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A bigger field as a vector space over a smaller one: the degree as a dimension, the tower law that multiplies degrees, and why every intermediate degree divides the total.
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 read a field extension as a vector space and name a basis, compute the degree of a simple extension, apply the tower law to a compound one, say why every intermediate degree divides the total, explain why a finite extension is algebraic while the converse fails, and prove the tower law by building a basis of products.
The first course built quotients of polynomial rings and found that $F[x]/(f)$ is a field when $f$ is irreducible. Linear algebra measured a vector space by its dimension. Neither was aimed at the other. This lesson points them together: a field containing $K$ is a vector space over $K$, and its dimension turns out to be the single most useful number attached to it.
$L/K$ is a field extension when $K$ is a subfield of $L$. The degree $[L : K]$ is the dimension of $L$ as a $K$-vector space; the extension is finite when that is finite. $\alpha \in L$ is algebraic over $K$ when some non-zero polynomial over $K$ has it as a root, and transcendental otherwise. A tower is a chain $K \subseteq F \subseteq L$.
The extension as a vector space. If $K \subseteq L$ are fields, then $L$ is a vector space over $K$: addition is the field's addition, and scalars from $K$ multiply by the field's multiplication. Every vector space axiom is a field axiom already available. Its dimension is written
$$[L : K] = \dim_K L$$
and called the degree of the extension. $[L : K] = 1$ exactly when $L = K$.
Examples. $[\mathbb{C} : \mathbb{R}] = 2$, with basis $\{1, i\}$ — which is why a complex number is written $a + bi$. $[\mathbb{Q}(\sqrt2) : \mathbb{Q}] = 2$, with basis $\{1, \sqrt2\}$. $[\mathbb{F}_{p^{n}} : \mathbb{F}_p] = n$. And $[\mathbb{R} : \mathbb{Q}]$ is infinite.
The tower law. For $K \subseteq F \subseteq L$,
$$[L : K] = [L : F] \cdot [F : K].$$
The proof is that the products of a basis of $F/K$ with a basis of $L/F$ form a basis of $L/K$; spanning is substitution, and independence is one grouping of a double sum.
The corollary that does the work. Every intermediate degree divides the total. A field strictly between $\mathbb{Q}$ and a degree-$4$ extension has degree $2$; there is no degree-$3$ intermediate field. That is exactly the shape of Lagrange's theorem, and it is what will make the Galois correspondence match subgroups with fields.
Finite implies algebraic. If $[L : K] = n$ and $\alpha \in L$, the $n+1$ elements $1, \alpha, \ldots, \alpha^{n}$ cannot be independent, so some non-trivial $K$-combination of them vanishes — and that combination is a polynomial over $K$ with $\alpha$ as a root.
The converse fails: the field of all algebraic numbers is algebraic over $\mathbb{Q}$ and has infinite degree, since it contains $\sqrt[n]{2}$ for every $n$.
Simple extensions. $K(\alpha)$ is the smallest field containing $K$ and $\alpha$. When $\alpha$ is algebraic, $K(\alpha) \cong K[x]/(m_\alpha)$ and $[K(\alpha) : K] = \deg m_\alpha$ — which is the next lesson.
A second course in algebra counts before it constructs. An orbit size divides the order of the group, a Sylow count is pinned between a divisibility and a congruence, and a degree multiplies up a tower — and each of those numbers rules out structures nobody has to go looking for.
Another way: picture
A tower of fields is a tower of vector spaces, each sitting inside the next as the scalars. Going up one floor multiplies the number of coordinates needed to describe a point; going up two floors multiplies twice. So the total number of coordinates from the ground floor is the product, and no floor can require a number of coordinates that fails to divide the total.
Another way: steps
To find $[L : K]$: 1. Write $L$ as $K$ with generators adjoined, one at a time. 2. For each step, find the minimal polynomial of the new generator over the current field; its degree is that step's degree. 3. Multiply the step degrees. 4. Check: every degree that appears must divide the answer. 5. If a generator turns out to lie in the field already, its step has degree $1$ and contributes nothing.
Ruling things out. This is the main use. A number of degree $3$ cannot lie in an extension of degree $4$, because $3 \nmid 4$. That one line settles the doubling of the cube, and versions of it settle the other classical impossibilities.
Computing compound degrees. $[\mathbb{Q}(\sqrt2, \sqrt3) : \mathbb{Q}]$: the first step is $2$; the second is $2$ provided $\sqrt3 \notin \mathbb{Q}(\sqrt2)$, which is checked by supposing $\sqrt3 = a + b\sqrt2$ and squaring. So the degree is $4$, and $\{1, \sqrt2, \sqrt3, \sqrt6\}$ is a basis.
Finding hidden equalities. $\mathbb{Q}(\sqrt2 + \sqrt3)$ looks smaller than $\mathbb{Q}(\sqrt2, \sqrt3)$ and is not. Let $\alpha = \sqrt2 + \sqrt3$; then $\alpha^{3} = 11\sqrt2 + 9\sqrt3$, and solving the two linear equations recovers $\sqrt2$ and $\sqrt3$ separately. So the two fields are equal, and $\alpha$ has degree $4$ — which is why its minimal polynomial is the quartic $x^{4} - 10x^{2} + 1$.
Bounding what a step can be. If $[L : K]$ is prime, there are no intermediate fields at all, because any would have degree dividing a prime. So a degree-$5$ extension is as indecomposable as a group of order $5$.
A warning about the second step. The degree of $\beta$ over $K(\alpha)$ can be strictly smaller than its degree over $K$, because the bigger base field offers more polynomials to factor with. $\sqrt3$ has degree $2$ over $\mathbb{Q}$ and degree $1$ over $\mathbb{Q}(\sqrt2, \sqrt3)$. Assuming the degrees stay the same is the commonest error in these computations.
Adding degrees instead of multiplying. The tower law is a product; it comes from a basis of products.
Using the degree over the wrong base. In a tower, the second step's degree is taken over the enlarged field, not over the original.
Assuming $[K(\alpha, \beta) : K]$ is the product of the two individual degrees. It divides that product and can be smaller: $\mathbb{Q}(\sqrt2, \sqrt8) = \mathbb{Q}(\sqrt2)$ has degree $2$, not $4$.
Confusing the degree of the extension with the degree of a polynomial that happens to have $\alpha$ as a root. Only the minimal polynomial gives the degree; $x^{4} - 4$ has $\sqrt2$ as a root and degree $4$.
Thinking algebraic implies finite. The algebraic numbers are a counterexample.
Forgetting that $[L:K] = 1$ means equality. It is not a way of saying very close.
$[\mathbb{Q}(\sqrt2) : \mathbb{Q}] = 2$, since $x^{2} - 2$ is irreducible over $\mathbb{Q}$.
The first step.
Is $\sqrt3 \in \mathbb{Q}(\sqrt2)$? If $\sqrt3 = a + b\sqrt2$ then squaring gives $3 = a^{2} + 2b^{2} + 2ab\sqrt2$, forcing $ab = 0$ and then $3 = a^{2}$ or $3 = 2b^{2}$, neither soluble in $\mathbb{Q}$.
So the second step is genuine.
$x^{2} - 3$ is therefore irreducible over $\mathbb{Q}(\sqrt2)$ and the second step has degree $2$: the tower law gives $2 \times 2 = 4$, with basis $\{1, \sqrt2, \sqrt3, \sqrt6\}$.
Degree four.
Suppose $[L : K] = 5$ and $K \subseteq F \subseteq L$. The tower law gives $[F : K] \cdot [L : F] = 5$.
The degrees multiply to a prime.
So one factor is $1$ and the other is $5$.
Only two ways to split it.
A degree of $1$ means equality, so $F = K$ or $F = L$. There is nothing in between — exactly as a group of prime order has no proper subgroups.
No intermediate field.
The first step $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt2)$ has degree $2$.
A quadratic step.
$\sqrt[3]{2}$ has degree $3$ over $\mathbb{Q}$, and $3$ must divide the total; so must $2$.
Both degrees divide the answer.
The total is at most $2 \times 3 = 6$ and divisible by both $2$ and $3$, so it is exactly $6$. The second step therefore has degree $3$: $x^{3} - 2$ stays irreducible over $\mathbb{Q}(\sqrt2)$.
Each row is a tower $K \subseteq F \subseteq L$ with the two step degrees given. Fill in the degree of $L$ over $K$.
| Degree of $F$ over $K$ | Degree of $L$ over $F$ | Degree of $L$ over $K$ | |
|---|---|---|---|
| The first tower | 2 | 3 | |
| The second tower | 2 | 2 | |
| The third tower | 3 | 2 | |
| The fourth tower | 4 | 2 | |
| The fifth tower | 1 | 5 |
$K \subseteq F \subseteq L$ with $[F : K] = 3$ and $[L : F] = 2$. What is $[L : K]$?
Answer:
Select every statement that is true of a field extension $L/K$.
This task has no paper form; do it on a device.
Match each extension to its degree over the base field.
| Degree $2$, with basis $\{1, \sqrt2\}$ | Degree $3$ | Degree $4$ | Degree $2$, with basis $\{1, i\}$ | Degree $6$ | |
|---|---|---|---|---|---|
| $\mathbb{Q}(\sqrt2)$ over $\mathbb{Q}$ | |||||
| $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$ | |||||
| $\mathbb{Q}(\sqrt2, \sqrt3)$ over $\mathbb{Q}$ | |||||
| $\mathbb{C}$ over $\mathbb{R}$ |
Put in order the steps of computing $[\mathbb{Q}(\alpha, \beta) : \mathbb{Q}]$.
Number the steps in order (write the number in the box):
Build the proof of the tower law: $[L : K] = [L : F][F : K]$ for $K \subseteq F \subseteq L$.
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.
Each row is a tower $K \subseteq F \subseteq L$ with the two step degrees given. Fill in the degree of $L$ over $K$.
| Degree of $F$ over $K$ | Degree of $L$ over $F$ | Degree of $L$ over $K$ | |
|---|---|---|---|
| The first tower | 2 | 3 | |
| The second tower | 2 | 2 | |
| The third tower | 3 | 2 | |
| The fourth tower | 4 | 2 | |
| The fifth tower | 1 | 5 |
You can compute the degree of a compound extension with the tower law and say what the answer is a dimension of. Say in your own words why the second step of a tower is measured over the enlarged field. Next: the polynomial that supplies each step's degree.
9. Your turn: what is $[\mathbb{Q}(\sqrt2, \sqrt[3]{2}) : \mathbb{Q}]$?, step 3