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If $m>1$ is a square-free integer, $3\nmid m$, and $m\not\equiv \pm 1(\mathrm{mod}9)$, then $1,\sqrt[3]{m},\sqrt[3]{m^2}$ is an integral basis of the pure cubic number field, with field discriminant $-27m^2$. The discriminant-index formula for an integral lattice restricts a possible index to primes dividing $3m$. At primes dividing $m$ the different exponent is $2$ by tame ramification. A shifted Eisenstein polynomial gives total wild ramification at $3$, so its different exponent is at least $3$. These exponents exhaust the polynomial discriminant and force index one.

An order in a number field $K$ is a subring $A\subseteq {\mathcal{O}}_K$ containing $1$ and having full rank $[K:\mathbb{Q}]$ as an additive abelian group. Its finite index in the ring of integers of a number field is related to its discriminant by the discriminant-index formula for an integral lattice.

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

For a finite extension $E/K$ of number fields, the inverse different is the fractional idealThe different ideal is its inverse. The trace pairing is nondegenerate because extensions of number fields are separable, so this definition gives a full fractional ideal. The inclusion ${\mathcal{O}}_E\subseteq {\mathfrak{D}}_{E/K}^{-1}$ shows that ${\mathfrak{D}}_{E/K}$ is integral.

Write its prime ideal factorization as ${\mathfrak{D}}_{E/K}={\prod }_{\mathfrak{P}}{\mathfrak{P}}^{d_{\mathfrak{P}}}$. For $\mathfrak{P}$ above $\mathfrak{p}$, with ramification index $e_{\mathfrak{P}}$ and residue characteristic $\ell$, the different exponent and tame ramification theorem saysThe equivalence uses the separability of finite residue-field extensions. In particular, $d_{\mathfrak{P}}=0$ precisely at unramified prime ideals; in the wild case $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}$. This theorem does not require a Galois extension. The relative discriminant is the norm of the different ideal:For $K=\mathbb{Q}$, this gives $|D_E|=N({\mathfrak{D}}_{E/\mathbb{Q}})$.

Now take ${\alpha }^3=m$ and $K=\mathbb{Q}(\alpha )$. For any prime number $q\mid m$, the square-free integer hypothesis makes $X^3-m$ an Eisenstein polynomial at $q$. It is therefore irreducible, and $K$ is a pure cubic number field of degree $3$. Since $\alpha$ is an algebraic integer, $A=\mathbb{Z}[\alpha ]$ is an order in a number field. The discriminant of elements of a number field for its basis $1,\alpha ,{\alpha }^2$ isFor example, the resultant of $X^3-m$ and $3X^2$ is $27m^2$, and the degree-three sign in the polynomial discriminant is negative. If $I=[{\mathcal{O}}_K:A]$, the discriminant-index formula for an integral lattice givesOnly prime numbers dividing $3m$ can therefore divide $I$.

For $q\mid m$, the Eisenstein polynomial gives a totally ramified extension of ${\mathbb{Q}}_q$ of degree $3$. Thus $K$ has a unique prime ideal $\mathfrak{P}$ above $q$, with ramification index $3$ and residue-field degree $1$. Since $q\ne 3$, this is tame ramification, and the different exponent and tame ramification theorem gives $d_{\mathfrak{P}}=2$. Hence $v_q(D_K)=2$. Comparing with $v_q(-27m^2)=2$ in the discriminant-index formula for an integral lattice yields $v_q(I)=0$.

At $3$, use the shifted Eisenstein polynomial of $\beta =\alpha -m$:Because $3\nmid m$, the constant term is divisible by $3$. Moreover,when $3\nmid m$: the factors $m-1$ and $m+1$ cannot both be divisible by $3$. The hypotheses therefore give $v_3(m^3-m)=1$, so the translated polynomial is an Eisenstein polynomial at $3$. The resulting completion is a totally ramified extension of degree $3$, with residue-field degree $1$, and its ramification index is divisible by the residue characteristic. Thus its different exponent is at least $3$.

It follows that $v_3(D_K)\ge 3$. But $v_3(-27m^2)=3$, soforces $v_3(I)=0$ and $v_3(D_K)=3$. No prime number divides $I$. We have proved the integral basis of a nonexceptional pure cubic field:Using local Eisenstein polynomials here only establishes the local ramification indices; it does not assume that $\mathbb{Z}[\alpha ]$ is already the full ring of integers of a number field.

Let $K=\mathbb{Q}(\sqrt{-31})$. Its ring of integers of a quadratic field is $\mathbb{Z}[(1+\sqrt{-31})/2]$, and its fundamental field discriminant is $-31$.

First compute the class number. The ideal-form correspondence for imaginary quadratic fields identifies the ideal classes with proper equivalence classes of primitive positive definite binary quadratic forms of discriminant $-31$. Each has a unique reduced positive definite binary quadratic form representative under the convention $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. From $31=4ac-b^2\ge 3a^2$ we get $1\le a\le \sqrt{31/3}<4$. Checking these three possible leading coefficients gives exactlyFor $a=3$, neither odd residue choice $b=\pm 1,\pm 3$ makes $(b^2+31)/12$ integral. The listed forms are primitive and satisfy the reduction convention, soThis computes the ideal class group of Q of square root minus thirty-one.

Now take a root $\theta$ of $F(T)=T^3+T-1$, and let $H$ be its splitting field over $\mathbb{Q}$. There is no rational root: the only candidates are $\pm 1$, and neither vanishes. Thus the cubic is irreducible. Its polynomial discriminant isSince this is nonsquare, the Galois group of an irreducible cubic is $S_3$. Its quadratic subfield is $K$: the product of pairwise root differences squares to $-31$ and changes sign under odd permutations. Thus $H/K$ is cyclic of degree three, and $H=K(\theta )$. Indeed the cubic field and $K$ have coprime degrees, so their compositum already has the full splitting-field degree six. The squarefree polynomial discriminant also shows, by the discriminant-index formula for an integral lattice, that $\mathbb{Z}[\theta ]$ is the full integer ring of the cubic field.

It remains to prove that $H/K$ is unramified everywhere. At any rational prime $\ell \ne 31$, the reduction of $F$ is separable. Over a finite residue extension in which it splits, all residue roots are simple and lift by Hensel lemma to the corresponding unramified extension of ${\mathbb{Q}}_{\ell }$. Therefore the local splitting field is unramified over ${\mathbb{Q}}_{\ell }$, and its extension over a completion of $K$ is unramified as well.

At the only possible ramified prime, direct reduction givesThe simple root lifts to $r\in {\mathbb{Q}}_{31}$. Write $F(T)=(T-r)Q(T)$, with $Q$ monic quadratic over ${\mathbb{Q}}_{31}$. The discriminant product formula givesHere $Q(r)=F^{\prime }(r)$ is a unit, so the quadratic discriminant has square class $-31$. Hence the entire local splitting field isIt is quadratic over ${\mathbb{Q}}_{31}$, but its extension over the completion of $K$ is trivial. Thus the prime over $31$ splits completely in $H/K$; in particular it is not ramified there. This is the local splitting-field argument behind an unramified cyclic cubic extension from a prime-discriminant cubic. There are no real places of the imaginary quadratic field $K$, so no infinite-place ramification remains to check.

The Hilbert class field theorem identifies the maximal everywhere unramified abelian extension with an extension of degree $h_K$. Our cyclic unramified extension already has that degree, so it must be the whole Hilbert class field. ThereforeEquivalently, it is the splitting field of $T^3+T-1$, of degree six over $\mathbb{Q}$ and degree three over $K$. Replacing $\theta$ by ${\theta }^{-1}$ gives the alternative defining cubic $T^3-T^2-1$. This is the Hilbert class field of Q of square root minus thirty-one.

Now let $n=m=p$ be prime and put $\alpha =\sqrt[p]{p}$ and $A=\mathbb{Z}[\alpha ]$. The Eisenstein criterion applied at $p$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the discriminant of elements of a number field of its power basis iswhere $f(X)=X^p-p$. By the discriminant-index formula for an integral lattice, every prime dividing $[{\mathcal{O}}_L:A]$ must therefore be $p$.

Suppose $A\ne {\mathcal{O}}_L$. The finite abelian group ${\mathcal{O}}_L/A$ is then a nontrivial finite p-group, so it contains an element of order $p$. Consequently there is an $x\in {\mathcal{O}}_L\setminus A$ such thatwith some $c_k$ not divisible by $p$.

The polynomial $X^p-p$ is Eisenstein, so total ramification from an Eisenstein polynomial gives a unique prime ideal $P$ above $p$ with normalized discrete valuationLet $k_0$ be the least index for which $p\nmid c_{k_0}$. The term $c_{k_0}{\alpha }^{k_0}$ has valuation $k_0<p$; every earlier term has valuation at least $p$, and every later term with coefficient prime to $p$ has a distinct, larger valuation. The non-Archimedean valuation therefore givesThis contradicts $x\in {\mathcal{O}}_L$, since an algebraic integer has nonnegative valuation at every prime ideal. Thus $A={\mathcal{O}}_L$, and

The polynomial $T^3+3T+3$ is Eisenstein at $3$, so part (ii) givesFor a cubic $T^3+aT+b$, the polynomial discriminant is $-4a^3-27b^2$. HereThe discriminant-index formula for an integral lattice saysAny prime dividing the index must therefore occur to at least the second power in $351$; the only possibility is $3$, which has already been excluded. The index is one, and hence

PutUsing ${\alpha }^3=5\alpha -8$, multiplication by $\beta$ on the ordered basis $(1,\alpha ,\beta )$ is determined byIts matrix, with these coordinate vectors as columns, isThe characteristic polynomial isBy the Cayley-Hamilton theorem, this monic integer polynomial annihilates $\beta$, so $\beta$ is an algebraic integer.

LetThe element $\alpha$ is also integral because it satisfies the monic polynomial $f$. Hence $A\subseteq {\mathcal{O}}_K$. Sincethe lattice $\mathbb{Z}[\alpha ]$ has index two in $A$. The given power-basis discriminant isso the discriminant-index formula for an integral lattice gives

If $m=[{\mathcal{O}}_K:A]$, thenBecause $307$ is prime, the square $m^2$ can divide $307$ only when $m=1$. Thusso $(1,\alpha ,\beta )$ is an integral basis of the cubic field of discriminant minus 307.

Finally, the relation ${\beta }^2=-4-\alpha +3\beta$ givesConsequently ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$, which will allow direct use of Dedekind's theorem.

Let ${\omega }_1,\dots ,{\omega }_n$ be an integral basis of ${\mathcal{O}}_K$. Since each ${\alpha }_j$ is an algebraic integer, there is an integer matrix $A=(a_{ij})$ such thatThe ${\alpha }_j$ form a $\mathbb{Q}$-basis, so $\det A\ne 0$.

Products of algebraic integers are algebraic integers, and their traces are rational algebraic integers, hence ordinary integers. Therefore every entryis integral. Part (i) makes the trace Gram matrix nonsingular, so $\Delta ({\alpha }_1,\dots ,{\alpha }_n)$ is a nonzero integer.

Let $G_{\omega }$ and $G_{\alpha }$ be the two trace Gram matrices. A change of basis givesTaking determinants yields the discriminant-index formula for an integral latticewhere $d_K=\det G_{\omega }$ is the field discriminant. Since $\det A$ is a nonzero integer,Equality holds exactly when $|\det A|=1$, which is equivalent to $A$ being unimodular and to ${\alpha }_1,\dots ,{\alpha }_n$ being a $\mathbb{Z}$-basis of ${\mathcal{O}}_K$. Thus the minimum is the positive integer

For a prime number $p$ and $\alpha =\sqrt[p]{p}$, the polynomial $X^p-p$ is Eisenstein at $p$, andThe discriminant-index formula for an integral lattice reduces the proof to the prime $p$. At the unique prime ideal above $p$, one has $v(\alpha )=1$ and $v(p)=p$, which rules out an integral element $g(\alpha )/p$ with $g\notin p\mathbb{Z}[X]$.