Algebraic and Transcendental Elements

Once you have a field extension E/F, the elements of the big field E split into two profoundly different kinds, and the whole of Galois theory hinges on the difference. Some elements are secretly tame: they satisfy a polynomial equation with coefficients from the base field F. Others are wild: no such equation exists, anywhere, ever. The tame ones are called algebraic, the wild ones transcendental — and this page is about telling them apart and pinning down the single polynomial that captures an algebraic element exactly.

Here is the definition, stripped to its bones. An element \alpha \in E is algebraic over F if there is a nonzero polynomial p(x) \in F[x] with p(\alpha) = 0. If no nonzero polynomial over F kills \alpha, it is transcendental over F. Note the phrase "over F" doing quiet but decisive work: being algebraic is always relative to the base field whose coefficients you are allowed to use.

The showpiece example: \sqrt 2 is algebraic over \mathbb{Q}, because it satisfies x^2 - 2 = 0 — a polynomial with rational (indeed integer) coefficients. By contrast \pi is transcendental over \mathbb{Q}: no polynomial with rational coefficients has \pi as a root, a deep fact proved by Lindemann in 1882 and the reason you cannot square the circle.

The minimal polynomial

If \alpha is algebraic, it is a root of many polynomials — x^2 - 2, but also 3x^2 - 6, x^3 - 2x, (x^2 - 2)(x^5 + 1), and infinitely many more. Among this whole crowd there is a distinguished champion: the minimal polynomial of \alpha over F, written m_\alpha(x) or \mathrm{min}_F(\alpha). It is the monic polynomial of least degree in F[x] having \alpha as a root. "Monic" means the leading coefficient is 1, which pins down a unique representative among the scalar multiples.

Its two structural properties are what make it indispensable. First, the minimal polynomial is irreducible over F — it cannot be factored into lower-degree polynomials with coefficients in F. (If it did factor, one factor would already vanish at \alpha, contradicting minimality.) Second, it divides every polynomial that kills \alpha: if p(\alpha) = 0 then m_\alpha(x) \mid p(x). In one stroke this says m_\alpha is the generator of the ideal of all polynomials vanishing at \alpha.

Let \alpha \in E be algebraic over F. There is a unique polynomial m_\alpha(x) \in F[x], the minimal polynomial, with:

Monic and minimal. m_\alpha is monic, has m_\alpha(\alpha) = 0, and no nonzero polynomial of smaller degree kills \alpha.
Irreducible. m_\alpha cannot be factored into lower-degree polynomials over F.
Divides all annihilators. For any p \in F[x], p(\alpha) = 0 if and only if m_\alpha \mid p.
Builds the extension. F(\alpha) \cong F[x] / (m_\alpha(x)), and its degree over F equals \deg m_\alpha.

A gallery of minimal polynomials

The fastest way to internalise the definition is to compute a handful. In every case below the base field is \mathbb{Q}. The recipe: find some polynomial that kills \alpha, then check it is irreducible and make it monic.

\sqrt 2: square it to get 2, so m(x) = x^2 - 2. Irreducible over \mathbb{Q} (no rational root), degree 2.
i: i^2 = -1, so m(x) = x^2 + 1, degree 2.
The golden ratio \varphi = \tfrac{1 + \sqrt 5}{2} satisfies \varphi^2 = \varphi + 1, so m(x) = x^2 - x - 1, degree 2.
\sqrt[3]{2}: cube it to get 2, so m(x) = x^3 - 2. Irreducible over \mathbb{Q} by Eisenstein at p = 2 — degree 3.
A primitive cube root of unity \omega = e^{2\pi i/3} satisfies \omega^3 = 1 but \omega \neq 1, so it is a root of \tfrac{x^3 - 1}{x - 1} = x^2 + x + 1 — degree 2, not 3.

The picture below plots p(x) = x^3 - 2. Its single real root is exactly the place where the curve crosses the axis — the point x = \sqrt[3]{2} \approx 1.26. An algebraic number is, quite literally, a place where its minimal polynomial vanishes.

Why "almost all" numbers are transcendental

Algebraic numbers feel like the whole world — every root, every surd, every solution of every polynomial you have met. Yet they are vanishingly rare. Each algebraic number is a root of some polynomial with integer coefficients; there are only countably many such polynomials, each with finitely many roots, so the algebraic numbers form a countable set. The real numbers are uncountable. Therefore almost every real number — in the precise sense of cardinality — is transcendental, even though named examples like \pi and e were spectacularly hard to find. Hermite proved e transcendental in 1873; Lindemann settled \pi in 1882. The generic number is untameable, but the specimens are elusive.

The single most common slip is to grab the first polynomial that kills \alpha and call it minimal. The minimal polynomial must be monic, of least degree, and irreducible over the specific base field — all three. So 2x^2 - 4 kills \sqrt 2 but is not monic; x^4 - 4 kills it but is not minimal degree (and factors as (x^2 - 2)(x^2 + 2)). The genuine minimal polynomial is x^2 - 2.

And the base field is decisive. \sqrt 2 has minimal polynomial x^2 - 2 of degree 2 over \mathbb{Q} — but over \mathbb{R} it is already an element of the base field, so its minimal polynomial is just x - \sqrt 2, of degree 1! "The minimal polynomial of \alpha" is meaningless until you say over which field.

Astonishingly, nobody knows. We know e and \pi are each transcendental, but the sum e + \pi and the product e\pi have never been proved transcendental — nor algebraic. There is a sly trick, though: they cannot both be algebraic. If they were, then so would be the roots of (x - e)(x - \pi) = x^2 - (e + \pi)x + e\pi, forcing e and \pi to be algebraic — a contradiction. So at least one of e + \pi, e\pi is transcendental. We just cannot say which.