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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.

Such a field embedding pulls the p-adic absolute value back to a finite place of a number field above $p$. It extends to an embedding of the corresponding completion, whose image is a finite extension of the P-adic numbers.

For a finite field extension $L/K$, multiplication by $\alpha \in L$ is a $K$-linear map $m_{\alpha }:L\to L$. Its determinant is the field norm:This does not depend on the chosen basis. If $L/K$ is a separable field extension, thenwhere the product runs over all $[L:K]$ field embeddings into an algebraic closure of $K$. Embeddings are counted even when several give the same value on $\alpha$: for $r\in K$, the result is $r^{[L:K]}$. This avoids ambiguity in the phrase “product of algebraic conjugates”.^[ref]

For $L=\mathbb{Q}(\sqrt{2})$ over $K=\mathbb{Q}$, write $\alpha =a+b\sqrt{2}$. In the basis $1,\sqrt{2}$, multiplication by $\alpha$ has columns $(a,b)$ and $(2b,a)$, henceThe determinant calculation and the product of the two field embeddings give the same answer. In particular,A field norm is an arithmetic invariant; it is not a nonnegative norm measuring length, and need not be positive on nonzero elements.

The field norm is multiplicative, $N(\alpha \beta )=N(\alpha )N(\beta )$, because $m_{\alpha \beta }=m_{\alpha }{m}_{\beta }$ and determinants are multiplicative. It vanishes exactly at zero. For a number field over $\mathbb{Q}$, it takes algebraic integers to integers, and an algebraic integer is a unit exactly when its field norm is $1$ or $-1$.^[ref][ref]

This makes the field norm a useful test for divisibility. If nonzero $\alpha ,\beta \in {\mathcal{O}}_L$ and $\alpha$ divides $\beta$, then the integer $N(\alpha )$ divides $N(\beta )$. For example, $3+\sqrt{2}$ has field norm $7$, so it cannot divide $2$, whose field norm is $4$. It does divide $7$, with quotient $3-\sqrt{2}$. This necessary condition can reject a proposed factor without solving for its quotient.

A number field is a finite field extension $K/\mathbb{Q}$ of the rational numbers. Its degree of a field extension $[K:\mathbb{Q}]$ is its dimension as a vector space over $\mathbb{Q}$. Every element is an algebraic number; the primitive element theorem also says that $K=\mathbb{Q}(\theta )$ for some single algebraic number $\theta$.

For example,The basis $1,\sqrt{2}$ makes arithmetic explicit: use $(\sqrt{2}{)}^2=2$ to reduce products. Every nonzero element has its inverse in the same field:The denominator cannot vanish unless $a=b=0$, since $\sqrt{2}$ is not a rational number. For instance, $(3+\sqrt{2}{)}^{-1}=(3-\sqrt{2})/7$ belongs to the number field, although it is not an algebraic integer.

A degree-$n$ number field has exactly $n$ distinct field embeddings into the complex numbers fixing $\mathbb{Q}$, because every finite field extension in characteristic zero is a separable field extension. For $\mathbb{Q}(\sqrt{2})$ these are the two real mapsBoth preserve addition and multiplication. Keeping both values, rather than just the positive choice of $\sqrt{2}$, is what makes the field norm and the Minkowski embedding of a number field useful.

Let $d_K$ denote the field discriminant, and put $n=[K:\mathbb{Q}]\ge 2$. Let $r_1$ be the number of real field embeddings and $r_2$ the number of conjugate pairs of nonreal field embeddings, so $n=r_1+2r_2$. The Minkowski bound for ideal classes provides a nonzero integral ideal $A$ satisfyingRearranging gives the Minkowski lower bound for a number-field discriminantBecause $2r_2\le n$ and $\pi /4<1$, this is at leastApply the arithmetic-geometric mean inequality to $1,2,\dots ,n$:It follows thatwhere $n/(n+1)\ge 2/3$ for $n\ge 2$ was used. ThusThe source writes $|\Delta (K{)}^2|>1$. If $\Delta (K)$ denotes the embedding determinant, its square is $d_K$; if it denotes the field discriminant, the displayed result also implies the stated inequality. The conclusion therefore does not depend on that notational convention.

To compute the Ideal class group of Q of square root thirty-five, put $\theta =\sqrt{35}$. The ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$, and the field discriminant is $4\cdot 35=140$. Both field embeddings are real, so the Minkowski bound for ideal classes isEvery ideal class therefore has an integral ideal representative of ideal norm at most five. By prime ideal factorization and multiplicativity of the ideal norm, the ideal class group is generated by prime ideals of ideal norm at most five.

Apply the Dedekind factorization theorem to $X^2-35$. At two and five it has a repeated linear factor, givingAt three, $X^2-35\equiv X^2+1$ has no root, so $(3)$ itself is a prime ideal of ideal norm nine, outside the bound. These are all the relevant prime ideals: a prime ideal of ideal norm four would have to lie above two with residue degree two, but the factorization above two has only the single factor of degree one. Consequently $[P_2]$ and $[P_5]$ generate, and each has order of a group element dividing two.

The algebraic integer $5+\theta$ has field norm $25-35=-10$. It belongs to both $P_2$ and $P_5$. Hence its principal ideal is contained in $P_2\cap P_5=P_2{P}_5$. Both have ideal norm ten, soIt follows that $[P_5]=[P_2{]}^{-1}=[P_2]$, and the ideal class group is generated by one element of order of a group element at most two.

Finally, $P_2$ is not a principal ideal. If $P_2=(a+b\theta )$ with $a,b\in \mathbb{Z}$, the equality of ideal norm and absolute field norm would implyModulo five this requires $a^2\equiv 2$ or $3$, whereas squares modulo five are $0,1,4$. Thus $[P_2]\ne 1$, provingThis computes the ordinary ideal class group, with no positivity restriction on generators.

Let $\alpha =\sqrt[3]{3}$. The polynomial $X^3-3$ is Eisenstein at three, so the number field has degree of a field extension three. Use the permitted ring of integers $R=\mathbb{Z}[\alpha ]$, whose integral basis is $1,\alpha ,{\alpha }^2$. The field traces $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\alpha )=\mathrm{Tr}({\alpha }^2)=0$, $\mathrm{Tr}({\alpha }^3)=9$, and $\mathrm{Tr}({\alpha }^4)=0$ giveThere is one real field embedding and one conjugate pair of nonreal field embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class has an integral ideal representative of ideal norm at most four, so it is enough to show that every prime ideal of ideal norm at most four is a principal ideal.

Modulo two,where the quadratic factor is an irreducible polynomial over the finite field ${\mathbb{F}}_2$. The Dedekind factorization theorem givesNowThus $(\alpha -1)=P_2$, by containment and equality of ideal norms, and cancellation in the second identity gives $({\alpha }^2+\alpha +1)=Q_2$. This explicitly handles the prime ideal of ideal norm four, as well as the one of ideal norm two.

Modulo three the polynomial is $X^3$, soThese exhaust the prime ideals of ideal norm at most four. Every bounded representative factors into the listed principal ideals, and hence is a principal ideal. Therefore the Class group of Q of cube root of three is

Interpret the map as the action on ${\mathbb{C}}^n$. A smooth representation requires every vector to have an open stabilizer in the locally profinite group ${\mathrm{GL}}_n(F)$. Take the first standard basis vector $e_1$. Every identity neighbourhood contains matrices $d_a=\mathrm{diag}(1+a,1,\dots ,1)$ with $a\ne 0$ sufficiently small in the local field. We may require $|a|<1$, so $1+a\ne 0$. Injectivity of the field embedding givesThus no identity neighbourhood fixes $e_1$, and its stabilizer is not open. Consequently the representation is not smooth for any $n\ge 1$. This includes $n=1$, when a nonzero vector's stabilizer is just the identity. The field embedding need not be continuous, but no continuity assumption would repair this stabilizer obstruction. The existence of the stated embedding itself forces characteristic zero; in positive characteristic no field embedding into $\mathbb{C}$ exists.

A formally real field is a field in which $-1$ is not a finite sum of squares. The theory of formally real fields consists of the field axioms and, for each $m\ge 1$, the sentenceA real closed field is a formally real field with no proper formally real algebraic extension. An equivalent first-order axiomatization of the Theory of real closed fields adds: every $a$ or its negative is a square, and every odd-degree polynomial has a root. Explicitly the square axiom isand the polynomial axioms say that every monic polynomial of degree $2d+1$ has a root, for each $d\ge 0$.

The Artin-Schreier ordering criterion says that every formally real field can be ordered, and the real closure theorem embeds each ordered field into an algebraic real closed field extension. Consequently every FRF model embeds into an RCF model. Conversely every RCF model is itself an FRF model, so the reverse embedding requirement is automatic.

It remains to establish model completeness. A real closed field has a unique order, definable in the field language byAny field embedding between real closed fields preserves this order: positive elements are nonzero squares, and negative elements have positive negatives. The quantifier elimination for ordered real closed fields theorem makes every such ordered embedding elementary, by the preceding part. Restricting to formulas of the field language makes the original embedding elementary as well. Thus RCF is model-complete in the field language.

The embedding conditions and model completeness proveThe quantifier elimination invoked here is in the ordered language. In the unordered field language, model completeness still holds, but full quantifier elimination does not follow from forgetting the order.

A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.

Apply the Poisson summation formula for a Euclidean lattice to ${\Lambda }_{\mathfrak{b}}$, whose covolume is $1$, and use the preceding description of its dual lattice. The anisotropic theta functional equation isFor comparison, the unnormalized Gaussian theta sumhas the functional equationand its corresponding limit is $C_{\mathfrak{b}}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta (y,\mathfrak{b})=\theta (C_{\mathfrak{b}}^{-2/n}y,\mathfrak{b})$.

For the small-parameter asymptotic of a lattice theta sum, $y\to 0$ means that every coordinate tends to zero. In $\Theta (y^{-1},{\mathfrak{b}}^{\vee })$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\le 1$, all these terms are bounded by the summable Gaussian theta sum ${\sum }_{\lambda \in {\Lambda }_{{\mathfrak{b}}^{\vee }}}{e}^{-\pi |\lambda {|}^2}$. The dominated convergence theorem therefore gives $\Theta (y^{-1},{\mathfrak{b}}^{\vee })\to 1$. Using the anisotropic theta functional equation,The condition that every coordinate tends to zero matters; $\Vert y\Vert \to 0$ alone would not justify this argument.

Here a divisor is a modulus of a number field, rather than an arbitrary real-weighted divisor. Writewhere ${\mathfrak{c}}_0$ is a nonzero integral ideal, the finite multiplicities $m_{\mathfrak{p}}$ are nonnegative integers, and ${\mathfrak{c}}_{\infty }$ is a set of real Archimedean places. The multiplicity of a place in a modulus is $m_v(\mathfrak{c})=m_{\mathfrak{p}}$ at the finite place corresponding to $\mathfrak{p}$, is $1$ at a real place in ${\mathfrak{c}}_{\infty }$, and is $0$ at every other infinite place. In particular complex places have multiplicity $0$.

Let $I_{\mathfrak{c}}$ be the group of nonzero fractional ideals prime to ${\mathfrak{c}}_0$. PutLet $P_{\mathfrak{c}}$ consist of the principal ideals $(a)$ with $a\in K_{\mathfrak{c},1}^{\times }$. The generalized ideal class group is the ray class groupLet $U={\mathcal{O}}_K^{\times }$ be the unit group, let $U_{\mathfrak{c}}=U\cap K_{\mathfrak{c},1}^{\times }$ be the group of ray units, and define the residue and signature group of a modulusThe first factor is omitted when ${\mathfrak{c}}_0={\mathcal{O}}_K$. There is a group homomorphism $\rho :U\to G_{\mathfrak{c}}$ recording the unit's finite residues and its signs. Its kernel is $U_{\mathfrak{c}}$.

Let $P^{\mathfrak{c}}$ consist of all principal fractional ideals prime to ${\mathfrak{c}}_0$, and put $R_{\mathfrak{c}}=P^{\mathfrak{c}}/P_{\mathfrak{c}}$. The two short exact sequences areandFor the first short exact sequence, weak approximation for number fields realizes every choice of finite unit residues and real signs by some $a\in K^{\times }$ prime to ${\mathfrak{c}}_0$. Send that data to the class of $(a)$ in $R_{\mathfrak{c}}$. Changing $a$ without changing its residues and signs multiplies it by an element of $K_{\mathfrak{c},1}^{\times }$, so the map is well-defined. Its kernel consists exactly of data arising from units: if $(a)=(b)$ with $b\in K_{\mathfrak{c},1}^{\times }$, then $a/b\in U$. This gives $R_{\mathfrak{c}}\cong G_{\mathfrak{c}}/\rho (U)$. For the second short exact sequence, forget the ray conditions. Its kernel is $R_{\mathfrak{c}}$, and weak approximation for number fields gives a representative prime to ${\mathfrak{c}}_0$ for every ideal class. These are the two parts of the ray class exact sequence.

For $K=\mathbb{Q}(\sqrt{15})$, the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{15}]$. The finite modulus is trivial and both real places occur, so $H_{\mathfrak{c}}$ is the narrow ideal class group and $G_{\mathfrak{c}}=\{\pm 1{\}}^2$. The two field embeddings send $\sqrt{15}$ to $\sqrt{15}$ and $-\sqrt{15}$. The given unit $ϵ=4+\sqrt{15}$ is positive at both places, since $4-\sqrt{15}>0$; the unit $-1$ is negative at both. Thus the unit signature map has imageThe given ideal class group has order $2$. The second short exact sequence therefore givesTo determine the group structure, retain the ideal $\mathfrak{p}=(2,1+\sqrt{15})$ that generates the ordinary ideal class group. We have ${\mathfrak{p}}^2=(2)$: all generators $4$, $2(1+\sqrt{15})$ and $(1+\sqrt{15}{)}^2$ lie in $(2)$, whileputs $2$ in ${\mathfrak{p}}^2$. Since $2$ is totally positive, $[\mathfrak{p}{]}^2=1$ in the narrow ideal class group. Its image in the ordinary ideal class group is nontrivial, so $[\mathfrak{p}]$ has order exactly $2$.

The class of the principal ideal $(\sqrt{15})$ is a nontrivial element of $R_{\mathfrak{c}}$. Its two signs are $(+,-)$, and multiplying by a unit can only reverse both signs or neither, so no generator of this ideal is totally positive. Its square is $(15)$, which does have a totally positive generator. Thus $[(\sqrt{15})]$ is another element of order $2$, distinct from $[\mathfrak{p}]$ because its ordinary ideal class is trivial. These two elements are independent and generate all four classes. ConsequentlyThe nontrivial ordinary ideal class already has a lift of order $2$, so the extension in the second short exact sequence splits; it cannot be cyclic of order $4$.

A suitable version of the Hensel lemma is this: if $R$ is a complete discrete valuation ring, $h\in R[X]$, and $\overline{a}$ is a root of $\overline{h}$ with ${\overline{h}}^{\prime }(\overline{a})\ne 0$, there is a unique $a\in R$ reducing to $\overline{a}$ with $h(a)=0$. Starting with any lift, Newton iteration over a valued field $a_{j+1}=a_j-h(a_j)/h^{\prime }(a_j)$ keeps the derivative a unit and doubles the error valuation, giving convergence. Uniqueness follows by factoring $h(u)-h(v)$ when two roots have the same residue.

The field in the rest of this question is not assumed complete, so that version cannot simply be applied to it. Instead use the stated uniqueness of the extension of an absolute value. Let ${\alpha }_i,{\alpha }_j$ be any two roots of a polynomial $f$, which is irreducible. The field embedding sending ${\alpha }_i$ to ${\alpha }_j$ identifies $K({\alpha }_i)$ with $K({\alpha }_j)$. Pulling the latter field's absolute value back gives an extension on $K({\alpha }_i)$, which must coincide with the given one. More generally, for every $h\in K[X]$,This proves equal absolute values of algebraic conjugates without assuming separability or a transitive action of a Galois group; repeated roots cause no difficulty.

Every root of the monic $f\in {\mathcal{O}}_K[X]$ has absolute value at most one. Indeed, if $|\alpha |>1$, then each lower term satisfies $|c_j{\alpha }^j|\le |\alpha {|}^j<|\alpha {|}^n$. The ultrametric inequality makes the leading term dominate their sum, contradicting $f(\alpha )=0$. Thus

All roots lie in the valuation ring ${\mathcal{O}}_M$, so their residues are defined in the residue field $k_M$. Let $\phi \in k_K[X]$ be the monic minimal polynomial of an algebraic element of ${\overline{\alpha }}_1$; this residue is algebraic because it satisfies the monic $\overline{f}$. Choose a coefficientwise lift $\Phi \in {\mathcal{O}}_K[X]$. Then $|\Phi ({\alpha }_1)|<1$, and the conjugacy argument gives $|\Phi ({\alpha }_i)|<1$ for every $i$. Hence every ${\overline{\alpha }}_i$ is a root of $\phi$.

Reduction of the splitting-field factorization, with multiplicities, givesEvery monic irreducible factor of $\overline{f}$ over $k_K$ has a root among these residues. That root also satisfies $\phi$, so its minimal polynomial is $\phi$. Unique factorization therefore proves the pure-power reduction of a monic irreducible polynomial:This does not assert that $\phi$ is separable or that the reduction is square-free.

Finally, factor the monic $g$ over $K$ as $g={\prod }_j{h}_j^{e_j}$ with distinct monic irreducible polynomials $h_j$. The same leading-term argument bounds every root of $g$ by one. Coefficients of any $h_j$ are elementary symmetric polynomials in some of these roots, so they lie in ${\mathcal{O}}_K$. By the preceding result each ${\overline{h}}_j$ is a power of a single monic irreducible residue polynomial ${\phi }_j$.

Since the prescribed ${\overline{g}}_1,{\overline{g}}_2$ are coprime, every ${\phi }_j$ occurs on exactly one side. Assign the entire factor $h_j^{e_j}$ to that side and form the two products. Comparing irreducible-factor multiplicities in the reduction proves the coprime factor lifting from valuation-extension uniqueness:Empty products are one, covering constant prescribed factors. The construction proves the required factor-lifting form of the Hensel lemma directly from extension uniqueness, without adding completeness as an unstated hypothesis.

Write $\alpha =\sqrt[n]{m}$ and $\zeta =e^{2\pi i/n}$. The $n$ field embeddings of $L$ into $\mathbb{C}$ send $\alpha$ to ${\zeta }^j\alpha$, so the field trace isFor the trace pairing $T(x,y)={\mathrm{Tr}}_{L/\mathbb{Q}}(xy)$, the basis ${\beta }_i={\alpha }^i$ therefore has dual basis

Every element of $\mathbb{Z}[\alpha ]$ is an algebraic integer. If $x\in {\mathcal{O}}_L$ and $z\in \mathbb{Z}[\alpha ]$, then $xz$ is an algebraic integer and its trace is an integer. Thus $x$ belongs to the trace-dual lattice of $\mathbb{Z}[\alpha ]$, which is generated over $\mathbb{Z}$ by the displayed dual basis. Since $1/n=m/(nm)$, every such linear combination lies in $(nm{)}^{-1}\mathbb{Z}[\alpha ]$. Hence

Suppose otherwise and choose a relationwith the least possible number $m$ of nonzero coefficients. After relabelling and scaling, take every ${\lambda }_i\ne 0$ and ${\lambda }_m=1$. Since the field embeddings ${\sigma }_1$ and ${\sigma }_m$ are distinct, there is a $y\in K$ such that ${\sigma }_1(y)\ne {\sigma }_m(y)$.

Apply the relation to $xy$ and subtract ${\sigma }_m(y)$ times the relation at $x$. Because each field embedding is multiplicative, this givesThe $m$th term has disappeared, while the coefficient of ${\sigma }_1$ is nonzero. This is a shorter nontrivial relation, contradicting minimality. Hence distinct embeddings are linearly independent as functions. This is the Artin independence theorem.

A field extension $L/K$ is a splitting field for $f\in K[X]$ when $f$ splits into linear factors over $L$ and $L$ is generated over $K$ by the roots of $f$. Equivalently, it is the smallest extension inside an algebraic closure over which $f$ splits.

To prove existence, choose an irreducible factor of $f$, adjoin one of its roots, and factor $f$ over the resulting finite extension. If a nonlinear factor remains, adjoin one of its roots and repeat. Each step lowers the total degree of the unsplit factors, so after finitely many finite extensions the polynomial splits. The field generated by the roots obtained is therefore a splitting field.

The precise uniqueness theorem says that if $L$ and $L^{\prime }$ are splitting fields of the same polynomial $f$ over $K$, then there is a field embedding isomorphismwhose restriction to $K$ is the identity. More generally, an isomorphism of base fields carrying one polynomial to another extends to an isomorphism of their splitting fields. Thus splitting fields are unique up to base-field isomorphism; inside a fixed algebraic closure, the field generated by all the roots is literally unique. These facts are summarized by existence and uniqueness of splitting fields.

LetThe roots of $X^3-2$ are $\alpha ,\omega \alpha ,{\omega }^2\alpha$, sois the splitting field of x cubed minus two. As a subfield of $\mathbb{C}$ it is unique, because any splitting field in $\mathbb{C}$ must contain all three roots and hence must contain both $\alpha$ andminimality then forces it to equal $\mathbb{Q}(\alpha ,\omega )$.

Because $\mathrm{char}K=0$, the finite extension $M/K$ is separable. By the primitive element theorem, write $M=K(\theta )$. The minimal polynomial of $\theta$ over $K$ has a root in the algebraically closed field $L$, so sending $\theta$ to that root defines a $K$-field embedding $\iota :M\hookrightarrow L$.

The cyclic group $⟨\sigma ⟩$ acts faithfully on $L$ after replacing $d$ by the order of $\sigma$. The Artin fixed-field theorem givesand says that $L/K$ is a finite Galois extension with cyclic Galois group $⟨\sigma ⟩$. The image $E=\iota (M)$ is an intermediate field. Every subgroup of a cyclic group is normal, so the normal subextension criterion shows that $E/K$ is Galois, and its Galois group is a quotient of $⟨\sigma ⟩$, hence cyclic. Transporting this structure through $\iota$ proves that $M/K$ is Galois with cyclic Galois group. This is the finite extensions of the fixed field of a finite-order automorphism of an algebraically closed field theorem.

Algebraic closedness is necessary. Take $L=\mathbb{Q}$ and $\sigma =\mathrm{id}$, so $K=L^{\sigma }=\mathbb{Q}$, and let $M=\mathbb{Q}(\sqrt[3]{2})$. The extension has degree three but is not normal, since the two nonreal roots of $X^3-2$ do not belong to $M$. It is therefore not Galois.

A quadratic number field is a number field of degree of a field extension two, necessarily of the form $\mathbb{Q}(\sqrt{d})$ for a square-free integer $d\ne 0,1$. Its two field embeddings send $a+b\sqrt{d}$ to $a\pm b\sqrt{d}$. If $d>0$, both images lie in the real numbers, giving a totally real number field; if $d<0$, they form a pair related by complex conjugation, giving an imaginary quadratic field. The ring of integers of a quadratic field depends on $d$ modulo four.^[ref]

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

A nonzero element of a number field is totally positive when its image is positive under every real field embedding. In a totally real number field, the principal fractional ideals having a totally positive generator form the subgroup used to define the narrow ideal class group. Signs of different generators are related by the unit signature map.

The unit signature map records the signs of a unit under all real field embeddings. Its kernel consists of totally positive units. Its image determines the kernel of the homomorphism from the narrow ideal class group to the ordinary ideal class group: this kernel is $\{\pm 1{\}}^{r_1}/\mathrm{sgn}({\mathcal{O}}_K^{\times })$.