Field Extensions

In this lesson you’ll learn what it means to extend a field, why the degree of an extension is a vector-space dimension, how degrees multiply in a tower, and how those facts settle three geometry problems that stood open for two thousand years.

$K$ is an extension of $F$, written $K/F$, if $F$ is a subfield of $K$. So $\mathbb{C}/\mathbb{R}$, $\mathbb{R}/\mathbb{Q}$, and $\mathbb{Q}(\sqrt2)/\mathbb{Q}$ are all extensions.

Why bother? Because you often need a root that isn’t there. $x^2 - 2$ has no root in $\mathbb{Q}$ and $x^2 + 1$ has none in $\mathbb{R}$. The fix is to build a bigger field containing one, which is exactly the quotient construction from two lessons ago:

$$\mathbb{Q}[x]/\langle x^2 - 2\rangle \cong \mathbb{Q}(\sqrt2), \qquad \mathbb{R}[x]/\langle x^2+1\rangle \cong \mathbb{C}$$

Here’s the idea that makes the whole theory work. If $F \subseteq K$, then $K$ is a vector space over $F$: you can add elements of $K$, and you can scale them by elements of $F$, and all the vector-space axioms follow from the field axioms.

The degree $[K : F]$ is the dimension of that vector space.

$$[\mathbb{C} : \mathbb{R}] = 2, \quad \text{basis } \{1, i\}$$ $$[\mathbb{Q}(\sqrt2) : \mathbb{Q}] = 2, \quad \text{basis } \{1, \sqrt2\}$$

The second says every element of $\mathbb{Q}(\sqrt2)$ is uniquely $a + b\sqrt2$ with $a,b \in \mathbb{Q}$. It’s a two-dimensional space over $\mathbb{Q}$, with $\sqrt2$ playing the role $i$ plays for $\mathbb{C}$.

An extension is finite if the degree is finite. $[\mathbb{R} : \mathbb{Q}]$ is infinite, which is why real analysis and field theory feel like different subjects.

For $\alpha$ algebraic over $F$ (meaning it’s a root of some polynomial with coefficients in $F$), the minimal polynomial $m(x)$ is the monic polynomial of least degree in $F[x]$ having $\alpha$ as a root.

It is automatically irreducible: a factorization $m = gh$ would give $g(\alpha)h(\alpha) = 0$, so one factor would already vanish at $\alpha$ and have smaller degree.

$$[F(\alpha) : F] = \deg m(x)$$

with basis $\{1, \alpha, \alpha^2, \dots, \alpha^{n-1}\}$ where $n = \deg m$. Higher powers reduce using $m(\alpha) = 0$, which is exactly the reduction step you did in $GF(4)$ using $a^2 = a+1$.

Examples over $\mathbb{Q}$:

• $\sqrt2$ has minimal polynomial $x^2 - 2$, so $[\mathbb{Q}(\sqrt2):\mathbb{Q}] = 2$.
• $\sqrt[3]2$ has minimal polynomial $x^3 - 2$, so the degree is 3.
• $i$ has minimal polynomial $x^2+1$, degree 2.
• $\sqrt2$ over $\mathbb{R}$ has minimal polynomial $x - \sqrt2$, degree 1, since it’s already there.

The last one is the reminder that a minimal polynomial depends on the base field.

Tower law. If $F \subseteq K \subseteq L$ then

$$[L : F] = [L : K] \cdot [K : F]$$

Degrees multiply, because a basis for $L$ over $F$ is obtained by multiplying each basis element of $L/K$ by each basis element of $K/F$: an $m \times n$ grid of products, giving $mn$ elements.

For $\mathbb{Q}(\sqrt2,\sqrt3)$ over $\mathbb{Q}$: two steps of degree 2 give total degree 4, with basis $\{1, \sqrt2, \sqrt3, \sqrt6\}$. Four elements, exactly as the tower law predicts.

The tower law is the tool that makes impossibility proofs work, because it constrains which degrees can occur inside a tower. If $[L:F] = 4$, no intermediate field can have degree 3 over $F$, since 3 does not divide 4.

The Greeks asked whether three constructions were possible with straightedge and compass. The answers, all negative, came in the nineteenth century, and all three are corollaries of the tower law.

The key fact: a straightedge-and-compass construction can only produce lengths lying in a tower of degree-2 extensions of $\mathbb{Q}$. A straightedge gives linear equations, a compass gives quadratics, and intersecting them never produces worse than a square root. So any constructible number has degree $2^k$ over $\mathbb{Q}$ for some $k$.

Doubling the cube. Building a cube of twice the volume needs length $\sqrt[3]{2}$, whose minimal polynomial $x^3 - 2$ is irreducible over $\mathbb{Q}$ by Eisenstein at $p=2$. So the degree is 3, and 3 is not a power of 2. Impossible.

Trisecting the angle. Trisecting 60° requires $\cos 20°$, which satisfies $8x^3 - 6x - 1 = 0$, irreducible over $\mathbb{Q}$. Degree 3 again. Impossible. Note some angles can be trisected; the claim is only that no method works for all of them.

Squaring the circle. A square of area equal to the unit circle needs $\sqrt{\pi}$. Lindemann proved in 1882 that $\pi$ is transcendental, so it satisfies no polynomial over $\mathbb{Q}$ at all and has no finite degree. Impossible.

Three questions open for two millennia, closed by a dimension count. That is as good an advertisement for abstraction as the subject has.

The splitting field of $f \in F[x]$ is the smallest extension in which $f$ factors completely into linear factors. Every polynomial has one, and it is unique up to isomorphism.

Galois theory studies the group of automorphisms of a splitting field that fix $F$, and it matches subgroups of that group against intermediate fields. The headline application: a polynomial is solvable by radicals exactly when its Galois group is solvable, meaning it can be broken apart by a chain of abelian quotients.

For degree 5 the generic Galois group is $S_5$, which contains the simple group $A_5$, and a simple non-abelian group cannot be broken apart that way. Hence no quintic formula. The question that started the whole subject is answered by a statement about one group of order 60.

Example 1: Find $[\mathbb{Q}(\sqrt3) : \mathbb{Q}]$.

Solution. $\sqrt3$ satisfies $x^2 - 3 = 0$, which is irreducible over $\mathbb{Q}$ since $\sqrt3$ is irrational and the polynomial is quadratic.

Minimal polynomial has degree 2, so $[\mathbb{Q}(\sqrt3):\mathbb{Q}] = 2$, with basis $\{1, \sqrt3\}$.

Every element is $a + b\sqrt3$. Inverting one shows the field is closed:

$$\frac{1}{a+b\sqrt3} = \frac{a - b\sqrt3}{a^2 - 3b^2}$$

which is again of the form $c + d\sqrt3$, and the denominator is nonzero because $\sqrt3$ is irrational.

Example 2: Find $[\mathbb{Q}(\sqrt[4]2) : \mathbb{Q}]$.

Solution. $\alpha = \sqrt[4]2$ satisfies $x^4 - 2 = 0$, irreducible over $\mathbb{Q}$ by Eisenstein with $p = 2$: it divides the constant $-2$ but $4 \nmid 2$.

$$[\mathbb{Q}(\sqrt[4]2) : \mathbb{Q}] = 4$$

with basis $\{1, \alpha, \alpha^2, \alpha^3\}$.

Note $\mathbb{Q}(\sqrt2)$ sits inside, since $\alpha^2 = \sqrt2$, and the tower law checks out:

$$4 = [\mathbb{Q}(\sqrt[4]2) : \mathbb{Q}(\sqrt2)] \cdot [\mathbb{Q}(\sqrt2):\mathbb{Q}] = 2 \cdot 2$$

Example 3: The tower law in use.

Find $[\mathbb{Q}(\sqrt2, \sqrt[3]2) : \mathbb{Q}]$.

Solution. $[\mathbb{Q}(\sqrt2):\mathbb{Q}] = 2$ and $[\mathbb{Q}(\sqrt[3]2):\mathbb{Q}] = 3$. Both must divide the total degree, so $6 \mid [\mathbb{Q}(\sqrt2,\sqrt[3]2):\mathbb{Q}]$.

Adjoining $\sqrt[3]2$ to $\mathbb{Q}(\sqrt2)$ raises the degree by at most 3, so the total is at most 6.

$$[\mathbb{Q}(\sqrt2,\sqrt[3]2) : \mathbb{Q}] = 6$$

with basis $\{1, \sqrt2\} \times \{1, \sqrt[3]2, \sqrt[3]4\}$, six products in all.

Example 4: Why $\sqrt[3]2$ is not constructible.

Solution. Every straightedge-and-compass construction produces numbers in a tower

$$\mathbb{Q} = F_0 \subset F_1 \subset \dots \subset F_k$$

with each $[F_{i+1} : F_i] = 2$. By the tower law, $[F_k : \mathbb{Q}] = 2^k$.

If $\sqrt[3]2$ were constructible it would lie in some $F_k$, so by the tower law

$$[\mathbb{Q}(\sqrt[3]2) : \mathbb{Q}] = 3 \ \big| \ 2^k$$

But 3 does not divide any power of 2. Contradiction, so $\sqrt[3]2$ is not constructible and the cube cannot be doubled. ∎

Example 5: A degree that is not what it looks like.

Find $[\mathbb{Q}(\sqrt2 + \sqrt3) : \mathbb{Q}]$.

Solution. Let $\alpha = \sqrt2+\sqrt3$. Then $\alpha^2 = 5 + 2\sqrt6$, so $\sqrt6 = \frac{\alpha^2-5}{2} \in \mathbb{Q}(\alpha)$.

Also $\alpha^3 = \alpha \cdot \alpha^2 = (\sqrt2+\sqrt3)(5+2\sqrt6) = 5\sqrt2 + 5\sqrt3 + 2\sqrt{12} + 2\sqrt{18} = 11\sqrt2 + 9\sqrt3$.

From $\alpha = \sqrt2 + \sqrt3$ and $\alpha^3 = 11\sqrt2+9\sqrt3$ we can solve for $\sqrt2$ and $\sqrt3$ individually, so

$$\mathbb{Q}(\sqrt2+\sqrt3) = \mathbb{Q}(\sqrt2,\sqrt3)$$

and the degree is 4, not 2. Its minimal polynomial is $x^4 - 10x^2 + 1$.

The lesson: adjoining a sum can generate more than the sum suggests.

Example 6: Finite fields as extensions.

Express $GF(8)$ as a field extension and give its degree.

Solution. $GF(8) = \mathbb{F}_2[x]/\langle x^3+x+1\rangle$, so it is $\mathbb{F}_2(\alpha)$ where $\alpha$ is a root of the irreducible cubic $x^3+x+1$.

$$[GF(8) : \mathbb{F}_2] = 3$$

with basis $\{1, \alpha, \alpha^2\}$. As a vector space it is $\mathbb{F}_2^3$, which has $2^3 = 8$ elements, confirming the order.

In general $[GF(p^n) : \mathbb{F}_p] = n$, and this is the cleanest way to see why finite field orders must be prime powers: the size of a vector space of dimension $n$ over $\mathbb{F}_p$ is $p^n$.

Cryptography. Elliptic-curve cryptography chooses curves over $GF(p)$ or $GF(2^n)$, and the extension degree is a security parameter. Pairing-based cryptography, which enables identity-based encryption and short digital signatures, works with towers of extensions and relies on degree arithmetic to be efficient.

Coding theory. BCH and Reed–Solomon codes are designed by choosing a splitting field for a chosen polynomial. The code’s error-correcting power is determined by the degrees of the minimal polynomials of the roots used, so the design problem is a field-extension problem.

Computer algebra. Representing an algebraic number exactly means storing its minimal polynomial plus enough data to pin down which root. Systems like SageMath and Mathematica do arithmetic in number fields this way, and degree bounds control the cost of every operation.

Signal processing. The discrete Fourier transform needs a primitive $n$-th root of unity, so implementing it over a finite field requires an extension containing one. Number-theoretic transforms used for exact large-integer multiplication choose the field precisely so a suitable root exists.

The impossibility results themselves. Beyond their historical weight, they are the standard example of using algebra to prove a negative. “No algorithm can do this” arguments in computer science have the same shape: find an invariant, show the target violates it.