Target Audience: Anyone who has met rings and fields and wants to see the abstraction settle problems that geometry alone could not.
Prerequisites: Rings and Fields for quotients by irreducible polynomials; Vector Spaces for bases and dimension; and Algebra, Polynomials and Complex Numbers for de Moivre and the roots of unity.
The Greeks asked for three constructions with a straightedge and compass: a cube of twice the volume of a given one, the trisection of any angle, and a square of the same area as a circle. For two thousand years nobody found them, and nobody could show they did not exist. The answer, when it came in the nineteenth century, was not geometric at all. It was a count of dimensions.
Sections 1 to 5 set up the count: a field sitting inside a bigger one is a vector space over it, and its dimension — the degree — can be read off a single irreducible polynomial. Section 6 shows how degrees multiply. Sections 7 and 8 turn constructions into degrees and settle the old problems, and Section 9 uses the same count to show that a finite field’s size is always a prime power.
The explorer in Section 4 finds the minimal polynomial of a number by exact linear algebra, then checks it two independent ways before it will report a degree.
If \(K\) is a field contained in a field \(L\), with the same addition and multiplication, then \(L\) is an extension of \(K\), written \(L/K\).
For a number \(\alpha\), \(\mathbb{Q}(\alpha)\) is the smallest field containing \(\mathbb{Q}\) and \(\alpha\): everything that can be built from rationals and \(\alpha\) by adding, subtracting, multiplying and dividing.
Sums and products of numbers \(a + b\sqrt{2}\) have the same form, since \((\sqrt{2})^2 = 2\). The only question is division, and it is answered by the conjugate:
The denominator is a non-zero rational whenever \((a, b) \neq (0, 0)\): \(a^2 = 2b^2\) with \(b \neq 0\) would make \(\sqrt{2} = |a/b|\) rational, which it is not.
If \(L/K\) is an extension, then \(L\) is a vector space over \(K\): add elements of \(L\), and multiply them by scalars from \(K\). Every vector space axiom is a field axiom of \(L\).
The degree \([L : K]\) is its dimension. The extension is finite when the degree is.
Any two bases of a vector space have the same size — the Invariance of Dimension. That page works in \(\mathbb{R}^n\), and notes that its argument uses nothing beyond adding and scaling vectors, so it holds verbatim over \(\mathbb{Q}\) or any other field. The degree therefore does not depend on which basis you find, and a single basis settles it.
A number \(\alpha\) is algebraic over \(\mathbb{Q}\) if it is a root of a non-zero polynomial with rational coefficients, and transcendental otherwise.
The minimal polynomial of an algebraic \(\alpha\) is the monic polynomial of least degree with rational coefficients having \(\alpha\) as a root. The degree of \(\alpha\) is the degree of that polynomial.
It divides every polynomial with \(\alpha\) as a root. The polynomials vanishing at \(\alpha\) form an ideal of \(\mathbb{Q}[x]\), and in \(F[x]\) every ideal is generated by its monic element of least degree. It is the argument the Minimal Polynomial page makes for matrices, now made for a number.
It is irreducible. If it factored as \(gh\) with both of lower degree, then \(g(\alpha)h(\alpha) = 0\) in a field, so one factor would vanish at \(\alpha\) — a polynomial of smaller degree, contradicting minimality.
If \(p\) is monic and irreducible with \(p(\alpha) = 0\), the minimal polynomial divides \(p\), and an irreducible polynomial has no monic divisor of positive degree except itself. So \(x^2 - 2\) is the minimal polynomial of \(\sqrt{2}\), and \(\sqrt{2}\) has degree \(2\). To find a degree, find one irreducible polynomial the number satisfies.
The algebraic numbers are countable, so uncountably many real numbers are transcendental — an existence proof that names none. Showing that a particular number is transcendental is much harder: Lindemann proved it for \(\pi\) in 1882, and that proof is beyond this page.
If \(\alpha\) is algebraic with minimal polynomial \(m\) of degree \(n\), then
with basis \(1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\).
Evaluation at \(\alpha\) is a ring homomorphism \(\mathbb{Q}[x] \to \mathbb{C}\). Its kernel is the polynomials vanishing at \(\alpha\), which is the ideal \((m)\), and its image is the set \(\mathbb{Q}[\alpha]\) of polynomial expressions in \(\alpha\). By the First Isomorphism Theorem, \(\mathbb{Q}[\alpha] \cong \mathbb{Q}[x]/(m)\), and because \(m\) is irreducible that quotient is a field. So \(\mathbb{Q}[\alpha]\) is already a field containing \(\alpha\), and it is the smallest one: \(\mathbb{Q}(\alpha) = \mathbb{Q}[\alpha]\).
For the basis: dividing any polynomial by \(m\) leaves a remainder of degree below \(n\), so \(1, \alpha, \ldots, \alpha^{n-1}\) span. They are independent, because a relation among them would be a non-zero polynomial of degree below \(n\) vanishing at \(\alpha\).
The powers \(1, \alpha, \alpha^2, \ldots\) stay linearly independent over \(\mathbb{Q}\) up to \(\alpha^{n-1}\), and \(\alpha^n\) is the first that is a rational combination of the ones before it. That dependence, rearranged, is the minimal polynomial. It is exactly how the minimal polynomial of a matrix is found, and it is what the explorer below computes.
Choose a number. The explorer computes its powers \(1, \alpha, \alpha^2, \ldots\) exactly, with rational coordinates, inside a ring where the number has an explicit form — \(\sqrt[3]{2}\) as \(x\) with \(x^3 = 2\), and a cosine through the roots of unity — and finds the first power that depends on the ones before it. Then it checks the result two independent ways: the polynomial must vanish at the number’s decimal value, and it must be irreducible, by the rational root test or an exact search for a factorisation into quadratics.
Compare \(\sqrt[3]{2}\) with \(\sqrt[4]{2}\): degree \(3\) against degree \(4\), and only one of them is a power of \(2\).
Section 3 reduced every degree calculation to one question: is this polynomial irreducible over \(\mathbb{Q}\)? For low degrees there is a finite test.
If \(\frac{p}{q}\), in lowest terms, is a root of \(a_n x^n + \cdots + a_1 x + a_0\) with integer coefficients, then \(p \mid a_0\) and \(q \mid a_n\).
Proof. Multiply the equation by \(q^n\): \(a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_0 q^n = 0\). Every term except the last is a multiple of \(p\), so \(p \mid a_0 q^n\), and since \(\gcd(p, q) = 1\), Euclid’s Lemma applied prime by prime gives \(p \mid a_0\). The same argument on the first term gives \(q \mid a_n\).
A polynomial of degree \(2\) or \(3\) that factors must have a factor of degree \(1\), and a rational linear factor means a rational root. So such a polynomial is irreducible over \(\mathbb{Q}\) exactly when none of the finitely many candidates from the Rational Root Theorem is a root.
\(x^3 - 2\): the candidates are \(\pm 1, \pm 2\), and none is a root. So \(x^3 - 2\) is irreducible and \(\sqrt[3]{2}\) has degree \(3\) — a sharper statement than its irrationality.
\(8x^3 - 6x - 1\): the candidates are \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}\), and none is a root. It is irreducible; Section 8 needs it.
\(x^4 + 4\) has no rational root — it is positive for every real \(x\) — yet it factors:
A quartic can split into two quadratics without a linear factor, so ruling that out needs its own argument. The explorer does it exactly for the quartics it meets.
If \(K \subseteq L \subseteq M\) are fields with finite degrees, then
Proof. Take a basis \(u_1, \ldots, u_m\) of \(L\) over \(K\) and a basis \(v_1, \ldots, v_n\) of \(M\) over \(L\). The \(mn\) products \(u_i v_j\) span \(M\) over \(K\): write \(z \in M\) as \(\sum_j b_j v_j\) with \(b_j \in L\), then each \(b_j\) as \(\sum_i a_{ij} u_i\) with \(a_{ij} \in K\). They are independent: if \(\sum_{i,j} a_{ij} u_i v_j = 0\), independence of the \(v_j\) over \(L\) forces each \(\sum_i a_{ij} u_i = 0\), and independence of the \(u_i\) forces every \(a_{ij} = 0\).
If \(\alpha\) lies in a finite extension \(L\) of \(\mathbb{Q}\), then \(\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq L\), so the degree of \(\alpha\) divides \([L : \mathbb{Q}]\). For example, \(\sqrt{2}\) is not in \(\mathbb{Q}(\sqrt[3]{2})\): its degree \(2\) does not divide \(3\). No computation with cube roots is needed.
First, \(\sqrt{3} \notin \mathbb{Q}(\sqrt{2})\). If \(\sqrt{3} = a + b\sqrt{2}\), squaring gives \(3 = a^2 + 2b^2 + 2ab\sqrt{2}\), so \(ab = 0\). Then either \(\sqrt{3} = a\) is rational or \(\sqrt{3} = b\sqrt{2}\) makes \(\sqrt{6} = 2b\) rational — both false. So \(x^2 - 3\) stays irreducible over \(\mathbb{Q}(\sqrt{2})\), and the tower law gives \(2 \times 2 = 4\), with basis \(1, \sqrt{2}, \sqrt{3}, \sqrt{6}\).
The single number \(\sqrt{2} + \sqrt{3}\) has degree \(4\) too, with minimal polynomial \(x^4 - 10x^2 + 1\), so it generates the whole field. Directly: \(\sqrt{2} = \frac{1}{2}\big((\sqrt{2} + \sqrt{3})^3 - 9(\sqrt{2} + \sqrt{3})\big)\).
Start with two points in the plane, \((0, 0)\) and \((1, 0)\). A construction may
A real number is constructible if it is a coordinate of a point that can be reached this way.
Sums and differences come from marking lengths along a line. Products and quotients come from similar triangles: lengths \(1\) and \(a\) on one side, \(b\) on the other, give \(ab\) by proportion.
For \(\sqrt{a}\), lay lengths \(1\) and \(a\) end to end and draw the semicircle on the combined segment. The perpendicular at the join meets the semicircle at height \(\sqrt{a}\): the angle in a semicircle is a right angle, by Thales, and the two similar right triangles give \(h^2 = 1 \cdot a\).
Suppose every point so far has coordinates in a field \(K \subseteq \mathbb{R}\). Then a new point has coordinates in \(K\) or in \(K(\sqrt{d})\) for some positive \(d \in K\).
So every constructible number lies at the top of a tower \(\mathbb{Q} = K_0 \subseteq K_1 \subseteq \cdots \subseteq K_r\) in which each step has degree \(1\) or \(2\).
By the tower law, \([K_r : \mathbb{Q}]\) is a product of \(1\)s and \(2\)s, so a power of \(2\). A constructible \(\alpha\) lies in \(K_r\), so its degree divides that power of \(2\), and is itself a power of \(2\).
The corollary only rules numbers out. A number of degree \(4\) is constructible only if it can actually be reached by square roots, and in general that needs more theory than this page develops. To show a number is constructible, exhibit the square roots.
A cube of volume \(2\) has side \(\sqrt[3]{2}\), which has degree \(3\) (Section 5). \(3\) is not a power of \(2\), so \(\sqrt[3]{2}\) is not constructible, and no ruler-and-compass construction doubles the cube.
A \(60^\circ\) angle is constructible — it is the angle of an equilateral triangle. Trisecting it would construct \(20^\circ\), and from it \(c = \cos 20^\circ\) by dropping a perpendicular. Take de Moivre with \(n = 3\) and compare real parts:
At \(\theta = 20^\circ\), \(4c^3 - 3c = \cos 60^\circ = \tfrac{1}{2}\), so \(c\) is a root of \(8x^3 - 6x - 1\) — irreducible, by Section 5. So \(\cos 20^\circ\) has degree \(3\), and \(60^\circ\) cannot be trisected. Some angles can be: \(90^\circ\) trisects to \(30^\circ\), whose cosine \(\frac{\sqrt{3}}{2}\) is built from a single square root.
Let \(\zeta = \cos\frac{2\pi}{7} + i\sin\frac{2\pi}{7}\) and \(y = \zeta + \zeta^{-1} = 2\cos\frac{2\pi}{7}\). The seventh roots of unity sum to zero; divide \(1 + \zeta + \cdots + \zeta^6 = 0\) by \(\zeta^3\) and group:
Since \(\zeta^2 + \zeta^{-2} = y^2 - 2\) and \(\zeta^3 + \zeta^{-3} = y^3 - 3y\), this is \(y^3 + y^2 - 2y - 1 = 0\), and with \(y = 2\cos\frac{2\pi}{7}\) the cosine is a root of \(8x^3 + 4x^2 - 4x - 1\). Its rational root candidates \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}\) all fail, so the degree is \(3\) and the regular heptagon is not constructible.
The same move with fifth roots gives \(y^2 + y - 1 = 0\) for \(y = 2\cos 72^\circ\), so
a number built from one square root. The regular pentagon is constructible, as Euclid knew. Gauss showed in 1796 that the regular \(17\)-gon is too — \(\cos \frac{2\pi}{17}\) has degree \(8\) — and the full answer, the Gauss–Wantzel theorem, is that the regular \(n\)-gon is constructible exactly when \(\varphi(n)\) is a power of \(2\). This page proves individual cases, like the heptagon, but not the general theorem in either direction.
A square with the area of the unit circle has side \(\sqrt{\pi}\). Every constructible number is algebraic, and if \(\sqrt{\pi}\) were algebraic so would \(\pi\) be. Lindemann’s 1882 theorem that \(\pi\) is transcendental therefore settles the third problem, though its proof needs analysis this page does not cover. The first two impossibilities were proved by Wantzel in 1837, with exactly the degree argument above.
Every finite field has \(p^n\) elements, for some prime \(p\).
Proof. A finite field has prime characteristic \(p\), since it is an integral domain and cannot have characteristic \(0\). The sums \(0, 1, 1 + 1, \ldots\) form a copy of \(\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}\) inside it, so the field is an extension of \(\mathbb{F}_p\) — a vector space over it, of some finite dimension \(n\). Choosing a basis, each element has \(n\) coordinates, each one of \(p\) values.
\(\mathbb{Z}/4\mathbb{Z}\) has \(4 = 2^2\) elements and is not a field, since \(2 \cdot 2 = 0\). The theorem says what size a field can have, not that every ring of that size is one. That a field of every prime-power order exists is also true, but needs more than this page proves.
The degree of an extension is one number, and it connects results from all over the site:
| On the site | As field theory |
|---|---|
| \(\sqrt[3]{2}\) is irrational | its degree is \(3\) — and so it is not constructible |
| The dimension of a vector space | the degree of an extension, well defined for the same reason |
| The minimal polynomial of a matrix | the minimal polynomial of a number, by the same ideal argument |
| \(F[x]/(f)\) is a field for irreducible \(f\), on the rings page | \(\mathbb{Q}(\alpha) \cong \mathbb{Q}[x]/(m)\) |
| De Moivre, and the roots of unity summing to zero | the cubics satisfied by \(\cos 20^\circ\) and \(\cos \frac{2\pi}{7}\) |
| Similar triangles and the angle in a semicircle | the constructible numbers form a field closed under square roots |
| The four- and nine-element fields | every finite field has prime-power order |
The Greek problems asked whether a certain point could be reached. Searching the constructions directly can never prove a negative, because there are infinitely many of them. Translating each construction step into a field extension of degree at most \(2\) turns “no construction exists” into “\(3\) does not divide a power of \(2\)” — a statement that can be checked.
The third Tier 5 unit, closing the abstract algebra arc begun with Groups and Rings and Fields. It is the first of the three where the abstraction answers a question the concrete material could not ask.
| Concept | Statement | Example |
|---|---|---|
| Degree | \([L : K] = \dim_K L\) | \([\mathbb{C} : \mathbb{R}] = 2\) |
| Minimal polynomial | monic, least degree, \(m(\alpha) = 0\); irreducible | \(x^3 - 2\) for \(\sqrt[3]{2}\) |
| Simple extension | \(\mathbb{Q}(\alpha) \cong \mathbb{Q}[x]/(m)\), degree \(\deg m\) | basis \(1, \sqrt{2}\) of \(\mathbb{Q}(\sqrt{2})\) |
| Rational Root Theorem | root \(\frac{p}{q}\) \(\Rightarrow\) \(p \mid a_0\), \(q \mid a_n\) | \(8x^3 - 6x - 1\) has none |
| Tower law | \([M : K] = [M : L][L : K]\) | \([\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4\) |
| Constructible | degree a power of \(2\) (necessary) | \(\cos 72^\circ = \frac{\sqrt{5} - 1}{4}\) |
| Not constructible | degree \(3\) | \(\sqrt[3]{2}\), \(\cos 20^\circ\), \(\cos \frac{2\pi}{7}\) |
| Finite field | order \(p^n\) | \(\mathbb{F}_4\), \(\mathbb{F}_9\); none of order \(6\) |