Two fields that every prime confuses

How a polynomial breaks modulo the primes found that the way a polynomial factors modulo a prime is a census of its symmetries. Reduce $x^3 - 2$ modulo every prime up to twenty thousand and it splits into three linear pieces a sixth of the time, into a line and a quadratic half the time, and stays whole a third of the time — the shares of the identity, the flips and the rotations among the six symmetries of a triangle. That essay ended on a remark: different polynomials can have the same census without being the same, because what the census records is a list of proportions, and different groups can share their proportions.

This essay is about a much stronger coincidence. Here are two polynomials that do not merely agree in proportion. They agree at every single prime: reduced modulo $p$ and factored, $x^8 - 3$ and $x^8 - 48$ break into pieces of exactly the same degrees, prime after prime, with no exception anywhere among the primes that divide neither 2 nor 3. And the two fields they define, $\mathbb{Q}(\sqrt[8]{3})$ and $\mathbb{Q}(\sqrt[8]{48})$, are nevertheless different fields — not the same field written twice, not related by any renaming of their numbers.

Factoring without finding a root #

The hero is drawn from an exact computation, and the method is worth stating because it is not a search. To factor a polynomial $f$ of degree eight modulo a prime $p$, there is no need to find its roots. The numbers in the field of $p^i$ elements are exactly the roots of $x^{p^i} - x$, so the greatest common divisor of $f$ with $x^{p^i} - x$ collects precisely those irreducible factors of $f$ whose degree divides $i$. Compute it for $i = 1$, divide those factors out, compute it for $i = 2$, divide those out, and so on. The powers $x^{p^i}$ are taken modulo $f$ by repeated squaring, so each step costs a few dozen multiplications of polynomials of degree below eight, whatever the size of $p$. This is distinct-degree factorisation, and it reads off the degrees of the factors without ever writing a factor down.

Run on both polynomials at each of the 9,590 primes below 100,000 other than 2 and 3, it returns two lists, and the lists agree every time. Modulo 5 both polynomials are irreducible; modulo 7 both split into four quadratics; modulo 11 both have two roots and three quadratic factors; modulo 17 both are irreducible again; modulo 73 both split into two quartics. The hero shows the first 46 columns, and the remaining nine and a half thousand look the same.

The primes 2 and 3 are left out for a reason. The discriminant of $x^8 - a$ is, up to sign, $8^8 a^7$, so modulo 2 and modulo a prime dividing $a$ the polynomial has repeated roots and its factorisation no longer reflects anything about the field. Modulo 2 both polynomials are $x^8$, and modulo 3 both are $x^8$ again, since 48 is a multiple of 3; agreement there is cheap. Everywhere else agreement is a fact about the fields, and it is the fact this essay is about.

Eight roots and thirty-two symmetries #

The census itself is the one the polynomial modulo the primes taught to expect, and computing it is the first step in seeing why two polynomials might share it. The roots of $x^8 - 3$ in the complex numbers are $\zeta^k \alpha$ for $k = 0, 1, \ldots, 7$, where $\alpha = \sqrt[8]{3}$ is the real positive root and $\zeta = e^{2\pi i/8}$ is a primitive eighth root of unity, a vertex of the regular octagon the equation $x^8 = 1$ forces. The field generated by all eight is $\mathbb{Q}(\alpha, \zeta)$, which has degree 32 over the rationals: degree 8 from $\alpha$, as the tower whose degrees multiply would put it, times degree 4 from $\zeta$, which brings in $i$ and $\sqrt 2$.

A symmetry of that field is decided by where it sends $\zeta$ and $\alpha$. It must send $\zeta$ to another primitive eighth root of unity, $\zeta^u$ with $u$ one of 1, 3, 5, 7, and $\alpha$ to another root, $\zeta^b \alpha$ with $b$ anything from 0 to 7. Thirty-two choices, and all of them occur. On the roots, the symmetry $(u, b)$ sends $\zeta^k\alpha$ to $\zeta^{uk}\zeta^b\alpha$, so in terms of the index it is the map $k \mapsto uk + b$, computed modulo 8. The Galois group is the group of these affine maps of the integers modulo 8, and permuting eight roots it falls into six cycle types: the identity alone, two maps that swap two pairs and fix four, eight that fix two and swap three pairs, five that swap four pairs, eight that make two four-cycles, and eight that make a single eight-cycle.

Chebotarev’s theorem says each cycle type turns up among the primes in proportion to how many symmetries have it, and the figure confirms it to within an eighth of a percentage point: a 32nd of the primes split completely, two 32nds split into four roots and two quadratics, and so on. Of the twenty-two ways to write 8 as a sum, the six cycle types are the only ones that ever occur — the same kind of exclusion that let twenty-seven million cubics be sorted by their symmetry — and a pattern such as $1 + 7$ or $3 + 5$ is never seen at any prime, because no affine map of the integers mod 8 has those cycles.

Now the second polynomial. Its roots are $\sqrt2\,\zeta^k\alpha$, since $48 = 16 \cdot 3$ and $16^{1/8} = \sqrt 2$. They lie in the same field — $\sqrt 2 = \zeta + \zeta^7$ is already there — so the same 32 symmetries permute them, and the census of $x^8 - 48$ is a census of the same group. That much explains why the two sets of bars look alike in proportion. It does not explain why they agree prime by prime, which is a much stronger statement.

The same cycles from different maps #

What a prime sees is a single symmetry. For each prime $p$ that does not divide the discriminant, the field has a Frobenius element: a symmetry, defined up to conjugation, that acts on the roots the way raising to the $p$-th power acts on their reductions modulo $p$. The cycles of the Frobenius on the eight roots are exactly the irreducible factors modulo $p$ — a cycle of length $d$ is a factor of degree $d$. So the factorisations of $x^8 - 3$ and $x^8 - 48$ modulo $p$ are the cycle types of one and the same symmetry acting on two different sets of eight roots.

How does one symmetry act on the second set? It sends $\sqrt 2$ to $\zeta^u + \zeta^{-u}$, which is $\sqrt 2$ when $u$ is 1 or 7 and $-\sqrt 2$ when $u$ is 3 or 5. And $-1 = \zeta^4$. So on the roots $\sqrt 2\,\zeta^k\alpha$ the symmetry $(u, b)$ acts by $k \mapsto uk + b$ when $u$ is 1 or 7, and by $k \mapsto uk + b + 4$ when $u$ is 3 or 5. The two actions are the same affine maps, except that half of them are shifted by four.

The figure draws one case. The symmetry that sends $\zeta$ to $\zeta^3$ and fixes $\alpha$ is $k \mapsto 3k$ on the left: it fixes 0 and 4 and swaps 1 with 3, 2 with 6 and 5 with 7. On the right it is $k \mapsto 3k + 4$: it fixes 2 and 6 and swaps 0 with 4, 1 with 7 and 3 with 5. Different maps, the same cycle lengths. A prime whose Frobenius is this symmetry factors both polynomials as two linear pieces and three quadratics, and it has no way to see that the two linear pieces come from different roots.

Thirty-two coincidences that are one theorem #

One coincidence could be luck. The grid checks all 32. In every cell the cycle type on the roots of $x^8 - 3$ equals the cycle type on the roots of $x^8 - 48$, and since every prime’s Frobenius is one of these 32 symmetries, that is exactly the statement that the two polynomials factor alike modulo every unramified prime. The 9,590 primes of the hero are not evidence for it; they are 9,590 instances of a fact about 32 permutations, which the grid proves outright.

The agreement has a name. Inside the Galois group $G$, the symmetries fixing $\alpha$ form a subgroup $H$ of order four — the maps $k \mapsto uk$ — and the field $\mathbb{Q}(\alpha)$ is the part of the big field fixed by $H$. The symmetries fixing $\sqrt2\,\alpha$ form another subgroup $H'$ of order four: $k \mapsto k$, $k \mapsto 7k$, $k \mapsto 3k + 4$ and $k \mapsto 5k + 4$. Fritz Gassmann observed in 1926 that two fixed fields have the same splitting at every prime exactly when $H$ and $H'$ meet every conjugacy class of $G$ in the same number of elements. Count, for each symmetry, how many of its conjugates lie in $H$, and how many in $H'$: here the counts agree class by class, and the agreement is equivalent to the agreement of cycle types in the grid.

Yet $H$ and $H'$ are not conjugate, and that is what makes the fields different. Conjugate subgroups fix conjugate fields, which are the same field up to renaming. The grid shows the obstruction directly: the symmetry $k \mapsto 7k$ belongs to $H'$ and fixes roots 0 and 4 on the left, while $k \mapsto 3k + 4$, also in $H'$, fixes roots 2 and 6 there. No root on the left is fixed by every element of $H'$, so $H'$ is not the stabiliser of any root of $x^8 - 3$, and no relabelling of the eight roots carries one action onto the other. Six cells of the grid show the two actions fixing different roots, and those six cells are where the fields part.

The prime that tells them apart after all #

The group argument says the fields differ. It is worth seeing the difference in arithmetic, because the way it shows itself is instructive: no prime distinguishes the fields by how it factors, but a prime can distinguish them by what it forbids.

If the two fields were the same, $\mathbb{Q}(\alpha)$ would contain a root $\gamma$ of $x^8 - 48$, and then $\gamma/\alpha$ would be a number in $\mathbb{Q}(\alpha)$ whose eighth power is 16. The eighth roots of 16 are $\pm\sqrt2$, $\pm\sqrt{-2}$ and $\pm1 \pm i$. The field $\mathbb{Q}(\alpha)$ sits inside the real numbers, so it contains no $\sqrt{-2}$ and no $i$, and the only candidate is $\sqrt 2$. So the question is whether $\sqrt 2$ lies in $\mathbb{Q}(\sqrt[8]{3})$.

Here a single prime settles it. Modulo 11, $x^8 - 3$ has the roots 2 and 9 — $2^8 = 256 = 23 \cdot 11 + 3$ — and a root modulo an unramified prime is a way of mapping the integers of $\mathbb{Q}(\alpha)$ onto the integers modulo 11, sending $\alpha$ to 2. Every number in the field that is an algebraic integer, $\sqrt 2$ included if it were there, would land on something whose square is 2 modulo 11. But 11 leaves 3 on division by 8, and for such primes 2 is not a square: the squares modulo 11 are 1, 4, 9, 5 and 3, the even powers of a primitive root, and 2 is an odd power. So $\sqrt 2$ is not in $\mathbb{Q}(\sqrt[8]3)$, the field contains no root of $x^8 - 48$, and the two fields are different. The table lists the next seven primes that do the same job, 59, 83, 107 and onward; at each of them $x^8 - 48$ also has exactly two roots, as the agreement demands. The prime does not see a difference in the factorisation. It sees a square root that one field would need and the residues refuse to supply.

The smallest instance is the Fano plane #

The eighth-root fields are not the smallest example. Robert Perlis proved in 1977 that two fields of degree six or less with the same splitting at every prime are always the same field, and that fields sharing their splitting share a great deal more — their degree, their discriminant, their Galois closure, their numbers of real and complex embeddings. The first degree at which the coincidence can happen is seven, and the group that makes it happen there is the most famous small group in combinatorics.

The Fano plane has seven points and seven lines, three points on every line and three lines through every point, and it is the projective plane over the field of two elements: its points are the seven non-zero vectors of $\mathbb{F}_2^3$ over the two-element field that the field with four elements is built on, and its lines are the planes through the origin. Its symmetries are the 168 invertible $3 \times 3$ matrices over $\mathbb{F}_2$, enumerated in the figure. Each permutes the seven points and, at the same time, the seven lines, and every one of the 168 has the same cycle type on the lines as on the points. The identity fixes seven of each. A collineation of order two fixes three points and three lines; one of order four, like the one drawn, fixes one point and one line; one of order three fixes one of each; one of order seven fixes none.

Yet points and lines are not the same thing to the group. A matrix that fixes a point fixes the line orthogonal to it under a different matrix — the inverse transpose — and no relabelling of points as lines is compatible with every collineation at once. So the stabiliser of a point and the stabiliser of a line, two subgroups of order 24, meet every conjugacy class equally and are not conjugate: Gassmann’s condition, in the smallest group that satisfies it. The fact that a symmetry of a finite plane fixes as many lines as points, which a student of projective geometry meets as a counting exercise, is the same fact as the existence of two different number fields of degree seven with identical primes.

A polynomial with the plane’s census #

The plane needs a polynomial to become a field, and one is known. Wolfgang Trinks found in 1968 that $x^7 - 7x + 3$, whose discriminant is $3^8 \cdot 7^8$, has the group of the Fano plane as its Galois group, acting on its seven roots the way the collineations act on the points. The figure factors it modulo the 9,590 primes below 100,000 other than 3 and 7 and sorts them, and the shares sit on the plane’s proportions within a third of a percentage point: a 168th split completely, 21 out of 168 into three roots and two quadratics, 42 into a root, a quadratic and a quartic, 56 into a root and two cubics, 48 irreducible. Only five of the fifteen ways to split seven ever occur.

Inside the splitting field of this polynomial sits a second field of degree seven, the field fixed by the stabiliser of a line rather than a point. It is generated by a root of another polynomial of degree seven, and that polynomial factors modulo every unramified prime exactly as $x^7 - 7x + 3$ does, by the argument of the grid, with points replaced by lines. The census of the figure is the census of both, prime by prime, and nothing in the factorisations will ever tell them apart.

There is a connection here nobody went looking for. In 1985 Toshikazu Sunada saw that Gassmann’s condition makes sense for any space a group acts on, not only for number fields: two quotients of one space by Gassmann-equivalent subgroups have the same spectrum, for the same reason the two fields have the same primes. Carolyn Gordon, David Webb and Scott Wolpert used exactly this — with the group of the Fano plane — to build the two differently shaped drums with identical frequencies that two graphs the eigenvalues cannot tell apart described. The drums that cannot be heard apart and the fields that cannot be told apart by primes are the same pair of subgroups, discovered first in number theory and sixty years later in geometry.

What a census can and cannot carry #

What fields with identical primes share is everything their zeta function records. The Dedekind zeta function of a field is built prime by prime from exactly the factorisation data in the hero, so two fields with the same factorisations have the same zeta function, and conversely. The zeta function knows the degree and the discriminant, and Perlis’s theorem says it knows the Galois closure: $\mathbb{Q}(\sqrt[8]3)$ and $\mathbb{Q}(\sqrt[8]{48})$ have the same splitting field, as they must. It also knows the class number and the regulator multiplied together, through the formula units that form a lattice ended on.

It does not know them separately. Bart de Smit and Perlis showed in 1994 that two fields with the same zeta function can have different class numbers, so that one factorises uniquely and the other does not, with the difference made up exactly by the regulators. Whether unique factorisation holds in a field is therefore not a property that its primes, counted with their splittings, can decide — a strange thing to learn about the one property most often motivated by asking how primes behave.

The computation shows only part of this. It checks agreement at the 9,590 primes below 100,000 and it checks Gassmann’s condition on all 32 symmetries, which is what turns the finite check into a statement about every prime; it does not verify Chebotarev’s theorem, which is why the bars sit near the ticks and not exactly on them. It finds the obstruction at 11 by trying all eleven residues, which is a proof, but the reduction from an isomorphism to $\sqrt2$ is an argument and not a calculation. And it says nothing about how often such pairs occur. Nothing in these pictures shows whether arithmetically equivalent fields are rare or common among fields of degree seven or eight, because the pictures exhibit pairs rather than count them.

Still open: when the census is the whole story #

For a few families the census decides the field. Perlis’s theorem settles every degree up to six, and there are groups of every size for which no Gassmann pair exists, so that a field with that group is determined by its primes. Which groups admit Gassmann pairs is a question in finite group theory with a large literature and no complete answer; the smallest examples are known, and so are infinite families, but a description of all of them is not.

The arithmetic side is less settled still. Given a number field, how many primes does it take to pin down its zeta function — to rule out every field that agrees with it on those primes and differs elsewhere? Under the generalised Riemann hypothesis the answer is a bound polynomial in the logarithm of the discriminant; unconditionally, the known bounds are far weaker, and they are the same bounds that limit how soon the census becomes fair. And the class number question has its own frontier: de Smit and Perlis showed the class numbers can differ, but how much they can differ — whether their ratio is bounded for arithmetically equivalent fields of a given degree, and by what — is known only in special cases. Two fields that every prime confuses can still disagree about the one thing number theory was invented to ask them.

Identical at every prime, different as fields #

$x^8 - 3$ and $x^8 - 48$ factor into pieces of the same degrees modulo every prime not dividing 6, because one group of 32 symmetries acts on their two sets of roots with the same cycle type element by element, and the fields $\mathbb{Q}(\sqrt[8]3)$ and $\mathbb{Q}(\sqrt[8]{48})$ are still different, because the two actions fix different roots. Gassmann’s condition — two subgroups meeting every conjugacy class equally without being conjugate — is the whole explanation, and in the smallest case it is the statement that a symmetry of the Fano plane fixes as many lines as points. The prime 11 cannot tell the two fields apart by factoring, and tells them apart anyway, by refusing 2 a square root. The same pair of subgroups, moved into geometry by Sunada, built the two drums nobody can hear apart.