Field extensions · Qual Corpus

Simple and algebraic extensions

For fields $K\subseteq L$ and $\alpha\in L$, $K(\alpha)$ is the smallest subfield of $L$ containing $K$ and $\alpha$, and $K[\alpha]$ is the smallest subring containing them. If $\alpha$ is algebraic over $K$ with minimal polynomial $f$, then $K[\alpha]=K(\alpha)\cong K[x]/\gens f$ and $[K(\alpha):K]=\deg f$. If $\alpha$ is transcendental over $K$, then $K[\alpha]\cong K[x]$ is a polynomial ring and $K(\alpha)\cong K(x)$ is its field of fractions. Every finite extension is algebraic, and by the primitive element theorem every finite separable extension is simple.

Simple extensions (Tag D-PUOGJ)

A field extension $L/k$ is simple if there exists $\alpha\in L$ such that $L = k(\alpha)$.

Subfield $K(\alpha)$ generated by an element of an extension field (Tag FD-J3HIA)

Let $F$ be a field, $K\subseteq F$ a subfield, and $\alpha\in F$. The subfield generated by $K$ and $\alpha$, written $K(\alpha)$, is the smallest subfield of $F$ containing $K$ and $\alpha$, that is, the intersection of all subfields of $F$ that contain $K\cup\theset{\alpha}$.

For $\alpha$ in an extension $F \supset K$, what is $K(\alpha)$?

Subfield $K(\alpha)$ generated by an element of an extension field (Tag FD-QPNDA)

Which subfield of $F$ is $K(\alpha)$?

Subring $K[\alpha]$ generated by an element of an extension field (Tag FD-3V3ZW)

Let $F$ be a field, $K\subseteq F$ a subfield, and $\alpha\in F$. The subring generated by $K$ and $\alpha$, written $K[\alpha]$, is the smallest subring of $F$ containing $K$ and $\alpha$, that is, the intersection of all subrings of $F$ that contain $K\cup\theset{\alpha}$.

$K[\alpha]=\theset{g(\alpha) \st g\in K[x]}$, the image of the evaluation homomorphism $K[x]\to F$, $g\mapsto g(\alpha)$: this image is a subring containing $K$ and $\alpha$, and every such subring contains each $g(\alpha)$.

For $\alpha$ in an extension $F \supset K$, what is $K[\alpha]$?

Primitive element theorem (Tag T-3HPWT)

Every finite separable field extension is generated by one element.Suggest a slogan

Let $L/k$ be a finite separable field extension. Then $L/k$ is simple: there exists $\alpha\in L$ with $L=k(\alpha)$.

Algebraic element of a field extension (Tag FD-2EVYB)

Let $L/K$ be a field extension. An element $\alpha \in L$ is algebraic over $K$ if there exists a nonzero polynomial $g\in K[x]$ with $g(\alpha) = 0$.

For a field extension $L/K$, when is $\alpha \in L$ algebraic over $K$?

Algebraic field extension (Tag FD-NS5RF)

A field extension $L/k$ is algebraic if every $\alpha \in L$ is algebraic over $k$, that is, a root of some nonzero polynomial $f\in k[x]$.

When is a field extension $L/k$ algebraic?

When is a field extension $L/K$ algebraic?

When is a field extension $L/K$ algebraic over $K$?

Finite extensions are algebraic (Tag T-NGBVC)

Finite vector-space dimension forces every field element to satisfy a polynomial.Suggest a slogan

Let $L/K$ be a field extension with $[L:K]<\infty$. Then every $\alpha\in L$ is algebraic over $K$.

Let $n=[L:K]$. The $n+1$ elements $1,\alpha,\ldots,\alpha^n$ of the $n$-dimensional $K$-vector space $L$ are linearly dependent, so there are $c_0,\ldots,c_n\in K$, not all zero, with $\sum_{i=0}^n c_i\alpha^i=0$. Thus $\alpha$ is a root of the nonzero polynomial $\sum_{i=0}^n c_ix^i\in K[x]$.

(Tower law and degree constraints)

For fields $K\subseteq L\subseteq M$, the tower law is \[ [M:K]=[M:L][L:K]. \] Hence an extension of prime degree has no intermediate fields other than its endpoints, and if $f\in\QQ[x]$ is irreducible with root $\alpha$, then $\deg f = [\QQ(\alpha):\QQ]$ divides the degree $[\SF(f):\QQ]$ of the splitting field, which is the order of the Galois group of $f$.

Finding a minimal polynomial

(Methods)

For $x\da \sqrt a + \sqrt b$ with $a,b\in\QQ$, isolate a radical and square: $(x-\sqrt a)^2 = b$ gives $x^2+a-b = 2x\sqrt a$, and squaring again gives $(x^2+a-b)^2 = 4ax^2$, a monic polynomial of degree $4$ with root $x$.
If $n \da [\QQ(\alpha):\QQ]$ is known, then $1, \alpha, \ldots, \alpha^n$ are $\QQ$-linearly dependent, and a dependence with $\alpha^n$ having coefficient $1$ gives the minimal polynomial.
For $x\da \sqrt a + \sqrt b$, write $1, x, x^2, x^3, x^4$ as vectors in the $\QQ$-spanning set $1, \sqrt a, \sqrt b, \sqrt{ab}$ of $\QQ(\sqrt a,\sqrt b)$. These are five vectors in a space of dimension at most $4$, and a $\QQ$-linear dependence among them is a polynomial of degree at most $4$ with root $x$.
If $\alpha,\beta\neq0$ and $\alpha\beta \in \QQ$, then $\alpha = (\alpha\beta)/\beta \in \QQ(\beta)$ and $\beta\in\QQ(\alpha)$, so $\QQ(\alpha)=\QQ(\beta)$.

Algebraic extensions

For fields $K\subseteq L\subseteq M$, if $L/K$ and $M/L$ are algebraic, then $M/K$ is algebraic.

An algebraic extension of an algebraic extension is algebraic (Tag PR-ABSJX)

Algebraicity is transitive in towers of field extensions.Suggest a slogan

Let $k \subseteq K \subseteq L$ be fields, with the extensions not necessarily finite. If $L/K$ and $K/k$ are algebraic, then $L/k$ is algebraic.

Quadratic extensions

If $[L:K]=2$, then $L=K(\alpha)$ for every $\alpha\in L\sm K$. If $\operatorname{char}K\neq2$, then $L=K(\sqrt d)$ for some $d\in K$ that is not a square in $K$. The quadratic extensions of $\QQ$ are the fields $\QQ(\sqrt d)$ for squarefree integers $d\neq 1$, and $\FF_p$ has a unique quadratic extension up to isomorphism, $\FF_{p^2}$.

Quadratic extensions in characteristic not $2$ (Tag PR-OFBRQ)

Every quadratic extension in characteristic not $2$ is $F(\sqrt a)$ and is automatically Galois.Suggest a slogan

Let $F$ be a field with $\ch(F)\neq 2$ and let $E/F$ be a field extension with $[E:F]=2$. Then $E/F$ is Galois, and $E=F(\sqrt{a})$ for some $a\in F$ that is not a square in $F$. For $F=\QQ$, $a$ can be taken to be a squarefree integer different from $1$.

Choose $\alpha\in E\setminus F$; then $E=F(\alpha)$ and the minimal polynomial of $\alpha$ over $F$ is $x^2+bx+c$ with $b,c\in F$. Since $2$ is invertible in $F$, put $\beta\coloneqq\alpha+b/2$ and $a\coloneqq b^2/4-c\in F$. Then $\beta^2=a$, $\beta\notin F$, and $E=F(\beta)=F(\sqrt a)$, so $a$ is not a square in $F$. The roots of $x^2-a$ are $\pm\beta\in E$, and $\beta\neq-\beta$ because $\beta\neq0$ and $\ch(F)\neq2$. Thus $E$ is the splitting field of the separable polynomial $x^2-a$, so $E/F$ is Galois. For $F=\QQ$, write $a=m^2 d/n^2$ with $m,n$ nonzero integers and $d$ a squarefree integer; then $\QQ(\sqrt a)=\QQ(\sqrt d)$, and $d\neq1$ because $a$ is not a square.

Quadratic extensions of $\QQ$ (Tag C-BOM4E)

Every quadratic extension of $\QQ$ comes from adjoining the square root of a squarefree integer.Suggest a slogan

Let $E/\QQ$ be a field extension with $[E:\QQ] = 2$. Then $E = \QQ(\sqrt q)$ for some squarefree integer $q \notin \theset{0, 1}$.

Choose $\alpha\in E\setminus\QQ$. Since $\QQ \subsetneq \QQ(\alpha) \subseteq E$ and $[E:\QQ]=2$, we have $E = \QQ(\alpha)$, and the minimal polynomial of $\alpha$ over $\QQ$ has the form $x^2 + bx + c$ with $b, c\in\QQ$. By the quadratic formula, $\alpha = \frac{-b \pm \sqrt d}{2}$ with $d \coloneqq b^2 - 4c\in\QQ$, so $E = \QQ(\sqrt d)$, and $d$ is not a square in $\QQ$ because $\alpha\notin\QQ$. Write $d = m/n$ with $m, n\in\ZZ$ and $n \neq 0$; then $\sqrt d = \sqrt{mn}/n$, so $E = \QQ(\sqrt{mn})$. Write $mn = s^2 q$ with $s\in\ZZ\setminus\theset 0$ and $q$ a squarefree integer; then $E = \QQ(\sqrt q)$, and $q\notin\theset{0,1}$ because $d$ is not a square in $\QQ$.

Quadratic extensions of $\FF_p$ are unique up to isomorphism (Tag PR-F2N4L)

$\FF_p$ has a unique quadratic extension up to isomorphism.Suggest a slogan

Let $p$ be a prime. Any two field extensions $E/\FF_p$ and $E'/\FF_p$ of degree $2$ are isomorphic.

Distinguished classes

A class of field extensions is distinguished if, for every tower $K\subseteq L\subseteq M$, $M/K$ is in the class if and only if $M/L$ and $L/K$ are, and if $E/K$ is in the class and $F/K$ is any extension with $E,F$ in a common field, then $EF/F$ is in the class. Separable extensions form a distinguished class. Normal extensions satisfy: if $M/K$ is normal and $K\subseteq L\subseteq M$, then $M/L$ is normal; normality is not transitive, by the example on Splitting fields and normal extensions.

Distinguished classes of field extensions (Tag D-JMATC)

A class $\mathcal{S}$ of field extensions is distinguished if it satisfies the following two conditions.

For every tower of fields $k\subseteq K\subseteq L$, $L/k \in \mathcal{S}$ if and only if $L/K \in \mathcal{S}$ and $K/k\in \mathcal{S}$:

If $K/k\in \mathcal{S}$ and $L/k$ is an extension such that $K$ and $L$ are subfields of a common field, then $LK/L \in \mathcal{S}$, where $LK$ is the compositum. The extension $LK/L$ is the lift of $K/k$ to $L$:

Let $\mathcal{S}$ be a distinguished class of field extensions. If $L/k\in\mathcal{S}$ and $K/k\in \mathcal{S}$, with $K$ and $L$ subfields of a common field, then $LK/k \in \mathcal{S}$:

By the lifting condition applied to $K/k\in\mathcal{S}$ and $L/k$, the extension $LK/L$ lies in $\mathcal{S}$. Applying the tower condition to $k\subseteq L\subseteq LK$, with $LK/L\in\mathcal{S}$ and $L/k\in\mathcal{S}$, gives $LK/k\in\mathcal{S}$.

Compositum and intersection of normal extensions (Tag C-4GQK3)

Normal extensions are closed under compositum and intersection.Suggest a slogan

Let $k$ be a field, let $\Omega$ be an algebraic closure of $k$, and let $E_1, E_2 \subseteq \Omega$ be extensions of $k$. If $E_1/k$ and $E_2/k$ are normal, then the compositum $E_1E_2/k$ and the intersection $(E_1 \cap E_2)/k$ are normal.

If $L/k$ is normal, then $L/K$ is normal for every intermediate field $K$ (Tag PR-MHOVR)

Normality descends from the base field to every intermediate field.Suggest a slogan

Let $L/k$ be a finite normal extension and let $K$ be a field with $k \subseteq K \subseteq L$. Then $L/K$ is normal.

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