ISEGORIA / MATH ENCYCLOPEDIA
Galois theory: the symmetry of equations
Why a group answers a question about equations. The roots of a polynomial can be permuted, but only in ways that respect every rational relation between them, and the permutations that survive form the Galois group. Its subgroups match the fields between the rationals and the splitting field one for one, and its size decides what ruler and compass can draw and which quintics have no formula.
Before you begin: Group theory, polynomials and complex numbers
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Which permutations of the roots are allowed?
A polynomial with rational coefficients has n roots in the complex plane. A field automorphism of the splitting field permutes them, but it cannot permute them arbitrarily: any relation between roots with rational coefficients, such as r₁ + r₃ = 0 for x⁴ − 2, must still hold after the permutation. The Galois group is the set of permutations that break no rational relation. Choose a polynomial, then step through the permutations of its roots with the slider or click a cell in the table on the right: the arrows show where each root is sent, and the readout names the first relation the permutation breaks, if any. For x⁴ − 2 only 8 of the 24 permutations survive, the dihedral group; for x⁵ − x − 1 every one of the 120 does.
Worked example. For \(x^4-2\) with roots \(r_1=\alpha,\ r_2=i\alpha,\ r_3=-\alpha,\ r_4=-i\alpha\), the relations \(r_1+r_3=0\) and \(r_2+r_4=0\) force a permutation to map the pair \(\{r_1,r_3\}\) to itself or to \(\{r_2,r_4\}\): the swap \((r_1\,r_2)\) sends \(r_1+r_3\) to \(r_2+r_3=(i-1)\alpha\ne0\), so it is excluded, while \((r_1\,r_2\,r_3\,r_4)\) is allowed. Eight permutations survive and \(|\operatorname{Gal}|=8=[\mathbb Q(\sqrt[4]2,i):\mathbb Q]\). For \(x^3-3x+1\) the roots satisfy \(r_2=r_1^2-2\), so \(r_1\) determines everything and only the 3 cyclic shifts survive; its discriminant \(81=9^2\) is a square, as it must be for a subgroup of \(A_3\).
Watch out. The lab tests each permutation against a short list of relations chosen for the polynomial, so it shows which permutations fail and, for the survivors, that no relation on the list objects. That the survivors are exactly the Galois group rests on theory, not on the test: the group order equals the degree of the splitting field, and for each polynomial here that degree is known (6, 8, 4, 4, 3 and 120). The quintic \(x^5-x-1\) is irreducible with group \(S_5\); it is the standard example of an equation with no solution in radicals, because \(S_5\) has no chain of normal subgroups with abelian quotients. Roots are computed numerically, and the relation checks use a tolerance, so they are evidence for the algebra, not a proof of it.
Why is the Galois group of a polynomial always a subgroup of the alternating group exactly when the discriminant is a square?
Let \(\delta=\prod_{i<j}(r_i-r_j)\), so \(\delta^2=\Delta\). A transposition of two roots changes the sign of \(\delta\), so an even permutation fixes \(\delta\) and an odd one sends it to \(-\delta\). Every \(\sigma\) in the Galois group fixes the rationals; if \(\delta\) is rational, every \(\sigma\) fixes it, so no odd permutation can be in the group. Conversely, if the group lies in \(A_n\), \(\delta\) is fixed by the whole group, and an element of the splitting field fixed by the whole group is rational by the fundamental theorem.
2. The Galois correspondence
Between the rationals and the splitting field K lie intermediate fields, and inside the Galois group lie subgroups. Galois’s fundamental theorem says they match exactly, and upside down: each subgroup H corresponds to the field of elements it fixes, larger subgroups fix smaller fields, and the degree of the fixed field is the index of H. Here both lattices are drawn for K = ℚ(⁴√2, i), whose group is the dihedral group of order 8, or for ℚ(³√2, ω) with group S₃. Click any subgroup or any field to light up its partner and the chain between them. The fixed fields are computed, not looked up: each candidate field is given by a generator written in the roots, and the lab checks numerically which permutations leave that generator unchanged.
Worked example. In \(\mathbb Q(\sqrt[4]2,i)\) the rotation \(\rho:\ \alpha\mapsto i\alpha,\ i\mapsto i\) generates a cyclic subgroup of order 4. It moves \(\alpha\), \(\alpha^2=\sqrt2\) goes to \(-\sqrt2\), but \(i\) is fixed, so the fixed field is \(\mathbb Q(i)\), of degree \(8/4=2\). The reflection \(\tau:\ \alpha\mapsto\alpha,\ i\mapsto-i\) fixes \(\alpha\) and hence all of \(\mathbb Q(\sqrt[4]2)\), degree \(8/2=4\). The subgroup \(\{1,\rho^2\}\) fixes \(\alpha^2\) and \(i\) but not \(\alpha\): its field is \(\mathbb Q(\sqrt2,i)\), degree 4, and since \(\rho^2\) is central the subgroup is normal and \(\mathbb Q(\sqrt2,i)/\mathbb Q\) is Galois with group \(G/\{1,\rho^2\}\cong C_2\times C_2\).
Watch out. The subgroups are found by closing every subset of the group under composition, and the fixed field is identified among a finite list of candidates by the numerical test; both steps are exact in principle but the second relies on the candidate list being complete, which the fundamental theorem guarantees since it says there are exactly as many fields as subgroups. The three non-normal subgroups of order 2 have fixed fields \(\mathbb Q(\sqrt[4]2)\), \(\mathbb Q(i\sqrt[4]2)\), \(\mathbb Q((1\pm i)\sqrt[4]2)\), which are not Galois over \(\mathbb Q\): each contains one root of \(x^4-2\) without the others. The correspondence needs the extension to be Galois, meaning normal and separable; for \(\mathbb Q(\sqrt[4]2)\) alone the group has only 2 elements and there are more subfields than subgroups.
Why does a larger subgroup fix a smaller field, and why is the correspondence a bijection rather than merely order-reversing?
If \(H\subseteq H^{\prime}\), an element fixed by every automorphism in \(H^{\prime}\) is in particular fixed by those in \(H\), so \(K^{H^{\prime}}\subseteq K^{H}\): more conditions, fewer elements. Bijectivity is Artin’s theorem: for a finite group \(H\) of automorphisms of \(K\), the degree \([K:K^H]\) equals \(|H|\) exactly, so starting from \(H\), taking its fixed field, and then taking the group of automorphisms fixing that field returns a group containing \(H\) with the same order, hence \(H\) itself. Going the other way uses that \(K/\mathbb Q\) is Galois, so that \(K\) is Galois over every intermediate field \(F\) and \(F\) is recovered as the fixed field of \(\operatorname{Gal}(K/F)\).
3. Ruler, compass and the degree of a field
A point can be constructed with ruler and compass exactly when its coordinates lie in a tower of quadratic extensions of ℚ, so every constructible number has degree a power of 2. That single fact settles three ancient problems. The regular n-gon needs cos(2π/n), of degree φ(n)/2, so Gauss and Wantzel’s criterion is that φ(n) be a power of 2, which happens exactly when n is a power of 2 times distinct Fermat primes 3, 5, 17, 257, 65537. Trisecting an angle θ means constructing cos(θ/3) from cos θ, and the relative degree of that extension is 1, 2 or 3; only 3 is fatal, and it is 3 for 60°, so the 20° angle cannot be drawn. Doubling the cube needs ³√2, of degree 3. Slide n or θ and read the degrees.
Worked example. \(n=17\): \(\varphi(17)=16=2^4\), so \(\cos(2\pi/17)\) has degree 8 and the 17-gon is constructible, Gauss’s discovery at nineteen. \(n=7\): \(\varphi(7)=6\), degree 3, impossible. \(n=9\): \(\varphi(9)=6\), impossible, which is the trisection of \(120^\circ\) in disguise. For \(\theta=60^\circ\), \(\cos\theta=\tfrac12\) has degree 1 and \(\cos20^\circ\) is a root of \(8c^3-6c-1\), irreducible over \(\mathbb Q\) since it has no rational root, so the relative degree is 3. For \(\theta=90^\circ\), \(4c^3-3c=0\) factors and \(\cos30^\circ=\sqrt3/2\) has degree 2: constructible.
Watch out. Degree a power of 2 is necessary for constructibility but not sufficient in general: the tower must consist of quadratic steps, and a quartic field need not contain a quadratic subfield. For \(\cos(2\pi/n)\) sufficiency holds because the extension is abelian, so its Galois group of order \(2^k\) has a chain of subgroups of index 2 and the fixed fields give the tower. The angles here are whole numbers of degrees, so every \(\cos\theta\) is algebraic and its degree is \(\varphi(m)/2\) with \(m=360/\gcd(\theta,360)\); for an angle with transcendental cosine the question of trisection is different, and a general angle can be trisected once cos θ is allowed as given, exactly when the relative degree is not 3. The five known Fermat primes are the only ones known; whether there are more is open, so the list of constructible polygons is complete only up to \(2^{32}+1\), which is composite.
Why does trisection reduce to a cubic, and why can a cubic with no rational root not be reached by quadratic steps?
The triple-angle formula \(\cos\theta=4\cos^3(\theta/3)-3\cos(\theta/3)\) makes \(c=\cos(\theta/3)\) a root of the cubic \(4c^3-3c-\cos\theta\) over \(\mathbb Q(\cos\theta)\). If the cubic has no root in that field it is irreducible there, so \(c\) has degree 3 over it. A tower of quadratic extensions has total degree \(2^k\), and by the tower law any subfield of it has degree dividing \(2^k\); 3 does not divide a power of 2, so \(c\) is in no such tower. The same argument kills \(\sqrt[3]2\), a root of the irreducible \(x^3-2\).