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Let a monic integer polynomial have distinct roots in its splitting field. Outside the finitely many primes dividing its polynomial discriminant or ramifying in that field, its roots reduce to distinct residue roots. The Frobenius automorphism permutes them by $a\mapsto a^p$; a reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed. Distinctness of reduction identifies this with a fixed root of the Galois permutation itself. Combining this observation with a derangement in a transitive group action and the Chebotarev density theorem proves that an irreducible polynomial of degree at least two has rootless reduction at a positive Dirichlet density of primes. The degree bound is essential: every monic linear integer polynomial has a root modulo every prime.

For the root-discriminant convention in the requested identity, defineThe Vandermonde determinant is $\det V(\alpha )={\prod }_{i<j}({\alpha }_j-{\alpha }_i)$. Squaring removes the signs, provingIt is nonzero because the roots are distinct. It lies in $F$, since its root expression is fixed by every Galois automorphism and $K^G=F$.

If $f$ is not monic and has leading coefficient $c$, the usual polynomial discriminant is $c^{2n-2}{\prod }_{i<j}({\alpha }_i-{\alpha }_j{)}^2$. Thus the printed Vandermonde equality uses the discriminant of the monic normalization $f/c$. The extra factor is itself a nonzero square in $F$, so it does not alter the square-class criterion in the next part.

We prove the Mahler alternative in full. Let $f(z)={\sum }_{n\ge 0}{z}^{l^n}$. Its series converges locally uniformly in the unit disc and satisfiesIt is transcendental as a function. At a root of unity $\zeta$ with ${\zeta }^{l^r}=1$, the radial tail ${\sum }_{n\ge r}{t}^{l^n}$ in $f(t\zeta )$ is real and tends to infinity as $t\uparrow 1$, while the initial terms remain bounded. These roots are dense on the unit circle. An algebraic function can have singularities only at the finitely many zeros of its leading coefficient and polynomial discriminant: elsewhere its defining polynomial and the implicit-function theorem give local analytic branches. Hence $f$ cannot be algebraic over $\mathbb{C}(z)$.

Assume now that $\alpha$ and $f(\alpha )$ are algebraic, with $0<|\alpha |<1$, and put both in a number field $K$ of degree $d$. For each $N$, the $(N+1{)}^2$ coefficients of a polynomial $P_N(X,Y)$ of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of $P_N(z,f(z))$. The equations have integer coefficients, so a nonzero integer solution exists after clearing denominators. Functional transcendence makes $R_N(z)=P_N(z,f(z))$ nonzero. Its order $T_N$ at zero satisfies $T_N\ge (N+1{)}^2-1$.

At ${\alpha }_r={\alpha }^{l^r}$ the functional equation gives $f({\alpha }_r)=f(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}\in K$. Thus ${\eta }_r=R_N({\alpha }_r)\in K$ is nonzero for all sufficiently large $r$ andUse the absolute logarithmic height $h$. The elementary inequalities for sums and polynomial evaluation giveHere $h(f({\alpha }_r))\le h(f(\alpha ))+(l^r-1)h(\alpha )/(l-1)+r\log 2$. The Liouville height inequality gives $\log |{\eta }_r|\ge -dh({\eta }_r)$. Divide by $l^r$ and let $r\to \infty$ to obtainThe left grows quadratically in $N$, the right only linearly. Choose $N$ large to contradict this. Hence every such algebraic $\alpha$ has transcendental $f(\alpha )$. This is the auxiliary-value height argument for the Fredholm series.

For the rational continuation, let $a_n={\sum }_{j=0}^n(p/q{)}^{l^j}$ and $Q_n=q^{l^n}$. This is a rational number with reduced denominator $B_n\le Q_n$, and its positive tail satisfies, for large $n$,A rational value $a/b$ would instead have distance at least $1/(b{Q}_n)$ from each distinct $a_n$, already impossible. If the value were algebraic irrational, choose $ϵ>0$ with $2+ϵ<\kappa$. The distinct truncations have unbounded reduced denominators, and the displayed errors are eventually smaller than $B_n^{-2-ϵ}$. This contradicts Roth theorem. Thus the requested rational values are transcendental by the Roth criterion for rational Fredholm values.

For completeness, the elliptic alternative's conclusion also has a short auxiliary-divisor certificate. If $\Phi$ is Frobenius on $E/{\mathbb{F}}_q$, then $\mathrm{deg}\Phi =q$ and $\mathrm{deg}(1-\Phi )=N=\#E({\mathbb{F}}_q)$: the latter map is separable and its kernel is exactly the rational points. The divisor identity for $x(P)-x(Q)$ on $E\times E$ gives the parallelogram rule $\mathrm{deg}(u+v)+\mathrm{deg}(u-v)=2\mathrm{deg}u+2\mathrm{deg}v$. Polarizing it yieldsAll these degrees are nonnegative. In particular $\mathrm{deg}(-t+2\Phi )=4q-t^2\ge 0$, so $|t|\le 2\sqrt{q}$. The Frobenius characteristic polynomial is $X^2-tX+q$; its complex roots have modulus $\sqrt{q}$, including the double-root case. The zeta function has numerator $1-tT+q{T}^2$, whose zeros consequently have modulus $q^{-1/2}$, the elliptic Riemann hypothesis. This supplements the fully proved Mahler branch with the degree-form proof of the Hasse bound; it is an auxiliary-divisor proof, not a claim that an unspecified transcendence theorem proves the bound.

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

The squared Vandermonde product defining the polynomial discriminant is symmetric in all roots. By the fundamental theorem of symmetric polynomials, it is a polynomial in the elementary symmetric functions, which are the signed coefficients of the monic polynomial. Hence it is a polynomial function of those coefficients over the base field.

For $f=X^3+pX+q$, $\Delta (f)=-\mathrm{Res}(f,f^{\prime })$. Eliminating the two leading rows in the Sylvester determinant reduces the resultant toThusFor $X^3-3X+1$, the only possible rational roots are $\pm 1$, and neither is a root. The cubic is therefore irreducible, so its Galois group is a transitive subgroup of $S_3$. Its discriminant is $81=9^2$. The Vandermonde product changes sign under odd permutations, and its square being a rational square implies that the product itself is rational; hence all Galois permutations are even. The only transitive subgroup of $A_3$ is $A_3$ itself. Consequently, writing $L$ for its splitting field,

Put $\alpha =44^{1/11}$ and $L=\mathbb{Q}(\alpha )$. The Eisenstein criterion at $11$ shows that $X^{11}-44$ is an irreducible polynomial, so $[L:\mathbb{Q}]=11$. Its derivative is $11X^{10}$, and the polynomial discriminant isThe discriminant-index formula for an integral lattice implies that the field discriminant has no prime divisor outside $\{2,11\}$. By the discriminant obstruction to ramification, every other prime is unramified.

To prove that neither candidate disappears through an index correction, let $\mathfrak{p}$ lie above $p\in \{2,11\}$ and normalize its valuation to have value group $\mathbb{Z}$. If $e=e_{\mathfrak{p}/p}$, the relation ${\alpha }^{11}=44$ givesIn both cases $11\mid e$. The fundamental identity for prime decomposition, ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}/p}{f}_{\mathfrak{p}/p}=11$, now forces exactly one prime ideal above each candidate, with $e=11$ and $f=1$. This is ramification forced by a coprime root valuation; at $2$ it avoids an invalid direct application of the Eisenstein criterion, since $4\mid 44$. Hence

The polynomial discriminant satisfiesA polynomial has no repeated roots over the algebraic closure exactly when it is coprime to its derivative. Consequently the assumed squarefree reduction has nonzero resultant with its derivative, so $p\nmid \mathrm{disc}(f)$. On the other hand the power basis is a basis of the order in a number field $\mathbb{Z}[\alpha ]$, and the discriminant-index formula for an integral lattice givesHere the field discriminant $d_L$ is an integer. ThereforeThis is squarefree reduction excludes a prime from the power-basis index; its argument establishes the index condition before applying the Kummer-Dedekind theorem.

Realize the two-dimensional irreducible representation $\rho$ as the standard representation of the symmetric group onwith $S_3$ permuting coordinates. It is the complement of the trivial summand in the three-point permutation representation. For each prime the local factor of an Artin L-function iswhere $I_p$ is the inertia group and Frobenius means any lift of the residue-field Frobenius automorphism. On $V^{I_p}$ the lift is independent of that choice. At an unramified prime, put $t=p^{-s}$. The Local Artin factors of the standard representation of S3 areThe polynomial discriminant of $X^3-3$ is $-243=-3^5$, so only $3$ can ramify in its splitting field. For $p\ne 3$, the squarefree factorization of $X^3-3$ modulo $p$ determines the Frobenius cycle type on its roots.

At $p=2$,The quadratic is irreducible, so Frobenius is a transposition andThus $a_2=0$, $a_4=1$ and $a_8=0$. At $p=5$,The quadratic discriminant is $3(\mathrm{mod}5)$, a nonsquare, so this is again a transposition. Consequently $L_5(\rho ,s)=(1-5^{-2s}{)}^{-1}$ and $a_5=0$.

At $p=7$, the possible cubes are $0,1,-1$, so $X^3-3$ has no root modulo $7$. Being cubic, it is irreducible; Frobenius is a three-cycle. Thereforeand $a_7=-1$.

The ramified factor at $3$ must be computed on inertia group invariants. Write $\alpha =3^{1/3}$ and set $\pi =({\zeta }_3-1)/\alpha$. Using $({\zeta }_3-1{)}^2=-3{\zeta }_3$ givesThe Eisenstein polynomial $X^6+3$ over ${\mathbb{Q}}_3$ has degree six, so ${\mathbb{Q}}_3(\pi )$ is a totally ramified degree-six subfield of ${\mathbb{Q}}_3({\zeta }_3,\alpha )$. The latter has degree at most $[F:\mathbb{Q}]=6$, so the fields agree. This Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3 shows $D_3=I_3=S_3$. Since the only permutation-invariant vectors in ${\mathbb{C}}^3$ have all coordinates equal, their intersection with $V$ is zero. ThusIn particular $a_3=a_9=0$; an unramified degree-two factor at $3$ would give the wrong coefficients.

Finally the Euler product makes its Dirichlet series coefficients multiplicative for coprime indices. Its constant coefficient is $a_1=1$, and $a_6=a_2{a}_3=0$, $a_{10}=a_2{a}_5=0$. Every integer up to ten is now covered, giving

The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.

We use the Dirichlet density form of the Chebotarev density theorem, the version directly proved by the permitted information at $s=1$. If $L/K$ is a finite Galois extension of number fields with group $G$ and $C$ is a conjugacy class, let $S_C$ be the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. ThenThe same conclusion holds for a union of conjugacy classes, replacing $|C|$ by the size of that union. The denominator may equivalently be the sum over all primes of $K$. Here and below $s>1$ is real. This specifies the density being proved; an unweighted prime-counting asymptotic is a stronger formulation and is not being inferred merely from a limit at one.

First establish the necessary pole for a general base field. The Dedekind zeta function as a permutation Artin L-function gives it using exactly the allowed assumptions. Take a finite normal closure $E$ of $K/\mathbb{Q}$, and let $W$ be the permutation representation on ${\mathrm{Hom}}_{\mathbb{Q}}(K,E)$. Its trivial constituent has multiplicity one: an invariant function on the transitive set of embeddings is constant. At a rational prime, a basis of $W^I$ consists of the sums over inertia orbits of embeddings. Frobenius automorphism permutes these orbit sums. Within the decomposition-group orbit belonging to a prime $\mathfrak{p}$ of $K$, the resulting cycle has length $f_{\mathfrak{p}}$, because the residue embeddings are permuted cyclically by finite-field Frobenius. A cycle of length $f$ has determinant $\det (1-T\rho (\mathrm{Frob}))=1-T^f$. Consequently the local factor of an Artin L-function givesincluding at ramified primes. Decomposing $W=1\oplus {⨁}_{\psi \ne 1}{m}_{\psi }\psi$ givesAll factors after $\zeta$ are holomorphic and nonzero at one by the allowed hypothesis. Thus ${\zeta }_K$ has a simple pole there andThis argument also explains why the trivial relative Artin L-function is ${\zeta }_K$, rather than always the Riemann zeta function.

Now let $\chi$ be an irreducible character of $G$, of degree $d_{\chi }$. At an unramified prime the Euler product has local factor $\det (1-{\rho }_{\chi }(g_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}{)}^{-1}$, where $g_{\mathfrak{p}}$ is a Frobenius representative. The eigenvalues of ${\rho }_{\chi }(g)$ are roots of unity, since $g$ has finite order. Expanding the logarithm for $s>1$ therefore giveswhere the finitely many ramified local factors contribute a function bounded as $s\downarrow 1$. They are nonzero and finite near one because their eigenvalues have modulus one while $(N\mathfrak{p}{)}^{-1}<1$.

The terms $m\ge 2$ are uniformly bounded. Indeed $|\chi (g)|\le d_{\chi }$, and for $N\mathfrak{p}\ge 2$ and $s\ge 1$,There are at most $[K:\mathbb{Q}]$ primes of $K$ above each rational prime and each has norm at least that prime. Hence ${\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-2}<\infty$. Isolating $m=1$ proves the character prime sums in the Chebotarev proof identityFor nontrivial $\chi$, the allowed holomorphy and nonvanishing at one give a local holomorphic logarithm and a bounded right side. The logarithm specified by the Euler series differs from that local branch by a constant multiple of $2\pi i$ on the real interval near one, so it too is bounded. For the trivial character, the right side is $\log (1/(s-1))+O(1)$ by the pole of ${\zeta }_K$. Thus

Expand the conjugacy-class indicator by character orthogonality. For $c\in C$,To see the coefficient, take its inner product with $\chi$: $|G{|}^{-1}{\sum }_{g\in C}\overline{\chi (g)}=|C|\overline{\chi (c)}/|G|$. The irreducible characters form an orthonormal basis of class functions, so this is the full expansion. Substituting the prime-sum estimates leaves only the trivial character's logarithmic term:Division by the diverging logarithm proves the theorem. The same trivial-character calculation shows that all primes of $K$ have density one; adding or removing finitely many primes does not change a Dirichlet density.

There is a necessary qualification for the final polynomial application: the printed claim is false for degree one. For example, $X$ is monic and irreducible in $\mathbb{Z}[X]$, yet has the root zero modulo every prime. We prove the intended assertion for degree $n\ge 2$.

Let $L$ be the splitting field of the monic irreducible polynomial over $\mathbb{Q}$, and let $\Omega$ be its $n$ roots. The Galois group $G$ acts transitively on $\Omega$. There is a derangement in a transitive group action: count the pairs $(g,\alpha )$ satisfying $g\alpha =\alpha$. The orbit-stabilizer theorem givesIf every $g$ fixed at least one root, the identity, which fixes $n\ge 2$, would make this sum at least $|G|+n-1$, a contradiction. Therefore some $g$ fixes no root, and neither does any conjugate of $g$.

For completeness, relate this group action to reduction without assuming the desired splitting result. All roots are algebraic integers, since the polynomial is monic. Exclude the finitely many primes ramifying in $L$ or dividing the nonzero polynomial discriminant. At a prime $\mathfrak{q}$ over any remaining $p$, the roots reduce to distinct elements of $k_{\mathfrak{q}}$ and give all roots of the reduced polynomial. Arithmetic Frobenius permutes them by $b\mapsto b^p$. A reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed by this map. Because reduction is injective on the original root set, this happens exactly when the Frobenius permutation fixes an original root. This proves the Frobenius permutation and roots modulo a prime criterion.

The Chebotarev density theorem assigns the conjugacy class of our fixed-point-free $g$ the positive Dirichlet density $|C_g|/|G|$. All but finitely many of these primes have rootless reduction, so there are infinitely many. More precisely, applying the same argument to the union of all fixed-point-free classes gives

A rational root of the monic cubic would be one of $\pm 1,\pm 3$, and substitution excludes all four. A reducible cubic over $\mathbb{Q}$ would have a rational root, so the polynomial is irreducible andIts polynomial discriminant is $-4(-1{)}^3-27(3{)}^2=-239$. The integer 239 is prime: it has no prime divisor among $2,3,5,7,11,13$, the primes not exceeding its square root.

The discriminant-index formula for an integral lattice for an integral primitive element isBoth discriminants are integers, and the left side is $-239$. Its squarefreeness forces the positive index to be one. Since $\alpha$ is an algebraic integer by its monic equation,

The coefficients of $\det (TI-A)$ are polynomial functions of the matrix entries. Its polynomial discriminant $\Delta (A)$ is therefore a polynomial in those entries, and it is nonzero exactly when all roots of the characteristic polynomial are distinct. Thus the required set is the principal open subset $D(\Delta )$ of affine space. It is nonempty: an algebraically closed field is infinite, so choose a diagonal matrix with $n$ pairwise distinct entries.

It is also affine as a variety. The map $A\mapsto (A,1/\Delta (A))$ identifies it with the closed hypersurfaceProjection to $A$ is its regular inverse. This proves the requested affineness while retaining the open embedding into matrix space.

The polynomial is an Eisenstein polynomial at $5$, so it is irreducible and $[K:\mathbb{Q}]=3$. Its polynomial discriminant isOne must remove the index at $2$ before using this as the field discriminant. Put $\beta =({\alpha }^2+\alpha )/2$. Direct reduction using ${\alpha }^3=5\alpha -20$ givesThus $B=\mathbb{Z}+\mathbb{Z}\alpha +\mathbb{Z}\beta$ is a ring finite over $\mathbb{Z}$, so its elements are integral. Its basis has discriminant of elements of a number field $-2575=-5^2\cdot 103$. The discriminant-index formula for an integral lattice says $\mathrm{disc}(B)=[{\mathcal{O}}_K:B{]}^2{\Delta }_K$, so only $5$ could divide the remaining index.

We use these standard different ideal facts: the norm of the different is the absolute field discriminant; its localization is the local different; an unramified prime has different exponent zero, and a tamely ramified prime of ramification index $e$ has different exponent $e-1$. The Eisenstein polynomial at $5$ gives total ramification of degree three, which is tame because $5\nmid 3$. Its contribution to $v_5({\Delta }_K)$ is therefore two. This excludes a further index factor $5$, proving ${\mathcal{O}}_K=B$ and $\boxed{{\Delta }_K=-2575}$.

Since the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]=2$ is prime to $5$ and $103$, Dedekind factorization theorem applies at both primes. Modulo $5$, $f=X^3$, giving a single prime $\mathfrak{p}=(5,\alpha )$ with residue degree one and ramification index three. Modulo $103$,Thus $\mathfrak{q}=(103,\alpha -6)$ has residue degree one and ramification index two; the other prime over $103$ is unramified. Both ramified primes are tame. No other prime ramifies because none divides ${\Delta }_K$. Consequently

The coordinate ring is $\mathbb{C}[W]=\mathrm{Sym}(W^{\ast })$. The induced action is contragredient on functions:It is a group action by degree-preserving algebra automorphisms, extending the dual action on linear polynomial functions. Its polynomial invariant ring isFor a monic polynomial $f(t)={\prod }_{i=1}^d(t-r_i)$, its polynomial discriminant is $\mathrm{disc}(f)={\prod }_{i<j}(r_i-r_j{)}^2$. This is symmetric in the roots, hence polynomial in the coefficients, and is zero exactly when a root is repeated.

For the alternating group action, let $e_1,\dots ,e_n$ be the elementary symmetric polynomials and put $\Delta ={\prod }_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any transposition, decomposeNormality and index two of $A_n$ show that $p_{+}$ is symmetric and $p_{-}$ transforms by the sign character of $S_n$. Any alternating polynomial vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_{-}=\Delta q$ with $q$ symmetric. The Fundamental theorem of symmetric polynomials yieldsThe sum is direct, since a polynomial that is both symmetric and alternating is zero in characteristic zero. Also ${\Delta }^2=D(e_1,\dots ,e_n)$, where $D$ is the discriminant polynomial ofThus, more precisely than the requested quotient assertion,Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The direct sum forces $a(e)=b(e)=0$, and algebraic independence of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.

Now let $G$ be finite with no nontrivial linear characters, and write $R=\mathbb{C}[W]$, $S=R^G$. The polynomial ring $R$ is a unique factorization domain. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal{O}$, choose representatives and form its orbit product of polynomial factorsFor every $g$, $g\cdot q_{\mathcal{O}}=\theta (g)q_{\mathcal{O}}$ for a nonzero scalar $\theta (g)$. Applying two group elements proves that $\theta :G\to {\mathbb{C}}^{\times }$ is a homomorphism. The hypothesis forces $\theta =1$, so $q_{\mathcal{O}}\in S$.

This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in \mathcal{O}$ divides $a$ or $b$ in $R$. Invariance of that chosen polynomial makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the integral domain $R$. Thus $q_{\mathcal{O}}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore $\mathbb{C}[W{]}^G$ is a unique factorization domain. This gives a direct proof, without assuming a divisor-class-group theorem.

For a failure in characteristic zero, let the cyclic group of order two act on ${\mathbb{C}}^2$ by $-I$. Its coordinate invariants areEvery invariant monomial has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^i{c}^j$ and $b{a}^i{c}^j$ map to distinct monomials. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, butgives two different irreducible factorizations. Hence this invariant ring is not a unique factorization domain. The nontrivial sign character is precisely the kind of character excluded in the preceding theorem.

For a direct check of unramified behavior, the cyclotomic polynomial ${\Phi }_{25}$ divides $X^{25}-1$. Modulo $7$, the latter has derivative $25X^{24}=4X^{24}$ and has no common root with that derivative. Thus ${\overline{\Phi }}_{25}$ is separable, so $7$ does not divide its polynomial discriminant. The field discriminant divides this polynomial discriminant, and therefore $7$ is unramified in $\mathbb{Q}({\zeta }_{25})$. The Frobenius automorphism satisfiesIndeed its residue action is seventh powering. The reduced $25$th roots of unity are distinct because $7\nmid 25$, so the residue congruence identifies the exponent uniquely modulo $25$.

The Galois group is $(\mathbb{Z}/25\mathbb{Z}{)}^{\times }$, of order $\phi (25)=20$. Since $7^2=49\equiv -1(\mathrm{mod}25)$, $7^4\equiv 1$ but $7^2\not\equiv 1$. Consequently the Frobenius automorphism has order four. An unramified decomposition group is generated by that element, and the orbit-stabilizer theorem givesThere are therefore five primes above $7$, each with residue field ${\mathbb{F}}_{7^4}$.

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 $p\equiv 1(\mathrm{mod}4)$.