The Galois correspondence

Galois extensions

Galois extension and Galois group (Tag D-5JYEI)

Let $L/k$ be a finite field extension. $L/k$ is a Galois extension if it is normal and separable. In this case the Galois group of $L/k$ is \[ \Gal(L/k) \coloneqq \Aut_{\Fieldsover{k}} (L). \]

Let $L/k$ be a finite field extension, and let $[L:k]_s$ be its separable degree. The following are equivalent:

$L/k$ is normal and separable.
The fixed field of $\Aut_{\Fieldsover{k}}(L)$ is $k$.
$L$ is a splitting field over $k$ of a separable polynomial $p\in k[x]$.
$L$ is a splitting field over $k$ of an irreducible separable polynomial $p\in k[x]$.
$\size \Aut_{\Fieldsover{k}} (L) = [L: k]_s = [L: k]$.

The separable degree counts the embeddings of $L$ into an algebraic closure that extend a fixed embedding of $k$:

Finite Galois extension (Tag FD-4GFTY)

Let $K/F$ be a finite field extension, and let $\Aut(K/F)$ be the group of field automorphisms of $K$ that restrict to the identity on $F$. $K/F$ is Galois if $\abs{\Aut(K/F)} = [K: F]$.

In terms of $\abs{\Aut(K/F)}$, when is a finite extension $K/F$ Galois?

Characterizations of finite Galois extensions (Tag FT-3WWKN)

Finite Galois means separable splitting, fixed field $F$, full automorphism count, or normal plus separable.Suggest a slogan

Let $K/F$ be a finite field extension. The following are equivalent:

$K$ is a splitting field over $F$ of a separable polynomial in $F[x]$.
The fixed field of $\Aut(K/F)$ is $F$.
$[K:F]=\abs{\Aut(K/F)}$.
$K/F$ is normal and separable.

A finite extension satisfying these conditions is Galois.

What four conditions each characterise a Galois extension $K/F$?

Conjugates of an algebraic element (Tag FD-YJEDU)

Let $L/K$ be a field extension and let $\alpha\in L$ be algebraic over $K$, with minimal polynomial $m_\alpha\in K[x]$, the monic irreducible polynomial in $K[x]$ having $\alpha$ as a root. A conjugate of $\alpha$ over $K$ is a root of $m_\alpha$, in a splitting field of $m_\alpha$ over $L$, other than $\alpha$.

What is a conjugate of a root $\alpha \in L$?

$\QQ(\sqrt[3]{2})/\QQ$ is not Galois (Tag FE-Z4VF2)

Let $\sqrt[3]{2}\in\RR$ be the real cube root of $2$. The extension $\QQ(\sqrt[3]{2})/\QQ$ has degree $3$ and trivial automorphism group, so it is not Galois.

The polynomial $x^3-2$ is irreducible over $\QQ$ by Eisenstein's criterion at $p=2$, so $[\QQ(\sqrt[3]{2}):\QQ]=3$. An automorphism $\sigma$ of $\QQ(\sqrt[3]{2})$ fixing $\QQ$ sends $\sqrt[3]{2}$ to a root of $x^3-2$ lying in $\QQ(\sqrt[3]{2})\subseteq\RR$. The only real root of $x^3-2$ is $\sqrt[3]{2}$, so $\sigma=\id$. Hence $\abs{\Aut(\QQ(\sqrt[3]{2})/\QQ)}=1\ne3$, and $\QQ(\sqrt[3]{2})/\QQ$ is not Galois by FT-3WWKN.

Give a field extension that is not Galois.

Galois subextensions of a Galois tower (Tag PR-QDRB4)

Intermediate Galois subextensions correspond to normal subgroups, with quotient Galois group upstairs.Suggest a slogan

Let $L/k$ be a finite Galois extension and let $F$ be an intermediate field, $k\subseteq F\subseteq L$. Then:

$L/F$ is Galois.
$F/k$ is Galois if and only if $\Gal(L/F) \normal \Gal(L/k)$.
In that case restriction $\sigma\mapsto\sigma|_F$ induces an isomorphism \[ \Gal(F/k) \cong \Gal(L/k)/\Gal(L/F). \]

$L$ is the splitting field over $k$ of a separable polynomial $f\in k[x]$, and $L$ is also generated over $F$ by the roots of $f\in F[x]$, so $L$ is the splitting field of $f$ over $F$ and $L/F$ is Galois.
and (c). Fix an algebraic closure $\bar k$ containing $L$. Every $k$-embedding $F\to\bar k$ extends to a $k$-embedding $L\to\bar k$, which has image $L$ by PR-OZYUC because $L/k$ is normal; so the $k$-embeddings $F\to\bar k$ are exactly the restrictions $\sigma|_F$ with $\sigma\in\Gal(L/k)$. For $\sigma\in\Gal(L/k)$ we have $\Gal(L/\sigma(F))=\sigma\Gal(L/F)\sigma^{-1}$, and by the Galois correspondence $\sigma(F)=F$ if and only if $\sigma\Gal(L/F)\sigma^{-1}=\Gal(L/F)$. Hence $\Gal(L/F)\normal\Gal(L/k)$ if and only if $\sigma(F)=F$ for every $\sigma\in\Gal(L/k)$, if and only if every $k$-embedding $F\to\bar k$ has image $F$, which by PR-OZYUC means $F/k$ is normal. Since $F/k$ is separable as a subextension of the separable extension $L/k$, this is equivalent to $F/k$ being Galois. In that case restriction is a homomorphism $\Gal(L/k)\to\Gal(F/k)$; it is surjective because every $k$-automorphism of $F$ extends to $L$, and its kernel is $\Gal(L/F)$, which gives (c).

An intermediate field $K$ is Galois over $k$ iff $\Gal(L/K) \normal \Gal(L/k)$ (Tag PR-3TYBE)

An intermediate extension is Galois exactly when its fixing subgroup is normal, with quotient the intermediate Galois group.Suggest a slogan

Let $L/k$ be a Galois extension and let $K$ be a field with $k \subseteq K \subseteq L$. Then \[ K/k \text{ is Galois} \iff \Gal(L/K) \normal \Gal(L/k), \] where $\normal$ denotes a normal subgroup, and in that case \[ \Gal(K/k) \cong \Gal(L/k) / \Gal(L/K). \]

The fundamental theorem

Fundamental theorem of Galois theory (Tag T-NLPZY)

Intermediate fields and subgroups correspond in reverse, with normal subgroups exactly the Galois intermediate fields.Suggest a slogan

Let $L/k$ be a finite Galois extension with Galois group $G \coloneqq \Gal(L/k)$. For a subgroup $H\leq G$, let $L^H$ be its fixed field, and for an intermediate field $k\subseteq F\subseteq L$, let \[ \Gal(L/F) \coloneqq \theset{ \sigma \in G \suchthat \sigma(\alpha) = \alpha \text{ for all } \alpha \in F}. \] Then the maps \[ \theset{\text{subgroups } H\leq G} \mapstofrom \theset{\text{intermediate fields } k\subseteq F\subseteq L}, \qquad H\mapsto L^H, \qquad \Gal(L/F) \mapsfrom F, \] are mutually inverse bijections, and they reverse inclusions. If $H$ and $F$ correspond under these bijections, then:

$[F: k] = [G: H]$ and $[L : F] = \abs{H}$.
$L/F$ is Galois and $\Gal(L/F) = H$.
$F/k$ is Galois if and only if $H \normal G$, and in that case $\Gal(F/k) \cong G/H$.

If $H_1,H_2$ correspond to $F_1,F_2$, then the compositum $F_1 F_2$ corresponds to $H_1 \intersect H_2$, and $F_1 \intersect F_2$ corresponds to the subgroup generated by $H_1$ and $H_2$, $\gens{H_1, H_2}$.

The group $\Gal(L/F)$ fixes $F$ pointwise. For $\sigma\in G$ one has $\sigma(L^H)=L^{\sigma H\sigma^{-1}}$, so the setwise stabilizer $\theset{\sigma\in G \suchthat \sigma(F) = F}$ of $F=L^H$ is the normalizer $N_G(H)$, which contains $H$ and equals $G$ exactly when $F/k$ is Galois.

The subgroup $\gens{H_1,H_2}$ contains the product set $H_1H_2$, and $H_1H_2$ is a subgroup if and only if $H_1H_2=H_2H_1$.

Fundamental theorem of Galois theory (Tag FT-GJ6NR)

Intermediate fields and subgroups correspond contravariantly; normal subgroups detect Galois subextensions.Suggest a slogan

Let $K/F$ be a finite Galois extension with Galois group $G=\Gal(K/F)$. The maps \[ \begin{aligned} \theset{E\suchthat F\subseteq E\subseteq K\text{ a field}}&\longleftrightarrow\theset{H\suchthat H\le G}\\ E&\mapsto\Gal(K/E)=\theset{\tau\in G\suchthat\tau|_E=\id_E}\\ K^H&\mapsfrom H \end{aligned} \] are mutually inverse bijections, where $K^H$ is the fixed field of $H$. If $E$ corresponds to $H$, and $E_1,E_2$ correspond to $H_1,H_2$, then:

$E_1\subseteq E_2$ if and only if $H_2\le H_1$.
$\abs H=[K:E]$ and $[E:F]=[G:H]$.
$K/E$ is Galois with $\Gal(K/E)=H$.
$E/F$ is Galois if and only if $H$ is normal in $G$, in which case $\Gal(E/F)\cong G/H$.
The intersection $E_1\cap E_2$ corresponds to the subgroup $\gens{H_1,H_2}$ generated by $H_1$ and $H_2$, and the compositum $E_1E_2$ corresponds to $H_1\cap H_2$.

State the fundamental theorem of Galois theory.

Finite normal extensions of perfect fields are Galois (Tag T-SGY3O)

Over a perfect field, finite normal extensions and polynomial splitting fields are Galois.Suggest a slogan

Let $k$ be a field.

If $\ch k = 0$ or $k$ is finite, then $k$ is perfect; in particular $\CC$, $\RR$, $\QQ$, and $\FF_p$ for $p$ prime are perfect.
If $k$ is perfect, then every finite normal extension of $k$ is Galois.
If $k$ is perfect, then the splitting field over $k$ of every nonconstant polynomial $f\in k[x]$ is Galois over $k$.

Galois group of the composite of a Galois extension with an arbitrary extension (Tag PR-FM5FN)

Base change cuts the Galois group down to the automorphisms fixing the field intersection.Suggest a slogan

Let $F/k$ be a finite Galois extension and $L/k$ an arbitrary extension, with $F$ and $L$ subfields of a common field. Then $FL/L$ is Galois, and restriction $\sigma \mapsto \sigma|_F$ is an isomorphism \[ \Gal(FL/L) \cong \Gal(F / F\intersect L) \leq \Gal(F/k). \]

(Degrees, indices, and normal subgroups)

Let $L/F$ be a finite Galois extension with $G=\Gal(L/F)$. The correspondence $K\mapsto\Gal(L/K)$ from intermediate fields to subgroups is inclusion-reversing, and if $K$ corresponds to $H$, then \[ [L:K] = \size H, \qquad [K:F] = [G:H], \qquad [L:F] = \size G. \] The extension $K/F$ is Galois if and only if $H$ is normal in $G$, and then $\Gal(K/F)\cong G/H$.

Criteria for Galois extensions

(Criteria)

Irreducibility of $f\in\QQ[x]$.

Eisenstein’s criterion, applied to $f(x)$, to $f(x+a)$ for some $a\in\ZZ$, or to the reversed polynomial $x^nf(1/x)$.
If $f\in\ZZ[x]$ has leading coefficient not divisible by a prime $p$ and $f\bmod p$ is irreducible over $\FF_p$, then $f$ is irreducible over $\QQ$.
A polynomial of degree $2$ or $3$ over a field is irreducible if and only if it has no root in the field.

Separability of $f$.

$f$ is separable if and only if $\gcd(f,f')=1$.
Over a perfect field, every irreducible polynomial is separable.
An irreducible $f$ is separable if and only if $f' \neq 0$.

Separability of an extension.

The splitting field of a separable polynomial is separable and normal, hence Galois.
Every algebraic extension of a perfect field is separable; in characteristic $0$ a finite extension is Galois if and only if it is normal.
A finite extension $L/k$ is separable if and only if its separable degree $[L:k]_s$ equals $[L:k]$, and separable extensions form a distinguished class.

Normality.

A finite extension $L/k$ is normal if and only if $L$ is the splitting field over $k$ of some polynomial.

Galois.

A finite extension is Galois if and only if it is normal and separable, if and only if it is the splitting field of a separable polynomial.
Every finite extension of finite fields is Galois: $\FF_{p^n}$ is the splitting field of the separable polynomial $x^{p^n}-x$ over every subfield.

Irreducibility

A monic integer polynomial irreducible modulo a prime is irreducible over $\QQ$ (Tag PR-PB6UE)

Irreducibility modulo one prime certifies irreducibility over $\QQ$ for monic integer polynomials.Suggest a slogan

Let $f\in \ZZ[x]$ be monic. If there exists a prime $p$ such that the reduction $\bar f\in\FF_p[x]$ is irreducible, then $f$ is irreducible in $\QQ[x]$.

Suppose $f=gh$ with $g,h\in\QQ[x]$ of degree at least $1$. By Gauss’s lemma we may take $g,h$ monic in $\ZZ[x]$. Reducing modulo $p$ gives $\bar f=\bar g\bar h$, and since $g$ and $h$ are monic, $\deg\bar g=\deg g\geq1$ and $\deg\bar h=\deg h\geq1$. This contradicts the irreducibility of $\bar f$.

The converse fails: $x^4+1$ is irreducible in $\QQ[x]$ but reducible modulo every prime $p$. Modulo $2$, $x^4+1=(x+1)^4$. For $p$ odd, at least one of $-1$, $2$, $-2$ is a square modulo $p$, since the product of two nonsquares is a square. If $i^2=-1$ then $x^4+1=(x^2-i)(x^2+i)$; if $c^2=2$ then $x^4+1=(x^2+cx+1)(x^2-cx+1)$; if $c^2=-2$ then $x^4+1=(x^2+cx-1)(x^2-cx-1)$. For a monic irreducible $f\in\ZZ[x]$, the Chebotarev density theorem shows that $\bar f$ is reducible for every prime $p$ exactly when the Galois group of $f$ over $\QQ$ contains no $(\deg f)$-cycle; the Galois group of $x^4+1$ over $\QQ$ is $(\ZZ/2\ZZ)^2$.

Over a finite field $\FF_p$, a polynomial of degree $d$ is irreducible if and only if it has no monic irreducible factor of degree at most $d/2$, which can be checked by dividing by each monic irreducible polynomial of degree at most $d/2$.

(Irreducibility modulo $2$)

$f(x) \da x^4 + x + 1$ is irreducible over $\QQ$. Modulo $2$, $f(0)=f(1)=1$, so $f$ has no linear factor, and the only monic irreducible quadratic over $\FF_2$ is $x^2+x+1$, which leaves remainder $1$ on dividing $f$; hence $f$ is irreducible over $\FF_2$.

Eisenstein's criterion (Tag T-CF6S3)

One prime dividing every lower coefficient but not the leading one, with $p^2\nmid a_0$, forces irreducibility over $\QQ$.Suggest a slogan

Let \[ f(x) = \sum_{i=0}^n a_i x^i = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in \ZZ[x] \] with $n\geq1$, and suppose there is a prime $p$ such that

$p$ divides $a_i$ for $0\leq i\leq n-1$,
$p$ does not divide $a_n$, and
$p^2$ does not divide $a_0$.

Then $f$ is irreducible in $\QQ[x]$. If moreover the coefficients of $f$ have no common prime factor, then $f$ is irreducible in $\ZZ[x]$.

Suppose $f$ is reducible in $\QQ[x]$. By Gauss’s lemma $f=gh$ with $g,h\in\ZZ[x]$ of degrees $r,s\geq1$. Reducing modulo $p$ gives $\bar a_nx^n=\bar g\bar h$ in $\FF_p[x]$ with $\bar a_n\neq0$. The leading coefficients of $g$ and $h$ multiply to $a_n$, so they are not divisible by $p$ and $\deg\bar g=r$, $\deg\bar h=s$. Since $\FF_p[x]$ is a unique factorization domain, $\bar g=bx^r$ and $\bar h=cx^s$ with $b,c\in\FF_p^\times$, so $p$ divides both constant terms $g(0)$ and $h(0)$, and $p^2$ divides $g(0)h(0)=a_0$, a contradiction. If the coefficients of $f$ have no common prime factor, a factorization of $f$ in $\ZZ[x]$ into nonunits has both factors of positive degree, which the first statement excludes.

Eisenstein's criterion (Tag FT-2P5VV)

One prime divides every lower coefficient, misses the leading coefficient, and its square misses the constant term: irreducible.Suggest a slogan

Let $f(x)=\sum_{i=0}^na_ix^i\in\ZZ[x]$ with $n\ge1$, and let $p$ be a prime such that

$p\nmid a_n$,
$p\divides a_i$ for $0\le i\le n-1$, and
$p^2\nmid a_0$.

Then $f$ is irreducible in $\QQ[x]$.

Suppose $f=gh$ with $g,h\in\QQ[x]$ nonconstant. By Gauss' lemma we may take $g,h\in\ZZ[x]$ with $\deg g,\deg h\ge1$. Reducing modulo $p$ gives $\overline g\,\overline h=\overline{a_n}x^n$ in $\FF_p[x]$ with $\overline{a_n}\ne0$. The leading coefficients of $g$ and $h$ multiply to $a_n$, so neither is divisible by $p$, and $\deg\overline g=\deg g\ge1$, $\deg\overline h=\deg h\ge1$. Since $\FF_p[x]$ is a unique factorization domain, $\overline g$ and $\overline h$ are nonzero multiples of positive powers of $x$, so $p$ divides the constant terms of $g$ and $h$. Then $p^2$ divides their product $a_0$, a contradiction.

State Eisenstein's criterion.

(Shifting)

For $a\in\ZZ$, $g(x)\mapsto g(x+a)$ is a ring automorphism of $\QQ[x]$, so if $f(x+a)$ satisfies Eisenstein’s criterion at $p$, then $f$ is irreducible. For monic $f$ of degree $n\geq2$, in that case $f(x+a)\equiv x^n \pmod p$, so $f \bmod p$ has a repeated root, and $p$ divides $\Delta_f = \Delta_{f(x+a)}$; the primes to test are among the prime divisors of the discriminant.

Galois group of an irreducible separable polynomial as a transitive subgroup of $S_n$ (Tag T-AILFB)

An irreducible separable degree-$n$ polynomial has a Galois group acting transitively on its $n$ roots.Suggest a slogan

Let $k$ be a field, let $f\in k[x]$ be irreducible and separable of degree $n$, let $L$ be a splitting field of $f$ over $k$, and let $G\coloneqq\Gal(L/k)$.

$L/k$ is Galois, and $\abs G=[L:k]$.
The action of $G$ on the $n$ roots of $f$ identifies $G$ with a transitive subgroup of $S_n$, well defined up to conjugation, and $n\divides\abs G\divides n!$.

$L$ is the splitting field of the separable polynomial $f$, so $L/k$ is Galois and $\abs G=[L:k]$ (PR-ZCKLJ).
An element of $G$ permutes the roots $\alpha_1,\ldots,\alpha_n$ of $f$ and is determined by this permutation because $L=k(\alpha_1,\ldots,\alpha_n)$; reordering the roots conjugates the image in $S_n$. For each $i$ there is a $k$-isomorphism $k(\alpha_1)\to k(\alpha_i)$ with $\alpha_1\mapsto\alpha_i$, since $f$ is irreducible, and it extends to an element of $G$, so the action is transitive. Then $n=[k(\alpha_1):k]$ divides $[L:k]=\abs G$, and $\abs G$ divides $\abs{S_n}=n!$.

These facts give a procedure for computing $G$. Check that $f$ is irreducible and separable, compute $[L:k]=\abs G$, and list the transitive subgroups of $S_n$ of that order. The remaining candidates are separated by the cycle types that occur in $G$, read off from factorizations of $f$ modulo primes when $k=\QQ$ (PR-XDWOP), or, for cubics, by the discriminant (PR-XPDHE).

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Galois theory