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A set $S$ of prime numbers has Dirichlet density $\delta$ if the following limit exists:Equivalently, the denominator can be ${\sum }_p{p}^{-s}$, since that sum is $\log (1/(s-1))+O(1)$. Adding or removing finitely many prime numbers leaves the Dirichlet density unchanged.

Put $\alpha =\sqrt[4]{2}>0$ and $L=\mathbb{Q}(\alpha ,i)$. This is the splitting field of $X^4-2$, whose four roots of a polynomial are $\alpha ,i\alpha ,-\alpha ,-i\alpha$. The polynomial is an Eisenstein polynomial at $2$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=4$. Since $\mathbb{Q}(\alpha )$ is contained in the real numbers, it does not contain $i$, giving $[L:\mathbb{Q}]=8$. Thus $L/\mathbb{Q}$ is a Galois extension of degree $8$. Its Galois group is the dihedral group of order $8$: the field automorphisms $r(\alpha )=i\alpha$, $r(i)=i$ and $s(\alpha )=\alpha$, $s(i)=-i$ satisfy $r^4=s^2=1$ and $srs=r^{-1}$.

For an odd prime number $p$, the condition $p\equiv 1(\mathrm{mod}4)$ means that the finite field ${\mathbb{F}}_p$ contains all fourth roots of unity. Under this condition, $2$ is a quartic residue exactly when $X^4-2$ has a root in ${\mathbb{F}}_p$; multiplying that root by $1,i,-1,-i$ gives all four distinct roots. Conversely, complete splitting of $X^4-2$ over ${\mathbb{F}}_p$ gives both a fourth root of $2$ and a primitive fourth root of unity, so it also forces $p\equiv 1(\mathrm{mod}4)$.

The polynomial discriminant of $X^4-2$ is $-2^{11}$. Hence every odd prime number is unramified in $L$, and the root-splitting criterion is equivalent to its Frobenius automorphism acting trivially on all the roots. Since the roots generate $L$, this is equivalent to the Frobenius automorphism being the identity, or to $p$ being a completely split prime of $L/\mathbb{Q}$. The Chebotarev density theorem says that the unramified prime numbers with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|\mathrm{Gal}(L/\mathbb{Q})|$. Here $C=\{1\}$, so

A splitting field of $f$ over $K$ is an extension generated over $K$ by all its roots in which $f$ splits into linear factors. A nonconstant irreducible polynomial is separable polynomial when its roots in a splitting field are distinct, equivalently when it is coprime to its formal derivative. For a general polynomial, the convention compatible with the printed assertion is that all its irreducible factors are separable; repeated occurrences of the same factor are allowed under this convention.

For a finite field of characteristic $p$, the finite-field Frobenius automorphism $x\mapsto x^p$ is injective and hence surjective. If an irreducible polynomial $g$ had zero derivative, every nonzero exponent would be divisible by $p$. Taking $p$th roots of its coefficients would give $g=h^p$, contradicting irreducibility. Thus $g^{\prime }\ne 0$, and irreducibility gives $\gcd (g,g^{\prime })=1$. Every irreducible factor is therefore separable: finite fields are perfect.

If instead one defines a separable polynomial to mean that the entire polynomial has no repeated roots, the printed claim for every $f$ is false: $(t-1{)}^2$ is a counterexample over any finite field. The proof above establishes the precise intended perfect-field assertion; a polynomial has entirely distinct roots if and only if it is also squarefree.

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection between subgroups $H\le \mathrm{Gal}(L/K)$ and intermediate fields $M$, by $H\mapsto L^H$ and $M\mapsto \mathrm{Gal}(L/M)$, for a finite Galois extension. One has $[L:L^H]=|H|$ and $[L^H:K]=[\mathrm{Gal}(L/K):H]$. The field $L^H$ is Galois over $K$ exactly when $H$ is normal; then its Galois group is the quotient by $H$.

Let $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The splitting field is $L=\mathbb{Q}(\alpha ,\zeta )$. A splitting field in characteristic zero is normal and separable, hence Galois. Eisenstein criterion at $2$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This field is real, whereas $\zeta$ is not, and $\zeta$ satisfies a quadratic polynomial. Thus $[L:\mathbb{Q}]=6$. The faithful permutation action on the three roots identifies the Galois group with $S_3$. The subgroup fixing $\alpha$ is generated by the transposition interchanging the other two roots. Its conjugates are different transposition subgroups, so it is not normal.

Label the distinct roots ${\alpha }_1,\dots ,{\alpha }_n$. Define $\tau (\sigma )$ by $\sigma ({\alpha }_i)={\alpha }_{\tau (\sigma )(i)}$. Composition respects this rule, so it is a group homomorphism. If its image permutation is the identity, $\sigma$ fixes all the roots; they generate $L$, so $\sigma$ is the identity. Thus $\tau$ is injective.

For $f=t(t^2-2)$, label the roots $0,\sqrt{2},-\sqrt{2}$. The splitting field is $\mathbb{Q}(\sqrt{2})$ and its nontrivial automorphism swaps precisely roots $2,3$. Hence $\boxed{\mathrm{im}\tau =⟨(23)⟩}$.

For $f=t^3-2$, Eisenstein criterion gives irreducibility. The real root $\alpha =\sqrt[3]{2}$ generates a degree-three real field. Adjoining a primitive cube root of unity ${\zeta }_3$ gives a further extension of degree two, since it is nonreal and satisfies $t^2+t+1=0$. Thus its splitting field has degree $6$. In characteristic zero this is a Galois extension, so its Galois group has order $6$. The injection into $S_3$ is therefore an isomorphism.

If $f$ is an irreducible polynomial, any two roots $\alpha ,\beta$ have the same minimal polynomial. Thus the map $\mathbb{Q}(\alpha )\to \mathbb{Q}(\beta )$ carrying $\alpha$ to $\beta$ is a field embedding. The extension theorem for field embeddings extends it to an embedding of the splitting field $L$ into an algebraic closure. Since $L/\mathbb{Q}$ is normal, its image is $L$, so it is an element of the Galois group.

Conversely, factor $f$ into irreducible factors over $\mathbb{Q}$. An automorphism preserves the vanishing of each factor, so it cannot send a root of one factor to a root of another. Since there are no repeated roots, different factors have disjoint root sets. Transitivity on all roots therefore forces a single irreducible factor. This proves the equivalence.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the Going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The Going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the Going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, let $G=\mathrm{Gal}(L/K)$, and normalize the discrete valuation by $v_L(L^{\times })=\mathbb{Z}$. The lower ramification numbering isHere $G_0$ is the inertia group and $G_1$ is the wild inertia group. For real $s\ge -1$, put $G_s=G_{\lceil s\rceil }$; thus $G_s=G_0$ for $-1<s\le 0$. Define the Herbrand function and its inverse byextending $\phi (s)=s$ on $[-1,0]$. The upper ramification numbering is $G^u=G_{\psi (u)}$ for $u\ge -1$.

Let ${\alpha }^p-\alpha =t^{1-p}$ over $K={\mathbb{F}}_p((t))$, with $t$ its uniformizer. The Artin–Schreier polynomial has derivative $-1$ in characteristic $p$, so all its roots are distinct. If $\alpha$ is one root and $\beta$ another, then $(\beta -\alpha {)}^p=\beta -\alpha$, so $\beta -\alpha \in {\mathbb{F}}_p$. Conversely every $\alpha +a$, $a\in {\mathbb{F}}_p$, is a root. Thus $L=K(\alpha )$ contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form ${\sigma }_a(\alpha )=\alpha +a$, and its Galois group embeds in the additive group of ${\mathbb{F}}_p$.

There is no root in $K$. If $v_K(z)\ge 0$, then $z^p-z$ cannot have negative valuation. If $v_K(z)<0$, the ultrametric inequality gives $v_K(z^p-z)=p{v}_K(z)$, divisible by $p$, whereas $v_K(t^{1-p})=1-p$ is not. The Galois group is consequently nontrivial. Its order divides the prime $p$, so

Let $e=e(L/K)$ be the ramification index. The equation forces $v_L(\alpha )<0$, and henceAs $\gcd (p,p-1)=1$, $p\mid e$. Since $e\le [L:K]=p$, we get $e=p$, residue-field degree one, and $v_L(\alpha )=1-p$. Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, $\sigma \in G_i$ exactly when $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$. Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at $p-1$, and the lower groups areThis includes $p=2$, where the break is one. There is no off-by-one shift: the condition is $v_L(\sigma {\pi }_L-{\pi }_L)\ge i+1$.

For the upper groups, compute the Herbrand function explicitly:Hence the upper break is also $p-1$ and, with the stated real-index convention,As a consistency check, ${\pi }_L$ satisfies the Eisenstein polynomial $X^p-t^{p-1}X-t$. Its derivative in characteristic $p$ is $-t^{p-1}$, giving different exponent $p(p-1)$, equal to ${\sum }_{i\ge 0}(|G_i|-1)=p(p-1)$ from the different exponent from ramification groups.

A splitting field of a nonzero polynomial $f$ over $K$ is an extension in which $f$ splits into linear factors and which is generated over $K$ by all its roots. For a constant nonzero polynomial the splitting field is $K$; the zero polynomial does not have a root-generated splitting field in this standard definition.

Given two splitting fields $L,L^{\prime }$, enumerate the roots generating $L$ and extend a $K$-embedding one root at a time. If an embedding is defined on $K({\alpha }_1,\dots ,{\alpha }_j)$, the minimal polynomial of the next root divides $f$ over that field. Its image under the embedding therefore divides $f$ over the corresponding subfield of $L^{\prime }$, and has a root in $L^{\prime }$. The quotient construction $K(\alpha )\cong K[t]/(m_{\alpha })$ extends the embedding to that root. The polynomial need not be a separable polynomial.

At the end this gives a $K$-embedding $L\to L^{\prime }$. It sends the distinct roots of $f$ injectively to its roots, hence permutes the finite set of distinct roots and contains every root in its image. Since these generate $L^{\prime }$, the embedding is onto. Thus the splitting field is unique up to a (K)-isomorphism.

Write $|F|=p^n$, where $n=[F:{\mathbb{F}}_p]$ since $F$ is a finite-dimensional vector space over its prime field. Every nonzero element satisfies $x^{p^n-1}=1$ by Lagrange theorem, so $F$ is exactly the splitting field of $X^{p^n}-X$ over ${\mathbb{F}}_p$. The derivative is $-1$, making it a separable polynomial. Thus the extension is a normal field extension and a separable field extension, hence a Galois extension.

The finite-field Frobenius automorphism $\sigma (x)=x^p$ fixes ${\mathbb{F}}_p$ and has ${\sigma }^n=1$. If ${\sigma }^d=1$ with $0<d<n$, all $p^n$ elements would be roots of the degree-$p^d$ polynomial $X^{p^d}-X$, impossible. It has order $n$, equal to the degree of the Galois extension. Therefore