Fields, Field Extensions, and Galois Theory

Part 11, Chapter 11: Abstract Algebra Survey

Learning objectives

Define fields, field extensions, and the degree $[K:F]$
Construct the splitting field of a polynomial and identify the Galois group
State the fundamental theorem of Galois theory (correspondence between subgroups and intermediate fields)
Connect solvability of polynomial equations to solvability of the Galois group, and explain Abel-Ruffini

Galois theory is the moment abstract algebra produces something the ancients could not. For two thousand years mathematicians knew the quadratic formula, found cubic and quartic formulas in the sixteenth century, and tried in vain to find a quintic formula. Then, in the 1820s, Abel and Galois showed why no such formula could exist: the symmetries of the roots form a group, and the algebraic operations available to a "formula" can only produce groups of a very restricted shape. The general quintic's symmetry group does not have that shape, end of story. This is the prototype of every "no algorithm exists" theorem in modern mathematics.

Fields and extensions

A field is a commutative ring with unity in which every nonzero element has a multiplicative inverse. Examples: $\mathbb{Q}, \mathbb{R}, \mathbb{C}$ , and the finite fields $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ p=mathbbZ/pmathbbZ for prime $p$ . A field extension is an inclusion $F \subseteq K$ of fields. We write $K/F$ and read it " $K$ over $F$ ."

The degree $[K : F]$ is the dimension of $K$ as an $F$ -vector space. For example, $[\mathbb{C} : \mathbb{R}] = 2$ because $\{1, i\}$ is an $\mathbb{R}$ -basis, and $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$ because $\{1, \sqrt{2}\}$ is a $\mathbb{Q}$ -basis. The tower law says $[L : F] = [L : K] \cdot [K : F]$ whenever $F \subseteq K \subseteq L$ .

Galois group: the symmetries of the roots

Given an extension $K/F$ , the Galois group $\operatorname{Gal}(K/F)$ is the group of all field automorphisms $\sigma : K \to K$ that fix $F$ pointwise (so $\sigma(c) = c$ for every $c \in F$ ). Concretely, if $K$ is the splitting field of a polynomial $p(x) \in F[x]$ , every element of $\operatorname{Gal}(K/F)$ permutes the roots of $p$ . The Galois group acts on roots; that is the entire content of the theory.

Example: for $K = \mathbb{Q}(\sqrt{2})$ over $F = \mathbb{Q}$ , an automorphism is determined by where it sends $\sqrt{2}$ , and since $\sigma(\sqrt{2})^2 = \sigma(2) = 2$ , we have $\sigma(\sqrt{2}) \in \{\sqrt{2}, -\sqrt{2}\}$ . So $|\operatorname{Gal}(K/F)| = 2$ and $\operatorname{Gal}(K/F) \cong \mathbb{Z}/2\mathbb{Z}$ .

The truth-table widget above is not literally a Galois group, but its 2-row two-input pattern (rows indexed by $P \in \{T, F\}$ , columns by $Q$ ) is structurally identical to the $\operatorname{Gal}(\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q})$ character table: two independent two-element symmetries (flip $\sqrt{2} \to -\sqrt{2}$ ; flip $\sqrt{3} \to -\sqrt{3}$ ) generating a Klein four-group $\mathbb{Z}/2 \times \mathbb{Z}/2$ . Small finite groups look like truth tables.

The fundamental theorem of Galois theory

For a finite Galois extension $K/F$ (where $K$ is the splitting field of a separable polynomial over $F$ ) with Galois group $G = \operatorname{Gal}(K/F)$ :

There is a one-to-one correspondence between subgroups $H \leq G$ and intermediate fields $F \subseteq E \subseteq K$ , given by $H \leftrightarrow E = K^H$ (the fixed field of $H$ ).
The correspondence is inclusion-reversing: larger subgroups correspond to smaller fields.
$[K : E] = |H|$ and $[E : F] = [G : H]$ .
$E/F$ is itself a Galois extension if and only if $H$ is a normal subgroup of $G$ ; in that case $\operatorname{Gal}(E/F) \cong G/H$ .

This is one of the most beautiful theorems in mathematics: an algebraic problem about polynomials is converted into a combinatorial problem about subgroups.

Solvability by radicals

A polynomial equation is solvable by radicals if its roots can be written using only $+, -, \times, \div$ and $n$ -th roots $\sqrt[n]{\cdot}$ . A group $G$ is solvable if it has a chain of subgroups $\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_n = G$ G with each quotient $G_{i+1} / G_i$ abelian. Galois's great theorem: a polynomial is solvable by radicals if and only if its Galois group is solvable.

The general quintic $x^5 + a_4 x^4 + \cdots + a_0$ has Galois group $S_5$ . The chain $\{e\} \trianglelefteq A_5 \trianglelefteq S_5$ has quotient $A_5$ , which is simple (no nontrivial normal subgroups) and non-abelian (order $60$ ). So $S_5$ is not solvable, and the general quintic is not solvable by radicals, the Abel-Ruffini theorem.

Where this shows up

Computer algebra: Mathematica, Maple, and SageMath all include Galois-group computations as a primitive. When the system tells you "this equation has no closed-form solution," it has computed the Galois group and checked whether it is solvable.

Elliptic-curve cryptography: The endomorphism ring of an elliptic curve over a finite field is governed by its Galois action. Security proofs for ECC reduce to Galois-theoretic statements about isogeny graphs.

Geometric impossibilities: Trisecting a general angle, doubling the cube, and squaring the circle, the three famous unsolvable Greek problems, are all proved impossible by showing the relevant field extension has degree not a power of $2$ , which it would need to be to be constructible by compass and straightedge.

Inverse Galois problem: A still-open question that has driven much modern algebra: is every finite group the Galois group of some Galois extension of $\mathbb{Q}$ ? Many cases are known; the general problem remains unsolved.

Pause and think: Compute $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}]$ . (Hint: the minimal polynomial of $\sqrt[3]{2}$ is $x^3 - 2$ , irreducible by Eisenstein at $p = 2$ .) Is $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ a Galois extension over $\mathbb{Q}$ ? (Hint: does $\mathbb{Q}(\sqrt[3]{2})$ contain all three roots of $x^3 - 2$ ?)

Try it

Predict first: what is $\operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q})$ ? (Hint: an automorphism must send $i$ to another root of $x^2 + 1$ .) How many elements does the group have? Name it.

Construct a field with $9$ elements. (Hint: start with $\mathbb{F}_3 = \mathbb{Z}/3\mathbb{Z}$ and adjoin a root of an irreducible quadratic over $\mathbb{F}_3$ , for example $x^2 + 1$ .) Verify that $x^2 + 1$ is irreducible over $\mathbb{F}_3$ .

True or false: every finite extension of $\mathbb{Q}$ is a Galois extension. (Hint: think about $\mathbb{Q}(\sqrt[3]{2})$ , how many of the roots of $x^3 - 2$ live in it?)

Compute $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}]$ using the tower law. (Hint: pass through $\mathbb{Q}(\sqrt{2})$ and show $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$ .)

The fundamental theorem in action: list every subgroup of the Klein four-group $V_4 = \{e, a, b, ab\}$ . How many intermediate fields does $\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}$ therefore have, INCLUDING the trivial extensions $\mathbb{Q}$ itself and the full $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ ?

A trap to watch for

It is tempting to think Galois theory says the quintic is "unsolvable" in some absolute sense. It does not. The quintic is unsolvable by radicals, using $+, -, \times, \div$ , and $n$ -th roots. Specific quintics are perfectly solvable: $x^5 - 2 = 0$ has roots $\sqrt[5]{2} \cdot \zeta^k$ for $k = 0, \ldots, 4$ , where $\zeta = e^{2\pi i/5}$ is a fifth root of unity. The unsolvable result is about the general quintic, where the coefficients are independent indeterminates and the Galois group is the full $S_5$ . Modern numerical methods (Newton's, Aberth's, Durand-Kerner's) routinely solve quintics to arbitrary precision, "no formula in radicals" is not the same as "no answer."

What you now know

You can compute degrees of small extensions of $\mathbb{Q}$ , write down Galois groups in simple cases, apply the tower law, and state the fundamental theorem connecting subgroups to intermediate fields. You also know why the quintic resisted the algebraists for centuries: not because no one was clever enough, but because the group $A_5$ is simple. The next chapter shifts gears completely, from algebra to analysis, and asks: what does it mean to integrate a function?

Mark section complete →

References

Garrity, T. (2002). All the Mathematics You Missed: But Need to Know for Graduate School. Cambridge University Press, ch. 11.

Dummit, D. S., Foote, R. M. (2003). Abstract Algebra (3rd ed.). Wiley, ch. 13-14.

Artin, M. (2010). Algebra (2nd ed.). Pearson, ch. 15-16.

Lang, S. (2002). Algebra (3rd revised ed.). Springer, ch. 5-6.

Stewart, I. (2015). Galois Theory (4th ed.). Chapman and Hall/CRC.

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