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The smallest extension in which a polynomial factors completely: building it by adjoining roots and re-factoring, why it exists even when no roots are available, and why it is unique.
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 a splitting field, build one by adjoining roots and re-factoring at each stage, compute its degree with the tower law, say why the degree divides $n!$ and rarely equals the degree of the polynomial, recognise that the answer depends on the base field, and prove that a splitting field exists by manufacturing a root.
The last lesson built $K(\alpha)$ from one root of an irreducible polynomial. But an irreducible cubic has three roots, and $\mathbb{Q}(\sqrt[3]{2})$ contains only one of them — the choice of which root to adjoin was arbitrary, and the resulting field is not symmetric in the roots. Adjoining all of them removes the arbitrariness, and that is what makes a Galois group possible.
A polynomial splits over $L$ when it factors into linear factors there. A splitting field of $f$ over $K$ is an extension $L$ in which $f$ splits and which is generated over $K$ by the roots — the smallest such field. Kronecker's construction manufactures a root of an irreducible $g$ as the class of $x$ in $K[x]/(g)$. An extension is normal when it is the splitting field of some family of polynomials.
The definition. $L$ is a splitting field of $f$ over $K$ when $f$ factors into linear factors over $L$, and $L = K(\alpha_1, \ldots, \alpha_n)$ where the $\alpha_i$ are those roots. The second condition is what makes it smallest: $\mathbb{C}$ is a field where $x^{2} - 2$ splits and is not its splitting field over $\mathbb{Q}$.
Existence. Induct on $\deg f$. Take an irreducible factor $g$; the ring $K[x]/(g)$ is a field containing a root of $g$, namely the class of $x$. Divide that root out and apply the hypothesis to what is left. No ambient field is needed — the root is built, which is the whole point of Kronecker's construction.
Uniqueness. Any two splitting fields of $f$ over $K$ are isomorphic by an isomorphism fixing $K$ pointwise. The proof runs the same induction on both at once. This is why one says the splitting field, and it is what makes the Galois group of a polynomial well defined.
Degree. $[L : K]$ divides $n!$ where $n = \deg f$: adjoining the first root costs at most $n$, the second at most $n-1$ over the enlarged field, and so on. It is usually much smaller, because adjoining one root often supplies others.
| $f$ | roots | splitting field | degree |
|---|---|---|---|
| $x^{2} - 2$ | $\pm\sqrt2$ | $\mathbb{Q}(\sqrt2)$ | $2$ |
| $x^{3} - 1$ | $1, \omega, \omega^{2}$ | $\mathbb{Q}(\omega)$ | $2$ |
| $x^{3} - 2$ | $\sqrt[3]{2}\omega^{k}$ | $\mathbb{Q}(\sqrt[3]{2}, \omega)$ | $6$ |
| $x^{4} - 1$ | $\pm 1, \pm i$ | $\mathbb{Q}(i)$ | $2$ |
| $x^{4} - 2$ | $\pm\sqrt[4]{2}, \pm i\sqrt[4]{2}$ | $\mathbb{Q}(\sqrt[4]{2}, i)$ | $8$ |
Why $x^{3} - 2$ needs two generators. $\mathbb{Q}(\sqrt[3]{2})$ sits inside $\mathbb{R}$ and two of the three roots are not real. So the field generated by one root contains exactly one — the extension is not normal, and the automorphism group is trivial. Adjoining $\omega$ as well repairs it, at the cost of a second step of degree $2$.
The word normal. A finite extension is normal exactly when it is the splitting field of some polynomial, equivalently when every irreducible polynomial over $K$ with one root in $L$ has all of them there. That equivalence, and the separability condition that goes with it, are the next unit's first lesson.
Another way: picture
Adjoining one root is like admitting one member of a family and leaving the others outside; the field can then tell that root apart from its siblings, and no symmetry of the situation survives. A splitting field admits the whole family at once, so nothing distinguishes one root from another — and the permutations of the roots that the field cannot detect are exactly the Galois group.
Another way: steps
To find a splitting field and its degree: 1. Factor $f$ over $K$ and pick an irreducible factor of degree above $1$. 2. Adjoin a root of it; the step degree is that factor's degree. 3. Re-factor over the new field — factors often split further. 4. Repeat until $f$ is a product of linear factors. 5. Multiply the step degrees; check the answer divides $n!$. 6. Name the field by the generators actually needed.
The mechanical part of building a splitting field is adjoining roots. The part that requires thought is asking, after each step, what has become reducible.
$x^{4} - 1$ over $\mathbb{Q}$. It factors as $(x-1)(x+1)(x^{2}+1)$ before anything is adjoined, so only $x^{2} + 1$ needs work; adjoining $i$ splits it and the degree is $2$, not $4$.
$x^{4} + x^{3} + x^{2} + x + 1$ over $\mathbb{Q}$. Irreducible, so the first step has degree $4$ — and it adjoins $\zeta_5$, whose powers are the other three roots. So one step is enough: degree $4$.
$x^{3} - 2$ over $\mathbb{Q}$. Irreducible, so the first step has degree $3$. Over $\mathbb{Q}(\sqrt[3]{2})$ the polynomial becomes $(x - \sqrt[3]{2})(x^{2} + \sqrt[3]{2}x + \sqrt[3]{4})$, and the quadratic is irreducible there because its roots are not real. Second step of degree $2$; total $6$.
$x^{3} - 2$ over $\mathbb{Q}(\omega)$. Now the base already has the cube roots of unity, so adjoining one root brings all three: a single step of degree $3$, total $3$. The same polynomial, a different base, a different answer.
The general lesson. Adjoining $\alpha$ adjoins every rational expression in $\alpha$, and whether that includes the other roots is a question about the polynomial rather than about the degree. A useful reflex: after each step, ask whether the roots can be written in terms of one another. For $x^{n} - 1$ they always can — the roots are the powers of one of them — which is why cyclotomic splitting fields are simple extensions.
Thinking $K(\alpha)$ for one root is the splitting field. It is only when the other roots happen to be expressible in $\alpha$. $\mathbb{Q}(\sqrt[3]{2})$ contains one root of $x^{3}-2$.
Expecting the degree to be $n$ or $n!$. It divides $n!$ and is usually strictly between $1$ and that.
Forgetting to re-factor. After adjoining a root, other factors may become reducible, and missing that inflates the degree.
Taking $\mathbb{C}$ as the splitting field of a rational polynomial. It is far too big; the splitting field is generated by the roots and nothing else.
Ignoring the base field. $x^{3} - 2$ has splitting degree $6$ over $\mathbb{Q}$ and $3$ over $\mathbb{Q}(\omega)$.
*Assuming uniqueness means literally the same set.* Two splitting fields inside different ambient fields are isomorphic, not equal. What is unique is the isomorphism type over $K$.
The roots are $\sqrt[3]{2}$, $\sqrt[3]{2}\omega$ and $\sqrt[3]{2}\omega^{2}$, with $\omega = \frac{-1 + \sqrt{-3}}{2}$.
All three roots.
$\mathbb{Q}(\sqrt[3]{2})$ has degree $3$ and lies in $\mathbb{R}$, so it misses two of them. Adjoining $\omega$ costs a further degree $2$, since $\omega$ satisfies $x^{2} + x + 1$ and is not real.
Two steps.
So the splitting field is $\mathbb{Q}(\sqrt[3]{2}, \omega)$ of degree $6$ — which divides $3! = 6$ exactly, so the Galois group will turn out to be all of $S_3$.
Degree six.
Over $\mathbb{F}_2$ the polynomial $x^{2} + x + 1$ has no root, and there is no bigger field lying around to look in.
Nothing to adjoin.
Kronecker: $\mathbb{F}_2[x]/(x^{2} + x + 1)$ is a field, and the class $\alpha$ of $x$ satisfies $\alpha^{2} = \alpha + 1$.
A root is manufactured.
Over that field, $x^{2} + x + 1 = (x - \alpha)(x - \alpha - 1)$ — the other root came free. So the splitting field is $\mathbb{F}_4$, of degree $2$.
The splitting field, built from nothing.
The roots are $\pm\sqrt2$ and $\pm\sqrt3$, so the splitting field is $\mathbb{Q}(\sqrt2, \sqrt3)$.
Adjoin both.
The first step has degree $2$, and $\sqrt3 \notin \mathbb{Q}(\sqrt2)$, so the second does too.
Two genuine steps.
The degree is $4$. It divides $4! = 24$ as it must, and the four automorphisms — flipping each square root independently — will make the Galois group the Klein four-group.
For each polynomial over $\mathbb{Q}$ the degree is given. Fill in the degree of its splitting field over $\mathbb{Q}$.
| Degree of the polynomial | Degree of the splitting field | |
|---|---|---|
| $x^{2} - 2$ | 2 | |
| $x^{2} + 1$ | 2 | |
| $x^{3} - 2$ | 3 | |
| $x^{3} - 1$ | 3 | |
| $x^{4} - 1$ | 4 | |
| $x^{4} - 2$ | 4 |
What is the degree over $\mathbb{Q}$ of the splitting field of $x^{5} - 1$?
Answer:
Select every statement that is true of the splitting field of a polynomial of degree $n$ over a field $K$.
This task has no paper form; do it on a device.
Put in order the steps of building the splitting field of a polynomial over $\mathbb{Q}$.
Number the steps in order (write the number in the box):
Match each polynomial over $\mathbb{Q}$ to its splitting field.
| $\mathbb{Q}(\sqrt2)$ | $\mathbb{Q}(i)$ | $\mathbb{Q}(\sqrt[3]{2}, \omega)$ | $\mathbb{Q}(\sqrt[4]{2}, i)$ | $\mathbb{Q}(\sqrt[3]{2})$ | |
|---|---|---|---|---|---|
| $x^{2} - 2$ | |||||
| $x^{2} + 1$ | |||||
| $x^{3} - 2$ | |||||
| $x^{4} - 2$ |
Build the proof that every polynomial over every field has a splitting field.
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.
For each polynomial over $\mathbb{Q}$ the degree is given. Fill in the degree of its splitting field over $\mathbb{Q}$.
| Degree of the polynomial | Degree of the splitting field | |
|---|---|---|
| $x^{2} - 2$ | 2 | |
| $x^{2} + 1$ | 2 | |
| $x^{3} - 2$ | 3 | |
| $x^{3} - 1$ | 3 | |
| $x^{4} - 1$ | 4 | |
| $x^{4} - 2$ | 4 |
You can build the splitting field of a small polynomial and compute its degree. Say in your own words why adjoining one root of a cubic need not bring the others. Next: the fields with finitely many elements, which are all splitting fields.
9. Your turn: the splitting field of $(x^{2} - 2)(x^{2} - 3)$ over $\mathbb{Q}$, step 3