Seven powers in a space of six

A number is algebraic when it is a root of a non-zero polynomial with whole-number coefficients. $\sqrt 2$ is, as a root of $x^2 - 2$; $\sqrt[3]{3}$ is, as a root of $x^3 - 3$. The question that sounds as if it should be easy is whether their sum is.

It is not obviously easy. The polynomials for $\sqrt 2$ and $\sqrt[3]{3}$ give no hint of a polynomial for $\sqrt 2 + \sqrt[3]{3}$, and trying to eliminate the radicals by hand — cube $x - \sqrt 2$, collect the terms with $\sqrt 2$ on one side, square — works for this pair and gives no method for the next. What settles the question for every pair at once is not algebra with radicals but the dimension of a field, and the same argument that proves the sum is algebraic also hands over its polynomial.

A field of dimension six #

The tower law says that adjoining a root of degree $m$ and then a root of degree $n$ gives a field of dimension at most $mn$ over the rationals. For $\sqrt 2$, of degree two, and $\sqrt[3]{3}$, of degree three, the field $\mathbb{Q}(\sqrt 2, \sqrt[3]{3})$ has dimension exactly six, since the dimension must be divisible by both two and three.

A basis is easy to write down: every product of a power of $\sqrt 2$ below two with a power of $\sqrt[3]{3}$ below three,

$$1,\quad \sqrt[3]{3},\quad \sqrt[3]{9},\quad \sqrt2,\quad \sqrt2\sqrt[3]{3},\quad \sqrt2\sqrt[3]{9}.$$

Every element of the field is a rational combination of those six, and multiplying two of them is bookkeeping: $\sqrt 2 \cdot \sqrt 2 = 2$, $\sqrt[3]{3} \cdot \sqrt[3]{9} = 3$, and the rest just combine exponents.

Seven vectors in a space of six #

Now take any element $\alpha$ of that field and list its powers: $1, \alpha, \alpha^2, \dots, \alpha^6$. Each is an element of the field, so each is a vector with six rational coordinates. Seven vectors in a six-dimensional space cannot be linearly independent. There are rational numbers $c_0, \dots, c_6$, not all nought, with

$$c_0 + c_1\alpha + c_2\alpha^2 + \cdots + c_6\alpha^6 = 0.$$

That is a polynomial with rational coefficients — clear the denominators and whole-number coefficients — of which $\alpha$ is a root. So $\alpha$ is algebraic, and its degree is at most six. The argument used nothing about $\alpha$ except that it lives in a six-dimensional field.

The polynomial comes out of an exact linear solve and it is worth reading. The coefficient of $\alpha^5$ is nought, which says the six roots of this polynomial add to nought — and they do, as the section on conjugates below shows. The constant term is 1, which says the six roots multiply to 1. And every coefficient is a whole number with leading coefficient 1, a fact that turns out to matter and is the subject of the integers a field contains.

This is not the method the existence proof suggests — it is the existence proof, run. There is no step where a polynomial is guessed or a radical is cleverly eliminated. The dependency among the powers is found by Gaussian elimination on a seven-by-six table of whole numbers, and the first power that depends on the earlier ones gives the minimal polynomial.

The rank climbs to the degree and stops #

The table above has a further structure that the argument predicts and the computation displays.

The rank climbs by one each time until some power first depends on the earlier ones, and then it never rises again: once $\alpha^d$ is a combination of lower powers, multiplying through by $\alpha$ shows $\alpha^{d+1}$ is too, and so on for ever. The height at which the rank stops is the degree of $\alpha$ — the dimension of the smallest field containing it — and that field sits inside the six-dimensional one, so by the tower law its dimension divides six. The possible degrees for anything in this field are 1, 2, 3 and 6, and the chart shows three of them.

Sums, products and quotients, all at once #

The argument proves more than was asked, and the general form is worth stating because it is the whole theorem.

Let $\alpha$ and $\beta$ be algebraic, of degrees $m$ and $n$. The field $\mathbb{Q}(\alpha, \beta)$ has dimension at most $mn$, by the tower law. It contains $\alpha + \beta$, $\alpha\beta$, $\alpha - \beta$ and, if $\beta \neq 0$, $\alpha/\beta$ — because it is a field. Every one of those is therefore an element of a finite-dimensional extension of the rationals, so its powers eventually become dependent, so it is algebraic, of degree at most $mn$.

The algebraic numbers form a field. No formula for the polynomial of a sum in terms of the polynomials of the summands was needed, and none would have been easy: the argument replaced a question about polynomials with a question about dimensions, which is exactly what the tower law was built to do.

Quotients deserve a sentence of their own, because “it is in the field” hides a small computation that the minimal polynomial makes explicit. The polynomial for $\alpha = \sqrt 2 + \sqrt[3]{3}$ says $\alpha^6 - 6\alpha^4 - 6\alpha^3 + 12\alpha^2 - 36\alpha = -1$. Factor an $\alpha$ out of the left side:

$$\alpha\,(\alpha^5 - 6\alpha^3 - 6\alpha^2 + 12\alpha - 36) = -1,$$

so $1/\alpha = -(\alpha^5 - 6\alpha^3 - 6\alpha^2 + 12\alpha - 36)$, a polynomial in $\alpha$ with whole coefficients. Dividing by an algebraic number is multiplying by a polynomial in it, and the constant term of the minimal polynomial is what makes that possible — it is non-zero, since otherwise the polynomial would have a factor of $x$ and not be minimal. The constant term being exactly $\pm1$, as it is here, means the reciprocal needs no fractions at all, which is a first sign of the integer structure the next essay is about.

The shortfalls in the table are instructive. $\sqrt[3]{3}$ and $\sqrt[3]{9}$ each have degree three, and the bound for their product is nine; the product is 3, of degree one, because the two cube roots were not independent — one is the square of the other. $\sqrt 2 + (1 - \sqrt 2)$ is 1. The bound is a bound on the size of the field the parts generate, and the degree falls short exactly when the parts generate less than the product of their separate fields. That can only be detected by computing, which is what the power table does.

Another way to the same polynomial #

There is a second route to a polynomial for $\alpha + \beta$, and comparing the two shows what each is good for.

If $p$ is a polynomial with root $\alpha$ and $q$ one with root $\beta$, then $x = \alpha + \beta$ is a value at which $p(y)$ and $q(x - y)$ share the root $y = \alpha$. Whether two polynomials share a root can be decided from their coefficients by a determinant, the resultant, and computing the resultant of $p(y)$ and $q(x - y)$ with respect to $y$ gives a polynomial in $x$ that vanishes at $\alpha + \beta$ — and at every sum of a root of $p$ with a root of $q$.

The resultant always has degree exactly $mn$. That is its strength and its weakness. It produces a polynomial by a single determinant, with no search, but when the degree of $\alpha + \beta$ is smaller than $mn$ the resultant is not the minimal polynomial: it is a power of it, or a product of it with polynomials for other sums of roots, and it has to be factored to extract the right piece. The power-table method finds the minimal polynomial directly, at the cost of working inside an explicit basis of the field. For $\sqrt[3]{3} + \sqrt[3]{9}$ the resultant has degree nine while the minimal polynomial, $x^3 - 9x - 12$, has degree three.

The six conjugates #

A polynomial with rational coefficients does not know which of its roots it was written for. $\sqrt 2 + \sqrt[3]{3}$ satisfies a degree-six polynomial, and so do five other numbers, and the polynomial treats all six identically.

The six conjugates are what the polynomial actually describes: every choice of sign for $\sqrt 2$ combined with every choice of cube root for $\sqrt[3]{3}$. Any rational expression built from $\sqrt 2$ and $\sqrt[3]{3}$ that happens to be rational would give the same rational value under all six choices — which is why the sum of the six roots is nought (the $\pm\sqrt 2$ cancel, and the three cube roots of 3 add to nought) and why the coefficients of the polynomial, being symmetric functions of the roots, are rational.

The picture also explains the degree without any computation. Six genuinely different numbers are forced to be roots of any rational polynomial that $\sqrt 2 + \sqrt[3]{3}$ satisfies, because swapping $\sqrt 2$ for $-\sqrt 2$ or $\sqrt[3]{3}$ for $\omega\sqrt[3]{3}$ preserves every rational relation. So the polynomial has at least six roots and degree at least six. The two real conjugates are worth a glance: $\sqrt 2 + \sqrt[3]{3} \approx 2.856$ and $-\sqrt 2 + \sqrt[3]{3} \approx 0.028$, one near three and one almost nought. Nothing about the polynomial prefers the first, which is the number the question was about; a reader handed only the polynomial could not say which real root had been meant. That swapping is the beginning of Galois theory, where the permutations of conjugates that preserve every rational relation form a group, and the structure of that group governs which equations can be solved.

Roots of algebraic equations are algebraic #

The dimension argument has one more consequence that sounds as if it should need something new.

Take a polynomial whose coefficients are themselves algebraic numbers — $x^3 - \sqrt 2\,x + \sqrt[3]{3}$, say — and let $\gamma$ be a root. Is $\gamma$ algebraic over the rationals? The coefficients generate a finite-dimensional field $K$. Adjoining $\gamma$ to $K$ gives an extension of $K$ of dimension at most three, since $\gamma$ satisfies a cubic over $K$. The tower law makes $K(\gamma)$ finite-dimensional over the rationals. So $\gamma$ is algebraic.

The algebraic numbers are closed under taking roots of polynomials with algebraic coefficients — they are algebraically closed — and nothing beyond counting dimensions was used. The field of all algebraic numbers, written $\overline{\mathbb{Q}}$, is the smallest algebraically closed field containing the rationals. It is also countable, because there are countably many polynomials with whole-number coefficients, each with finitely many roots — which is how Cantor knew there must be transcendental numbers before anybody could name one convincingly.

Cosines that are secretly cubic #

The dimension count reaches numbers that do not look algebraic at all, and the most familiar are values of trigonometric functions.

Let $\zeta = \cos(2\pi/7) + i\sin(2\pi/7)$, a seventh root of 1. It satisfies $x^6 + x^5 + \cdots + x + 1 = 0$, so the field $\mathbb{Q}(\zeta)$ has dimension at most six. The number $\cos(2\pi/7)$ is $(\zeta + \zeta^{-1})/2$, which lies in that field, so its powers are dependent after at most six steps: $\cos(2\pi/7)$ is algebraic. Running the count shows more. The powers of $\zeta + \zeta^{-1}$ only ever involve the combinations $\zeta^k + \zeta^{-k}$, of which there are three, so the rank stops at three, and

$$8x^3 + 4x^2 - 4x - 1 = 0$$

is the minimal polynomial of $\cos(2\pi/7)$. A cubic with no rational root, in a field that cannot be reached by square roots alone, which is the whole content of the regular heptagon’s refusal to be constructed: the degree three does not divide any power of two.

The same argument makes $\cos(2\pi/n)$ algebraic for every $n$, of degree half the number of residues coprime to $n$, and it makes $\sin$ and $\tan$ of every rational multiple of $\pi$ algebraic too. None of that needs a trigonometric identity. It needs the observation that each value lives in a field generated by a root of unity, and a count of that field’s dimension.

Finding the polynomial from the digits alone #

The power table needs an explicit basis for the field, which is available for numbers built from known radicals and unavailable for a number known only as a decimal. There is a numerical twin of the method that needs nothing but digits.

Given a number $x$ to fifty decimal places, form the seven numbers $1, x, x^2, \dots, x^6$ and search for whole numbers $c_0, \dots, c_6$, not too large, with $c_0 + c_1 x + \cdots + c_6 x^6$ extremely close to nought. That is an integer relation problem, and it is solved by lattice reduction — the same shortening of a basis that produces the two squares a prime is made of, carried out in seven dimensions — or by the related PSLQ algorithm of Helaman Ferguson and David Bailey. Fed $2.8564\ldots = \sqrt 2 + \sqrt[3]{3}$ to enough digits, either returns $x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 1$.

What it cannot do is prove. A relation found numerically is a relation that holds to the precision supplied, and a number can be approximated by roots of polynomials with small coefficients far better than chance would suggest without being algebraic itself — which is the phenomenon Liouville used to build a transcendental number by making it too well approximated. The exact table and the numerical search are the same linear algebra; only the exact one is a proof, and only the numerical one can start from a number nobody knows how to build.

What the tables cannot carry #

The power table is one element. The theorem that every sum of algebraic numbers is algebraic is proved by the dimension count, which works for all of them; the table shows the count producing an actual polynomial in one case, and it cannot show that the method never fails.

The rank chart depends on an exact basis, and the basis is correct only because the field has been proved to have dimension six. If $\sqrt 2$ lay in $\mathbb{Q}(\sqrt[3]{3})$ — it does not, since two does not divide three — the six products would not be independent and the whole table would be computing in a space that does not exist. The figures assume the tower law’s conclusion; they do not check it.

And the conjugates are found by a numerical root-finder and matched to the formula $\pm\sqrt2 + \omega^j\sqrt[3]{3}$ within a tolerance. That is strong evidence that the six numbers are the roots, and the proof that they are is the observation that each makes the polynomial vanish algebraically, which a floating-point match cannot establish.

Still open: whether familiar constants are algebraic #

The dimension argument says a number is algebraic exactly when it lies in some finite-dimensional extension of the rationals, and for numbers built from radicals that is easy to recognise. For numbers defined any other way, it can be extraordinarily hard to decide.

$\pi$ and $e$ are known to be transcendental, by Lindemann and Hermite. Nobody knows whether Euler’s constant $\gamma = 0.5772\ldots$ — the limit of $1 + \tfrac12 + \cdots + \tfrac1n - \ln n$ — is algebraic. It is not even known to be irrational. Computations show that if it were a ratio of whole numbers the denominator would need hundreds of thousands of digits, and if it were a root of a low-degree polynomial the coefficients would have to be enormous; neither amounts to a proof. The power-table method needs an explicit field to work in, and $\gamma$ comes with no field — there is no finite list of numbers from which it is known to be built — so the one argument that decides algebraicity so cleanly for $\sqrt 2 + \sqrt[3]{3}$ has nothing to count.

A polynomial as a linear dependency #

The question at the start was whether $\sqrt 2 + \sqrt[3]{3}$ satisfies a polynomial, and the answer turned out to be a statement about seven vectors. The polynomial is a linear dependency among powers, the degree is a rank, and the field in which everything lives provides the dimension that forces the dependency to exist.

That translation is what makes the algebraic numbers a field without anybody writing down how to combine polynomials, and it is also what makes the answer computable: every question about a number built from finitely many radicals becomes a question in linear algebra over the rationals, answered exactly in a table of whole numbers.