Field extension

A simple field extension is a field extension generated by one element: $F(\alpha )$ is the smallest subfield of an ambient field containing $F$ and $\alpha$. Its elements are rational expressions in $\alpha$ with coefficients in $F$ and nonzero denominators. If $\alpha$ is algebraic, its minimal polynomial $m_{\alpha }$ identifies $F(\alpha )\cong F[t]/(m_{\alpha })$, so $[F(\alpha ):F]=\mathrm{deg}{m}_{\alpha }$. The classes of $1,t,\dots ,t^{d-1}$ give a basis when $d=\mathrm{deg}{m}_{\alpha }$.

In particular, ${\alpha }^2\in F$ but $\alpha \notin F$ implies $[F(\alpha ):F]=2$. Conversely, in characteristic different from two, any degree-two extension is of this form: for $\beta \notin F$, express ${\beta }^2=a+b\beta$ in the basis $1,\beta$, then $\alpha =\beta -b/2$ has ${\alpha }^2=a+b^2/4\in F$ and generates the same extension. The restriction on the characteristic makes division by two possible.

If $K(\alpha )/K$ is a simple algebraic field extension of degree $n$, the displayed powers form a basis. The minimal polynomial of $\alpha$ reduces every higher power to a linear combination of them, while a nontrivial linear relation among the first $n$ powers would give a polynomial of degree below the minimal polynomial vanishing at $\alpha$. If $\alpha$ is an algebraic integer over the rational numbers, this is a basis of the order $\mathbb{Z}[\alpha ]$ as a free abelian group; it need not be an integral basis of the whole ring of integers.

The algebraic conjugates of an algebraic element over a field $K$ are the polynomial roots of its algebraic minimal polynomial over $K$ in an algebraic closure.

The algebraic conjugates of an algebraic number are the polynomial roots of its algebraic minimal polynomial, equivalently its images under field embeddings of its number field into an algebraic closure.

A field extension obtained by adjoining finitely many square roots in characteristic different from two. If the classes of $a_1,\dots ,a_r$ are independent in $K^{\times }/(K^{\times }{)}^2$, the extension has degree $2^r$ and Galois group $(\mathbb{Z}/2\mathbb{Z}{)}^r$: each square root's sign can be changed independently. The products of distinct square roots form a basis over $K$.

In characteristic different from two, a field generated by finitely many nonzero square roots is the splitting field of a collection of separable quadratic polynomials. Its Galois group embeds into the sign-vector group $(\mathbb{Z}/2{)}^n$. Every index-two subgroup is the kernel of a nontrivial sign character. Coordinate sign characters span the dual of this subgroup of $(\mathbb{Z}/2{)}^n$, by extending a linear functional to the ambient vector space. A product of the corresponding roots transforms by that character and generates precisely the quadratic fixed field.

A field extension $L/K$ is algebraic if every element of $L$ is an algebraic element over $K$. Every finite field extension is algebraic, but algebraic extensions can have infinite degree.

The degree $[L:K]$ of a field extension is the dimension of a vector space $L$ over $K$.

Algebraic independence means the absence of a nonzero polynomial relation over the base field. It extends the distinction between algebraic and transcendental elements to finite tuples.

There is a sequence of real numbers for which every finite initial segment consists of algebraically independent elements over the rationals. Given an independent initial segment, specialize each nonzero integer polynomial in one extra variable at those values. At least one coefficient remains nonzero, so each resulting polynomial has finitely many roots. There are countably many such polynomials by countability of integer polynomial rings. Choosing the next real outside the countable union of their roots continues the construction. Clearing denominators makes independence over the integers equivalent to independence over the rationals.

A transcendence basis of a field extension $L/k$ is an algebraically independent set $T$ such that $L/k(T)$ is algebraic. Its cardinality is the transcendence degree. For finitely generated extensions it is finite, and the remaining algebraic extension is finite.

Apply Zorn's lemma to subsets of a field extension with algebraic independence, ordered by inclusion. The union of a chain is independent because any polynomial relation involves only finitely many distinct elements, all of which lie in one member of the chain. A maximal independent set $B$ is a transcendence basis: if any element were transcendental over $K(B)$, adjoining it would preserve independence, by clearing denominators, contradicting maximality. Tuple entries must be distinct in the set-based independence definition; otherwise the nonzero polynomial $X_1-X_2$ vanishes on a repeated pair and excludes every nonempty set.

A transcendence basis $T$ for $L/k$ is separating if $L/k(T)$ is a separable field extension. In this case the $dt$, $t\in T$, form a basis of the Kähler differentials. In characteristic zero every transcendence basis is separating. In positive characteristic the condition on the chosen basis is essential.

For a finitely generated field extension $L/k$ in characteristic zero, a tuple $t_i$ is algebraically independent exactly when its differentials $d{t}_i$ are linearly independent in the Kähler differentials. A relation of minimum total degree gives a nontrivial differential relation; an independent tuple extends to a transcendence basis. In positive characteristic, derivatives of nonconstant relations can vanish identically.

The transcendence degree is the cardinality of a transcendence basis. It counts independent transcendental parameters. In a finitely generated separably generated field extension, it equals the dimension of the Kähler differentials as a vector space over the extension field.

Elements $x_1,\dots ,x_n$ of a field extension of $K$ are algebraically independent over $K$ when no nonzero polynomial in $K[X_1,\dots ,X_n]$ vanishes at $(x_1,\dots ,x_n)$.

The rational function field $K(x_1,\dots ,x_n)$ is the field of fractions of $K[x_1,\dots ,x_n]$ when the variables are algebraically independent over $K$.

Let $t$ be transcendental over a field $k$ and let $\tau$ be the field automorphism sending $t$ to $1/t$. The rational function $u=t+1/t$ is fixed, and $t$ satisfies $X^2-uX+1=0$. Thus the extension over $k(u)$ has degree at most two. It is not one because the nontrivial automorphism $\tau$ fixes $k(u)$. The fixed field of $\tau$ also has index two by the fixed-field theorem, so it equals $k(u)$.

No nonconstant polynomial in $k[t]$ is fixed by inversion. For a polynomial of positive degree, substitution of $1/t$ creates negative powers, whereas the original has positive powers. Uniqueness of the coefficients of a Laurent polynomial rules out equality. Consequently this fixed field cannot have a generator that is a polynomial in the original variable, even though it has the simple rational-function generator above. The statement is valid in characteristic two as well: $t$ and $1/t$ remain distinct transcendental rational functions.

If $f\in k[X]$ is irreducible, then $(f)$ is a maximal ideal andis a field extension of $k$. The class $\alpha =X+(f)$ is a root of $f$, so this quotient adjoins a root of $f$ to the base field.

For finite extensions $K\subseteq L\subseteq M$,

Inside a common overfield, the compositum $L_1{L}_2$ is the smallest field containing both $L_1$ and $L_2$.

The multiplication image of the tensor product of fields in a common overfield generates the compositum as a field. For algebraic extensions that image is already a field; in general it need not be. The compositum is therefore distinct from the tensor product algebra.

More generally, two field extensions inside a common overfield are linearly disjoint field extensions when the multiplication map from their tensor product of fields into that overfield is injective. The following basis criterion is the finite-degree case.

Finite extensions $L_1/K$ and $L_2/K$ are linearly disjoint when products of basis elements form a basis of their compositum. In particular,Extensions of coprime finite degrees are linearly disjoint.

An algebraic closure of a field $K$ is an algebraic extension $\overline{K}/K$ for which $\overline{K}$ is algebraically closed.

For a field extension $L/K$, an intermediate field is a field $M$ satisfying $K\subseteq M\subseteq L$.

A quadratic extension is a field extension of degree two. It is normal because the other root of the minimal polynomial $T^2+bT+c$ of any non-base-field element is $-b-\alpha$ and lies in the same extension.

For a field of characteristic different from two, adjoining a square root of a nonsquare gives every quadratic extension. If two such extensions are isomorphic over $K$, write $\sqrt{a}=x+y\sqrt{b}$; the vanishing of its cross coefficient forces $x=0$, giving $a/b=y^2$. This explains why one subtracts the identity class when counting extensions from the square-class group.

For a p-adic field with odd residue characteristic, the square map is an isomorphism on $1+{\mathfrak{m}}_K$ by Hensel lemma. The parity of the valuation and the square class of the residue unit therefore determine the square-class group of a field. Its three nontrivial classes give the three quadratic extensions: one unramified extension and two totally ramified extensions.

If $L/K$ and $K/F$ are algebraic field extensions, then $L/F$ is algebraic. In particular, an element algebraic over an algebraic extension of $F$ is algebraic over $F$.

This is a closure property of algebraic extensions, not an alternative name for an extension.