Field Extensions and Algebraic Elements

A field is a commutative ring with identity in which every nonzero element has an inverse: arithmetic in which division always works. Two questions organize the subject. Given a polynomial with no root in $F$, like $x^2+1$ over $\mathbb{R}$ or $x^2-2$ over $\mathbb{Q}$, is there a larger field in which it has a root? And given a larger field $K\supseteq F$, how much larger is it? The second question is answered by a single integer, the first by a quotient of a polynomial ring.

Characteristic and the prime subfield

Every field contains a smallest subfield. Repeatedly adding the identity $1_F$ to itself gives elements $1_F,1_F+1_F,1_F+1_F+1_F,\dots$, which either are all distinct or eventually return to $0$.

The characteristic is $0$ or prime. If $n=ab$ were composite with $n\cdot 1_F=0$, then $(a\cdot 1_F)(b\cdot 1_F)=0$, and because a field has no zero divisors one of the factors already vanishes, contradicting minimality. The map $\phi :\mathbb{Z}\to F$, $n\mapsto n\cdot 1_F$, is a ring homomorphism with kernel $\mathrm{ch}(F)\mathbb{Z}$, so its image is a copy of either $\mathbb{Z}$ (characteristic $0$) or $\mathbb{Z}/p\mathbb{Z}$ (characteristic $p$). Since $F$ is a field it contains the fractions of that image.

So $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ all have characteristic $0$ and prime subfield $\mathbb{Q}$; the finite field ${\mathbb{F}}_p$ and the rational function field ${\mathbb{F}}_p(x)$ both have characteristic $p$. Every field is an extension of its prime subfield, so the subject reduces to what sits between that subfield and a given field.

Extensions and degree

A field containing $F$ is in particular an $F$-vector space, and vector spaces have dimensions.

For example $\mathbb{C}=\mathbb{R}+\mathbb{R}i$ has $\{1,i\}$ as an $\mathbb{R}$-basis, so $[\mathbb{C}:\mathbb{R}]=2$. The notation $K/F$ is read $K$ over $F$ and is not a quotient; it only records that $F$ sits inside $K$.

Building an extension with a root

Given an irreducible $p(x)\in F[x]$, we want a field where it acquires a root. The construction reuses the fact that $F[x]$ is a principal ideal domain in which the ideal $(p(x))$ generated by an irreducible element is maximal.¹ Quotienting a ring by a maximal ideal produces a field, so the following works.

The elements of $K$ have a clean description.

Addition in $K$ is polynomial addition; multiplication is polynomial multiplication followed by reduction of the product modulo $p(x)$.

For $F=\mathbb{R}$ and $p(x)=x^2+1$, the quotient $\mathbb{R}[x]/(x^2+1)$ is a degree-$2$ extension with elements $a+b\theta$ and the rule ${\theta }^2=-1$; the map $a+bx\mapsto a+bi$ is an isomorphism onto $\mathbb{C}$. Taking $F=\mathbb{Q}$ and $p(x)=x^2-2$ gives a degree-$2$ field whose root is written $\sqrt{2}$, with

Over ${\mathbb{F}}_2$ the polynomial $x^2+x+1$ is irreducible, so ${\mathbb{F}}_2[x]/(x^2+x+1)$ is a field of four elements $\{0,1,\theta ,\theta +1\}$ with ${\theta }^2=\theta +1$.

Inverses by the Euclidean algorithm

Division in $F(\theta )$ looks harder than the other operations, but it reduces to the extended Euclidean algorithm. If $a(\theta )\ne 0$ then $\gcd (a(x),p(x))=1$, because $p$ is irreducible and does not divide the lower-degree $a$. The extended algorithm produces $u(x),v(x)$ with $u(x)a(x)+v(x)p(x)=1$; reducing modulo $p$ leaves $u(\theta )a(\theta )=1$, so $u(\theta )$ is the inverse.

Generated subfields and simple extensions

Intersecting all subfields of $K$ that contain $F$ and a set of elements gives a smallest such subfield.

The construction $F[x]/(p(x))$ and the concrete field $F(\alpha )$ are the same object whenever $\alpha$ is a root of $p$.

A useful consequence, proved the same way, is that the two roots of an irreducible are algebraically interchangeable: an isomorphism $\phi :F\to F^{\prime }$ carrying $p(x)$ to $p^{\prime }(x)$ extends to $F(\alpha )\cong F^{\prime }(\beta )$ sending any root $\alpha$ of $p$ to any root $\beta$ of $p^{\prime }$.⁵ The same lifting step drives the uniqueness of splitting fields.

Algebraic and transcendental elements

When $\alpha$ is algebraic, the polynomials it satisfies form a nonzero ideal of $F[x]$; being a principal ideal domain, that ideal has a single monic generator.

Combining the two theorems, the simple extension a single algebraic element generates has degree equal to the degree of its minimal polynomial:

The last row shows that degree depends on the base field: over $\mathbb{R}$ the cube root of $2$ already lives in the field, so its minimal polynomial is linear. The irreducibility claims for $x^n-2$ come from Eisenstein's criterion.

Finiteness and algebraicity of a single element are the same condition.

The tower law

Degrees multiply through a tower of fields, exactly as index does through a chain of subgroups.

An immediate corollary: if $L/F$ is finite and $F\subseteq K\subseteq L$, then $[K:F]$ divides $[L:F]$. This settles containment questions cheaply. Because $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ and $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$ for $\alpha$ a root of $x^3-3x-1$, and $2\nmid 3$, the number $\sqrt{2}$ cannot lie in $\mathbb{Q}(\alpha )$ — a fact awkward to check by hand. For a positive use, $\sqrt[6]{2}$ has degree $6$, and since $(\sqrt[6]{2}{)}^3=\sqrt{2}$, the field $\mathbb{Q}(\sqrt{2})$ sits inside $\mathbb{Q}(\sqrt[6]{2})$ with $[\mathbb{Q}(\sqrt[6]{2}):\mathbb{Q}(\sqrt{2})]=3$.

Finitely generated algebraic extensions

Adjoining several algebraic elements one at a time keeps degrees finite. Writing $F(\alpha ,\beta )=(F(\alpha ))(\beta )$ realizes $F({\alpha }_1,\dots ,{\alpha }_k)$ as a chain of simple extensions $F=F_0\subseteq F_1\subseteq \cdots \subseteq F_k$, and each step degree is at most the degree of the next generator over $F$.

The bound can be strict, since a later generator may already lie in the field a previous one built. For example $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{2})$ because the second generator adds nothing, so the naive product bound of $9$ collapses to $3$.

In this genuinely independent case the product bound is attained.

Two closure statements follow. First, sums, products, and quotients of algebraic elements are algebraic, because they all live in the finite extension $F(\alpha ,\beta )$; so the elements of any extension $L/F$ that are algebraic over $F$ form a subfield of $L$. Second, algebraicity is transitive.

Applied to $\mathbb{C}/\mathbb{Q}$, the elements of $\mathbb{C}$ algebraic over $\mathbb{Q}$ form the field $\overline{\mathbb{Q}}$ of algebraic numbers, an infinite algebraic extension since it contains $\sqrt[n]{2}$ for every $n$. A counting argument shows $\overline{\mathbb{Q}}$ is countable while $\mathbb{R}$ is not, so transcendental reals exist; $\pi$ and $e$ are the standard examples, though proving it for either is hard.