Field

Computation

A fixed set of operations, and the exact question of what it can and cannot build.
01−12√3⁄2−√3⁄2two points, 1 line and 2 circles, 4 new pointseach new point was checked to lie on two of the objects drawn before it

What two points can build

A compass and a straightedge are not a craft. They are two operations on a set of points, applied over and over, and writing them that way turns "can this be drawn?" into a question with an answer.

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.

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.

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.

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.

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.

0000010100111001011101112 codewords in the 3-cube, minimum distance 3 — 1 error corrected, 2 detectedthe 2 balls of radius 1 hold 4 words each and cover all 8 exactly once: the code is perfect

Distance is a picture

A message is a corner of a cube and an error is a step along an edge. Everything a code can do is decided by how far apart the corners it uses are — and that is a fact about a drawing.

four data bits, three parity bitsdataparityweight00000000000101130010110300111014010011140101100301100013011101041000101310011104101001141011000311000103110100141110100411111117sixteen codewords of length seven, minimum distance 3 — measured over all 120pairsweights: 1×0 · 7×3 · 7×4 · 1×7, and the lightest non-zero one weighs 3

Sixteen spheres that fill a cube

A hundred and twenty-eight seven-bit words, sixteen of them chosen, and a ball of eight around each. Sixteen times eight is a hundred and twenty-eight exactly — so the balls tile the space with nothing left over, and the code wastes nothing at all.

received1011010H111010001110101101001010synsyndromethe bit it names101bit 1111bit 2110bit 3011bit 4100bit 5010bit 6001bit 7000no error1011010 gives syndrome 010, which is column 6 — so bit 6 is wrong and 1011000 is thecodewordthe syndrome is three bits and the message is four: the check finds the error without everrecovering what was sent

Finding the error without reading the message

Three parity checks on a seven-bit word produce three bits. If they are all zero nothing is wrong; otherwise they are the number of the position that broke. The message is never consulted, because the answer does not depend on it.

012345678910012345678910sentf(0) = 4f(1) = 8f(2) = 7f(3) = 1f(4) = 1f(5) = 7f(6) = 8the message 4, 1, 3 read as 3x² + 1x + 4 over GF(11), evaluated at x = 0…6all 35 ways of keeping 3 of the 7 values were interpolated and every one returned the message

A polynomial through the gaps

Write the message as the coefficients of a polynomial and send its values instead. Any k of them determine the polynomial, so it does not matter which ones are lost — and it does not matter how many, as long as k survive.

+01αα+101αα+101αα+110α+1ααα+101α+1α10×01αα+101αα+1000001αα+10αα+110α+11α× mod 40000012302020321the 4 elements of GF(4), built as polynomials in α modulo α+1 + α² — every product of two non-zero elements is non-zerobeside it the integers mod 4, where 2 × 2 = 0 and 1 element has no reciprocal

The field with four elements

The integers modulo four are not a field: two times two is zero and two has no reciprocal. There is nevertheless a field with four elements, and building it means giving up on counting as the way to make arithmetic finite.

1αα²α³α+1α²+αα³+α²α³+α+1α²+1α³+αα²+α+1α³+α²+αα³+α²+α+1α³+α²+1α³+1GF(16)*15 elementspowerelementα⁰1α¹αα²α²α³α³α⁴α+1α⁵α²+αα⁶α³+α²α⁷α³+α+1α⁸α²+1α⁹α³+αα¹⁰α²+α+1α¹¹α³+α²+αα¹²α³+α²+α+1α¹³α³+α²+1α¹⁴α³+1every non-zero element of GF(16) is a power of α — the 15 of them, once each, around the wheel8 of the 15 elements generate the whole group, which is φ(15) = 8, and the product rule waschecked on all 225 pairs

Every element is a power of one of them

Pick the right element of a finite field and its powers run through every other non-zero element exactly once before returning to one. Multiplication becomes addition of exponents, and a table of q − 1 entries replaces the whole multiplication table.

001010100110101011111point on line?0123456L0L1L2L3L4L5L6seven points, seven lines, three points on every line and three lines through every pointthe drawing was checked against the algebra by searching all 5,040 relabellings — one of them carries GF(2)³ onto thispicture

Seven points, seven lines

A geometry with seven points, in which every two points lie on exactly one line and every two lines meet in exactly one point. There are no parallels, the whole thing is built out of the two-element field, and one of its lines has to be drawn as a circle.

01234567812 triples0 1 20 3 40 5 60 7 81 3 51 4 71 6 82 3 82 4 62 5 73 6 74 5 89 points, 12 triples, each point in 4 of them — and every one of the 36 pairs appears exactly oncefound by backtracking over the pairs, which decides existence rather than assuming it

A schedule where every pair meets once

Sort n people into groups of three so that every two of them share a group exactly once. Two divisions have to come out whole, that rules out most sizes — and at every size the divisions permit, a schedule exists.

order 6, symbols 0…5012345123450234501345012450123501234transversals of the cyclic squareorder 33order 40order 515order 60order 7133the cyclic square of order 6 has no transversal at all: all 720 placements were tried and every one repeats asymbola square with no transversal cannot have an orthogonal mate, so the search for one need never begin

The thirty-six officers

Six regiments send six officers each, one of every rank. Arrange all thirty-six in a square so that each row and each column holds every rank once and every regiment once. Euler could not, guessed why, and was wrong about the reason.

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