Concept

Degree of an extension

The dimension of one field regarded as a space of vectors over a smaller one inside it.

Named by 5 essays across one field — each of them below, with the objects they name alongside it.

dim 12ℚ(√2)dim 22ℚ(√2, √3)dim 41√2√3√61√2√3√61√2√3√6√22√62√3√3√633√2√62√33√262 square roots taken, one at a time, and the degree doubles at each: 1 → 2 → 4the 4×4 table is the closure check — every product of basis elements landed on a whole-numbermultiple of another

Every step is a square root

A line meets a line by solving a linear equation and a circle by solving a quadratic one. There is no third case, so the numbers a construction reaches can only ever double in complexity — and a doubling is a thing that can be counted.

computation · constructible numbers
x³ − 2candidatevalue thereroot?-2-10-1-31-126x³ − 2 has no rational root — all 4 candidates the theorem allows weretested and none is zeroa cubic with no rational root is irreducible over ℚ, so its roots have degree3

The cube that will not double

Doubling a cube needs an edge in the ratio of the cube root of two. That number satisfies an equation of degree three, three does not divide any power of two, and the oldest open problem in geometry closes in a line.

computation · constructible numbers
3 trisect · 14 do notcos θθrational root8/8y = 17/829°none6/841°none5/851°none4/860°none3/868°none2/876°none1/883°none0/890°y = 0-1/897°none-2/8104°none-3/8112°none-4/8120°none-5/8129°none-6/8139°none-7/8151°none-8/8180°y = -1cos θ = k/8 for k from 8 down to −8: 3 of 17 angles trisecteach verdict is the rational root theorem run to the end — 18–28 candidates per cubic, every onedivided out

The angle that will not divide by three

Halving an angle costs one circle. Cutting it in three means solving a cubic, and for sixty degrees that cubic has no rational root — but plenty of angles do trisect, and which ones is a question with a countable answer.

computation · constructible numbers
3 · 4 · 5 · 6 · 8 · 10 · 12 · 15 · 16 · 17 · 20 · 24 · …3φ24φ25φ46φ27φ68φ49φ610φ411φ1012φ413φ1214φ615φ816φ817φ1618φ619φ1820φ821φ1222φ1023φ2224φ825φ2026φ1227φ1828φ1229φ2830φ831φ3032φ1633φ2034φ1635φ2436φ1237φ3638φ1839φ2440φ1641φ4042φ1243φ4244φ2045φ2446φ2247φ4648φ1649φ4250φ2051φ3252φ2453φ5254φ1855φ4056φ2457φ3658φ2859φ5860φ1661φ6062φ3063φ3664φ3265φ4866φ2067φ6668φ3269φ4470φ2471φ7072φ2473φ7274φ3675φ4076φ3677φ6078φ2479φ7880φ3281φ5482φ4083φ8284φ2485φ6486φ4287φ5688φ4089φ8890φ2491φ7292φ4493φ6094φ4695φ7296φ3297φ9698φ4299φ60100φ40n = 3 to 100: 24 constructible, 74 notdecided twice — by the Fermat-prime criterion and by φ(n) being a power of two — and thetwo agreed at every one of the 98

Which polygons can be drawn

Three sides yes, seven no, seventeen yes. The list of constructible regular polygons is neither everything nor almost nothing, and the pattern in it is a fact about which numbers are one less than a power of two.

computation · constructible numbers
π = 3.141592654…coefficients from −5 to 5degreeclosest missvalue there1−x + 30.142110 tried2−2x² + 5x + 40.03121,210 tried3−x³ + 2x² + 2x + 50.016113,310 tried4−2x⁴ + 5x³ + 5x² − 4x + 30.00515146,410 tried161,040 integer polynomials of degree ≤ 4 with coefficients in [−5, 5], evaluated at π —none is zerothe same search finds x² − 2 for √2, so its silence about π is a report and not a proof

The circle that will not square

The other three impossibilities are a number having the wrong degree. This one is a number having no degree at all — and that is a claim no finite search can establish, which makes it the one place in this field where the picture has to admit what it is not doing.

computation · constructible numbers

Named alongside it

The objects these essays reach for when they reach for this one.

Constructible numberStraightedge and compassRational root theoremMinimal polynomialIrreducible polynomialOperation setAlgebraic numberBasisClosureCosineCyclic groupDoubling the cube

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