Translate geometric operations into quadratic extensions

Try

Start with a unit segment and rational coordinate axes. A straightedge joins known points; a compass draws a circle with a known center and constructible radius. New points arise from intersections of lines and circles. Predict the algebraic degree of the equations you must solve at each step.

Intersecting two nonparallel lines solves a linear system. Intersecting a line and a circle generally solves a quadratic. This is the bridge from geometric operations to field extensions.

Predict

If each genuine new step has degree two, predict the degree of the field after several steps. Then distinguish that total field degree from the degree of one coordinate inside it. The coordinate degree divides the total degree by the tower law; it need not equal the whole tower degree.

Experiment

Translate the three intersection types into equations with coefficients in the current real coordinate field F.

Intersection	Algebraic reduction	Degree needed
Two nonparallel lines	Nonsingular linear system	1
Line and circle	Substitute to obtain a quadratic	At most2
Two distinct circles	Subtract to get a line, then use a circle	At most2

Degenerate cases need separate treatment. Parallel lines may not meet, tangent intersections may add no square root, and concentric circles may have no discrete intersection. A construction only selects actual real intersection points.

Understand

Line equations use field operations on known coordinates. Circle equations have the form (x−a)²+(y−b)²=r², with coefficients in the current field. Substitution yields a quadratic whose solutions involve at most one square root. Subtracting two circle equations cancels the quadratic terms, reducing their intersection to the line-circle case when the subtraction gives a genuine line.

Therefore every constructible coordinate belongs to a finite tower of real extensions whose degrees are one or two. Conversely, field operations and adjoining square roots of positive constructible quantities can be realized by classical similar-triangle and geometric-mean constructions. Thus the precise algebraic criterion is containment in such a tower, not merely a numerical degree condition.

The tower law implies that the degree over Q of a constructible algebraic number is a power of two. To double a unit cube, the new side would be ∛2. Its polynomial x³−2 is irreducible by Eisenstein at two, so its degree is three. Three cannot divide the power-of-two degree of a quadratic tower. Hence no finite straightedge-and-compass construction from the unit segment produces that side length.

A related angle obstruction can be made explicit. Trisecting a constructible sixty-degree angle would construct twenty degrees and therefore u=2cos20°. The triple-angle formula gives u³−3u−1=0. The only possible rational roots of this monic cubic are ±1, and neither works, so it is irreducible over Q. The degree-three obstruction again applies. This does not say no angle can be trisected; it rules out a universal method by one constructible input.

Power-of-two degree is necessary but not sufficient in general. The full criterion requires a chain of quadratic extensions; an arbitrary real algebraic number of degree four or eight need not come with that chain. Proving sufficiency for a particular number requires constructing the chain or a geometric construction, rather than inspecting degree alone.

Exact regular polygons

For n≥3, constructing a regular n-gon on the unit circle amounts to constructing cos(2π/n) and sin(2π/n). This deeper section uses cyclotomic irreducibility and Galois groups and the Galois correspondence. The earlier cube and angle obstructions do not require this machinery.

Let ζ=exp(2πi/n), M=Q(ζ), c=ζ+ζ⁻¹=2cos(2π/n), and K=Q(c). The element ζ solves X²−cX+1 and is nonreal, whereas K is real. Thus [M:K]=2 and [K:Q]=φ(n)/2. Complex conjugation fixes c and has order two; the fixed-field degree comparison proves its fixed field is exactly K. Hence K is the maximal real subfield of M.

The cyclotomic group is abelian, so its conjugation subgroup is normal. Therefore K/Q is Galois with abelian group of order φ(n)/2. This extra structure matters: an arbitrary field of degree four or eight need not have a chain of quadratic subfields.

If the regular polygon is constructible, c is constructible, so its degree φ(n)/2 is a power of two. Thus φ(n) is a power of two. Conversely, suppose φ(n) is a power of two. A finite abelian 2-group has a chain of subgroups from the whole group to the identity with each successive index two. Indeed, a maximal proper subgroup has a nontrivial simple abelian quotient. Any nonidentity element of a simple abelian group generates the whole group; a cyclic group is simple only if its order is prime. That prime must be two here. Repeat inside the subgroup until reaching the identity.

Reversing this chain through Galois correspondence gives a tower from Q to K with quadratic steps. All fields in the tower are real. In a real quadratic step L=F(β), the monic minimal polynomial X²+bX+d has discriminant b²−4d=(2β+b)² strictly positive in the chosen real embedding. The quadratic formula constructs β by a positive square root and field operations. The trivial group gives K=Q and needs no steps.

Consequently c/2 is constructible, and the positive square root of 1−c²/4 gives sin(2π/n), since 0<2π/n<π. Repeated rotations use sums and products of these coordinates to give all vertices. We have proved: a regular n-gon, n≥3, is exactly straightedge-and-compass constructible if and only if φ(n) is a power of two. This is a field-tower existence proof; for a large n it does not itself list a convenient sequence of drawings.

Classify the permitted integers

Write n=2ᵏ times a product of odd prime powers pᵉ. The formula φ(n)=φ(2ᵏ)∏pᵉ⁻¹(p−1) shows the totient is a power of two exactly when each odd-prime exponent e equals one and each p−1 is a power of two. Here φ(1)=1 handles k=0; for k≥1, φ(2ᵏ)=2ᵏ⁻¹.

If an odd prime p=2ᵐ+1 and m has an odd factor d>1, write m=ad. Then (2ᵃ)ᵈ+1 has the proper divisor 2ᵃ+1, contradicting primality. Thus m is a power of two. A prime of the form 2^(2ʳ)+1 is called a Fermat prime; the formula alone does not guarantee primality.

The equivalent classification is therefore n=2ᵏ times a product of distinct Fermat primes, allowing the empty product. Repeating an odd prime introduces an odd factor into the totient. This explains why 15 works but 9 and 25 do not.

n	φ(n)	Degree of 2cos(2π/n)	Exactly constructible?
3	2	1	Yes
5	4	2	Yes
9	6	3	No
15	8	4	Yes
17	16	8	Yes
20	8	4	Yes
25	20	10	No

For the pentagon, divide 1+ζ+ζ²+ζ³+ζ⁴=0 by ζ². Put c=ζ+ζ⁻¹ and use ζ²+ζ⁻²=c²−2 to obtain c²+c−1=0. Since c=2cos72° is positive, c=(√5−1)/2. The coordinates are cos72°=(√5−1)/4 and sin72°=√(10+2√5)/4. Verify their squared sum is one. These expressions supply an explicit positive-square-root tower, not just a permitted degree.

A real quartic root that is not constructible

Consider f(X)=X⁴−X−1. Since f(1)=−1 and f(2)=13, continuity gives a real root α between 1 and 2. Its degree over Q is four. To prove this, reduce f modulo 2 to X⁴+X+1. Neither 0 nor 1 is a root. A reducible quartic without linear factors would be a product of two irreducible quadratics. Over F₂ the only monic irreducible quadratic is X²+X+1: checking constant term one and its two possible linear coefficients proves this assertion. Its square is X⁴+X²+1, not our polynomial. The reduction is irreducible, hence the original monic integer polynomial is irreducible over Q by Gauss’s lemma.

We need a stronger obstruction than degree four. Any real number in a quadratic tower lies in a finite Galois extension of Q whose degree is a power of two. Here is a proof, allowing that larger Galois field to contain nonreal numbers. Inductively suppose a current tower field F is contained in a finite Galois field N over Q of power-of-two degree. The next step is F(√d) for some nonzero d∈F. Adjoin to N square roots of every σ(d), for σ∈Gal(N/Q), obtaining N′. Each adjunction has degree at most two, so [N′:Q] is still a power of two.

The polynomial P(X)=∏σ(X²−σ(d)) has coefficients fixed by every automorphism of N/Q, so those coefficients belong to Q. Choose a polynomial over Q whose splitting field is N. Then N′ is the splitting field of its product with P: it contains N and all the new roots, and these generate N′. Thus N′/Q is Galois in characteristic zero. Starting from N=Q completes the induction. Repeated factors in P do not affect its splitting field. Any irreducible polynomial over Q having a root in N′ splits there by normality; therefore its splitting field has degree dividing a power of two.

Now let r₁,r₂,r₃,r₄ be the four roots of f in its splitting field L. Form the three pairings

Pairing	Element of L
12 with 34	u₁=r₁r₂+r₃r₄
13 with 24	u₂=r₁r₃+r₂r₄
14 with 23	u₃=r₁r₄+r₂r₃

Write e₁,e₂,e₃,e₄ for the elementary symmetric sums of the four roots: their sum, pair-product sum, triple-product sum and product, respectively. Comparing f with X⁴−e₁X³+e₂X²−e₃X+e₄ gives e₁=e₂=0, e₃=1, e₄=−1. Expanding the three pairing expressions gives

u₁+u₂+u₃=e₂,

u₁u₂+u₁u₃+u₂u₃=e₁e₃−4e₄,

u₁u₂u₃=e₃²+e₁²e₄−4e₂e₄.

The second identity follows by expanding e₁e₃: its four all-distinct products contribute 4e₄, and the remaining twelve terms are exactly the pairing products. For the third, the eight expanded pairing terms divide into the four squared triple products and e₄ times the sum of the four root squares. Use e₃²=(sum of squared triple products)+2e₂e₄ and (sum of root squares)=e₁²−2e₂ to obtain the displayed identity.

Consequently the u values are the roots of Y³+4Y−1. This cubic has no rational root: its only possible rational roots are 1 and −1, and neither vanishes. It is irreducible, so [Q(u₁):Q]=3. Since u₁∈L, the tower law implies 3 divides [L:Q].

If α were constructible, the quadratic-tower lemma would put α in a Galois extension of power-of-two degree. That field would contain every conjugate of α and hence L. This would make [L:Q] divide a power of two, contradicting its factor three. Thus α is a real algebraic number of degree four that is not constructible. No claim about the full Galois group of the quartic is needed for this obstruction.

Apply

Write an impossibility proof with three distinct steps: identify the target algebraic number, prove its minimal degree, and apply the quadratic-tower obstruction. An equation of degree three is not enough unless irreducibility has been established.

For a positive example, explain why √(2+√2) is constructible. First adjoin √2, then take the positive square root of the resulting positive quantity. This gives an actual tower rather than merely a permitted total degree.

Compare exact and approximate goals. A numerical routine can place nine nearly equally spaced vertices to any practical tolerance. That does not contradict the obstruction to a finite exact straightedge-and-compass construction from a unit segment. State the tools and exactness requirement before claiming impossibility. Reconstruct the totient proof for n=15 and n=25 without looking at the table, explaining which odd factor makes the difference.

For a final proof repair, locate the missing inference in “α has degree four, so it can be obtained by two square roots.” Compare the proved cyclotomic subgroup chain with the cubic pairing element above. A degree counts basis elements; it does not supply the required intermediate fields. Check the three pairing identities by expanding them before using the cubic obstruction.

Continue

Finite fields next provide another family of algebraic extensions. Their arithmetic is exact and bounded, making them useful for testing quotient calculations while still requiring proofs of irreducibility and field structure.

The general symmetric-polynomial theorem and Newton identities explain why symmetric root expressions can be written in coefficients. Keep the explicit pairing expansions above as a worked application of that general result.