x^{2} - 2$.
The lower step.
$\mathbb{Q}(\sqrt2) \subset \mathbb{Q}(\sqrt[4]{2})$ is normal too: it is the splitting field of $x^{2} - \sqrt2$ over $\mathbb{Q}(\sqrt2)$.
The upper step.
But $\mathbb{Q} \subset \mathbb{Q}(\sqrt[4]{2})$ is not normal, as the previous example showed. Normality at each step says nothing about the composite.
Not transitive.
It is the splitting field of $(x^{2} - 2)(x^{2} - 3)$ over $\mathbb{Q}$, so it is normal.
Normal.
The characteristic is zero, so every irreducible polynomial is separable and the extension is separable.
Separable for free.
So it is Galois, and the automorphism count must equal the degree: four automorphisms against a degree of four, which is exactly what was found two lessons ago.
Select every statement that is true of a finite Galois extension $L/K$.
This task has no paper form; do it on a device.
Is this extension Galois: $\mathbb{Q}(\sqrt[4]{2})$ over $\mathbb{Q}$?
The real field $\mathbb{Q}(2^{1/9})$ has degree $9$ over $\mathbb{Q}$, and $9$ is odd. How many automorphisms of that field fix $\mathbb{Q}$?
Answer:
For each extension the degree is given. Fill in how many automorphisms fix the base, and then write $1$ if the extension is Galois and $0$ if it is not.
| Degree | Automorphisms fixing the base | Galois: one or zero | |
|---|---|---|---|
| $\mathbb{Q}(\sqrt2)$ over $\mathbb{Q}$ | 2 | ||
| $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$ | 3 | ||
| The splitting field of $x^{3} - 2$ over $\mathbb{Q}$ | 6 | ||
| $\mathbb{Q}(\sqrt[4]{2})$ over $\mathbb{Q}$ | 4 | ||
| $\mathbb{F}_{16}$ over $\mathbb{F}_2$ | 4 |
Put in order the steps of deciding whether a finite extension is Galois.
Number the steps in order (write the number in the box):
Build the proof that over a field of characteristic zero, every irreducible polynomial has distinct roots.
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.
The real field $\mathbb{Q}(2^{1/7})$ has degree $7$ over $\mathbb{Q}$, and $7$ is odd. How many automorphisms of that field fix $\mathbb{Q}$?
Answer:
You can decide whether an extension is Galois and say which of the two conditions fails when one does. Say in your own words why normality is not transitive, with an example. Next: the correspondence that Galois extensions make possible.
9. Your turn: is $\mathbb{Q}(\sqrt2, \sqrt3)$ Galois over $\mathbb{Q}$?, step 3