A tower whose degrees multiply

Adjoining $\sqrt2$ to the rational numbers produces the set of all $a + b\sqrt2$ with $a$ and $b$ rational. Checking that this is closed under multiplication takes a line — $(a+b\sqrt2)(c+d\sqrt2) = (ac+2bd) + (ad+bc)\sqrt2$ — and checking it is closed under division takes two. So it is a field, and it sits inside the real numbers containing the rationals.

The productive way to look at it is not as a set of numbers but as a vector space over the rationals. It has a basis, $\{1, \sqrt2\}$; every element is a rational combination of those two; and so it has a dimension, which is two.

That dimension is called the degree of the extension. Nothing has been added by the word except the license to use linear algebra on a question about numbers, and that license turns out to settle problems that resisted for two thousand years.

Why the degrees multiply #

The whole method rests on one theorem, and its proof is a basis count.

The tower law. If $F \subseteq K \subseteq L$ are fields, then $[L:F] = [L:K]\,[K:F]$, where $[\,\cdot : \cdot\,]$ denotes the degree.

Take a basis $u_1, \dots, u_m$ of $K$ over $F$ and a basis $v_1, \dots, v_n$ of $L$ over $K$. The claim is that the $mn$ products $u_i v_j$ form a basis of $L$ over $F$.

They span: any element of $L$ is a $K$-combination of the $v_j$, and each $K$-coefficient is an $F$-combination of the $u_i$; substituting gives an $F$-combination of the products. They are independent: a vanishing $F$-combination $\sum_{i,j} c_{ij} u_i v_j = 0$ can be grouped as $\sum_j \bigl(\sum_i c_{ij} u_i\bigr) v_j = 0$, and independence of the $v_j$ over $K$ forces each inner bracket to vanish, and independence of the $u_i$ over $F$ then forces every $c_{ij}$ to be zero.

So the degree of a tower is the product of the degrees of its steps, and — the consequence everything else uses — the degree of each step divides the degree of the whole.

Degree, and the polynomial that produces it #

Where does a number’s degree come from? From the smallest polynomial it satisfies.

If $\alpha$ satisfies a polynomial with coefficients in $F$, the one of least degree it satisfies — normalised to be monic — is its minimal polynomial, and it is irreducible: a factorisation would give a polynomial of smaller degree with $\alpha$ as a root of one factor. The extension $F(\alpha)$ then has degree equal to that polynomial’s degree, with basis $1, \alpha, \alpha^2, \dots, \alpha^{d-1}$, because any higher power can be rewritten using the polynomial and no lower combination can vanish.

For $\sqrt2$ the minimal polynomial is $x^2 - 2$ and the degree is two. For the cube root of two it is $x^3 - 2$ and the degree is three, once irreducibility has been established — which for a cubic is the same as having no rational root, and the rational root theorem leaves only four candidates to test.

Degree three is a small number and it is about to do a great deal of work.

Three impossibilities, one argument #

Every construction with straightedge and compasses starts from two given points and produces new ones as intersections of lines and circles drawn through points already built.

Coordinates of an intersection are found by solving a linear or a quadratic equation with coefficients in the field already reached. So each step multiplies the degree by one or two, and after finitely many steps the degree of everything constructible is a power of two.

That single sentence disposes of three classical problems.

Doubling the cube asks for a segment of length $\sqrt[3]{2}$, whose degree is three. Three does not divide any power of two, so the cube cannot be doubled.

Trisecting an angle asks, for a general angle, for a root of $4x^3 - 3x = \cos\theta$, which for $\theta = 60°$ is a cubic with no rational root, hence degree three. So the angle cannot be trisected.

Squaring the circle asks for a segment of length $\sqrt\pi$. If it were constructible, $\pi$ would be algebraic — a root of some polynomial with rational coefficients — and Lindemann proved in 1882 that it is not. So the circle cannot be squared, on much deeper grounds than the other two.

The first two impossibilities are elementary once the degree argument is available, and neither was settled until Wantzel wrote it down in 1837. That is a gap of two millennia between the question and an argument that fits on a page, and the reason for the gap is that the argument needs the right object: not a cleverer construction, but a number attached to a field.

What the argument does not say #

A power of two is necessary, not sufficient. Degree four does not guarantee constructibility; what is needed is a tower of quadratic steps, which is a statement about the whole splitting field rather than about one number’s degree. The distinction matters: there are degree-four numbers that cannot be constructed, and telling which is which needs Galois theory.

The regular polygon case shows what the extra condition costs. A regular $n$-gon is constructible exactly when $\varphi(n)$ is a power of two, and the proof that this is sufficient is Gauss’s chain of quadratic steps rather than a degree count alone. Which polygons can be drawn is the essay for that.

Adding tools changes the answer. With a marked ruler, or with paper folding, cubics become solvable and the cube can be doubled. The impossibility is a statement about a particular toolset, and the degree argument is exactly the accounting of what that toolset can reach.

Reading a degree off a picture #

The abstraction earns its keep by being computable, and there are three ways to get at a degree without solving anything.

Count a basis. If a set of elements spans the extension and is independent, its size is the degree. That is what the multiplication tables above verify: eight products, all landing in the span of the same eight, so the dimension is eight and not more.

Find the minimal polynomial. Its degree is the degree of the extension the element generates, provided it is irreducible — and the practical procedure is to write down a polynomial the element obviously satisfies and then check it does not factor.

Use the tower. If $\alpha$ has degree three and $\beta$ has degree two, then $\mathbb{Q}(\alpha, \beta)$ has degree divisible by both, hence by six, and at most six; so it is exactly six. Divisibility from below and multiplicativity from above frequently pin a degree down with no computation at all.

The third method is the one that does the work in practice, and it is worth an example. Is $\sqrt2 + \sqrt[3]{2}$ of degree six? It lies in $\mathbb{Q}(\sqrt2, \sqrt[3]{2})$, which has degree six by the tower law; so its degree divides six. It is not rational, not of degree two — the cube root would then lie in a quadratic field, whose degree three does not divide two — and not of degree three by the same argument applied to the square root. Six is what is left. No polynomial was ever written down.

Building a tower on purpose #

Extensions are not only obstacles; they are the standard way of producing a field with a property one wants.

To make a field in which a chosen irreducible polynomial has a root, take the polynomials over $F$ and work with them modulo that polynomial: the result is a field of degree equal to the polynomial’s, and the class of $x$ is a root. Nothing needs to be found in advance — the root is manufactured, and Kronecker’s construction is how the field with four elements is built out of the field with two.

Iterating gives a field in which the polynomial splits completely, and the tower law measures the cost: at most $d!$ for a polynomial of degree $d$, since each root adjoined reduces the remaining degree by at least one. That splitting field is the object Galois theory is about, and its degree is the size of the group of symmetries that permute the roots — which is where the classification of solvable equations comes from.

The finite case is particularly tidy. Every finite field has $p^k$ elements for a prime $p$, and is an extension of degree $k$ of the field with $p$ elements; the subfields correspond exactly to the divisors of $k$, which is the tower law with nothing left over. The multiplicative structure being cyclic is the other half of the story, and between them the two facts describe finite fields completely.

Where the idea came from #

The vocabulary is late nineteenth century and the ideas are earlier and scattered.

Wantzel’s 1837 paper is the first place the degree argument appears, and it settles doubling the cube and trisecting the angle in a few pages using explicit manipulations of polynomials — no fields, no vector spaces, no word for what he was counting. His result was so little noticed that both problems continued to attract attempted solutions for another century, and his name is attached to nothing.

The abstraction came from Dedekind, who introduced the word field in the 1870s while editing Dirichlet’s lectures, and from Kronecker, whose construction of a root by working modulo an irreducible polynomial removed the need for the root to exist anywhere in advance. Steinitz’s 1910 paper set out the theory of fields as it is now taught, including the tower law and the classification of extensions, and it is the point at which the subject stopped being a collection of techniques about particular numbers.

What changed between Wantzel and Steinitz is not the mathematics but what counts as an object. Wantzel could compute the degree of a number; Steinitz could speak of the degree of an extension, prove it multiplies, and thereby get the divisibility argument in one line instead of a page of manipulation. That is the return on an abstraction, and it is why the same three impossibilities are now an exercise rather than a research problem.

Where it fails, and what it costs #

Not every extension has finite degree. The real numbers over the rationals have infinite degree, and the degree of a transcendental element is not a number at all — the powers $1, \pi, \pi^2, \dots$ are independent, so no finite basis exists. Everything above applies to algebraic extensions and says nothing otherwise.

Degree is coarse. Two numbers of the same degree can behave completely differently, and the degree alone does not determine the field: $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt3)$ both have degree two and are different fields, neither containing the other.

Irreducibility must be proved, not assumed. The whole method rests on the minimal polynomial being the right one, and establishing that a polynomial does not factor is often the hardest step. For a cubic it is the rational root test, because a cubic can only factor by shedding a linear piece; a quartic can factor into two quadratics with no rational root anywhere, so the test fails there and something else is needed. Beyond that it takes Eisenstein’s criterion, reduction modulo a prime, or genuine work — and a degree claimed on the strength of an unproved irreducibility is a degree that may be too large, which would make an impossibility argument prove nothing at all.

What the pictures cannot show #

The multiplication tables here are the closure check, and closure is what dimension means operationally. But a table is a finite object and a field is not: what the eight-by-eight table shows is that the eight basis elements multiply back into the span, which is the content, and the infinitely many elements of the field are all rational combinations that the table never displays.

Nothing here can show irreducibility. The rational root figure tests four candidates and finds none is a root, which settles the cubic case because a cubic factors only by having a linear factor. For higher degrees no finite table of candidates exists and the figures cannot help.

And the transcendence figure is the clearest case of a picture reporting a search rather than establishing a fact. A hundred and sixty thousand polynomials fail to vanish at $\pi$; that is evidence of nothing at all, since a polynomial with larger coefficients might. Lindemann’s theorem is not drawable, and the figure says so in its own caption.

The ladder from here #

Below: what two points can build, which is the toolset the degree argument accounts for, and every step is a square root, which is the doubling made concrete one construction at a time. Sideways: the three classical impossibilities that the tower law settles, and the field with four elements, where an extension is built rather than found. Above: Galois theory, where the tower of fields is matched against a tower of groups and the correspondence explains which equations can be solved by radicals.

What is worth carrying away #

The move is to replace a question about numbers with a question about dimensions, and it is worth extracting the general form because it recurs everywhere.

An extension of fields is a vector space, so it has a dimension. Dimensions of nested spaces multiply. Multiplication turns into divisibility, and divisibility is decidable by inspection. So a question of the form can this be reached from that? becomes a question of the form does this number divide that one?, and the second kind is answerable.

What made it possible was noticing that a field containing another is more than a set: it carries a linear structure over the smaller field, and that structure was invisible for as long as the numbers were regarded one at a time. The general lesson is the reliable one in algebra — when a problem resists, look for the structure the objects carry that nobody has been using.