Dedekind Domains and Ideal Classes Examples

By [F1], every nonzero proper ideal of $R$ is principal, hence remains principal after localising at any maximal ideal. The ring is Noetherian by [L1], so [L2] makes $R$ a Dedekind domain.

Now [L3] applies to the Dedekind domain $R$ and gives that its class group is trivial.

Let $I={\prod }_{i=1}^r{\mathfrak{m}}_i^{n_i}$ be a nonzero ideal by [L1]. For each $i$, choose $x_i\in {\mathfrak{m}}_i^{n_i}\setminus {\mathfrak{m}}_i^{n_i+1}$. By [L2], there exists $x\in R$ with $x\equiv x_i(\mathrm{m}\mathrm{o}\mathrm{d}{\mathfrak{m}}_i^{n_i+1})$ for each $i$. Then $x$ has valuation exactly $n_i$ at ${\mathfrak{m}}_i$ for every $i$, and there are no other primes to consider. Hence $(x)=I$ by uniqueness in [L1].

Every nonzero ideal of $R$ is therefore principal, so [L3] identifies $R$ as a PID.

The ideal $I$ is fractional because $35I=12\mathbb{Z}\subseteq \mathbb{Z}$, so [F1] applies.

In the PID $\mathbb{Z}$, the principal ideal generated by $12\mathrm{/}35$ records the usual prime factorization of the numerator and denominator. Hence the exponents are $2$ at $(2)$, $1$ at $(3)$, $-1$ at $(5)$, and $-1$ at $(7)$, with all others zero. By [L1] this is exactly the prime-ideal factorization of $I$.

An element $x\in \mathbb{Q}$ lies in $(R:I)$ exactly when $x(2\mathrm{/}3)\in \mathbb{Z}$, equivalently when $x\in (3\mathrm{/}2)\mathbb{Z}$. Thus $(R:I)=(3\mathrm{/}2)\mathbb{Z}$.

Multiplying the displayed generators gives $(2\mathrm{/}3)(3\mathrm{/}2)=1$, so $I(R:I)=\mathbb{Z}$. This agrees with the valuation description from [L1].

The module $M=R\oplus I$ is already displayed as a direct sum of two invertible ideal summands, so it is finite torsion-free and fits the decomposition pattern of [L1].

Under the identification of [L2], the free summand $R$ contributes the neutral Picard class and the other summand contributes exactly the class of $I$. Thus the rank-two module $M$ is labelled by the same ideal class $[I]$.

By [L1], the localisation $R_{\mathfrak{p}}$ is a DVR. Localising the ideal ${\mathfrak{p}}^n$ gives exactly ${\mathfrak{p}}^n{R}_{\mathfrak{p}}$ by the definition of localisation of ideals.

Comparing step 1.1 with [F1] shows that the corresponding valuation is $v_{\mathfrak{p}}({\mathfrak{p}}^n)=n$.

The theorem [L1] identifies a finite set of bad primes, namely those for which $v_{\mathfrak{p}}(a)>v_{\mathfrak{p}}(I)$, chooses local correction terms at those primes, and combines them by the Chinese remainder step in its proof.

The resulting element $b\in I$ has exactly the missing prime valuations, so $(a,b)=I$. This is the concrete content of the two-generator construction.

The generator $40\mathrm{/}9$ contributes prime exponents $3$ at $(2)$, $1$ at $(5)$, and $-2$ at $(3)$, so $\mathrm{div}(40\mathrm{/}9)$ is the displayed valuation vector.

Because $I$ is principal, its class is zero in the class group by [L1].

The inclusion $R\subseteq k[t]$ is integral because $t$ satisfies the monic equation $X^2-t^2=0$ with coefficient $t^2\in R$. The ring $k[t]$ is a one-variable polynomial ring over a field, hence a nonfield principal ideal domain and therefore one-dimensional. Thus [L2] gives $\dim R=1$. The same monic equation shows that $t$ is integral over $R$, but $t\mathrm{\notin }R$, so $R$ is not integrally closed and therefore is not Dedekind.

In the local ring $R_{\mathfrak{m}}$ one has ${\mathfrak{m}}_{\mathfrak{m}}^2=(t^4,t^5,t^6)R_{\mathfrak{m}}$. If $t^2$ lay in ${\mathfrak{m}}_{\mathfrak{m}}^2$, we could write $t^2=(a{t}^4+b{t}^5+c{t}^6)\mathrm{/}s$ with $a,b,c\in R$ and $s\mathrm{\notin }\mathfrak{m}$, so $s=t^2(a+bt+c{t}^2)\in \mathfrak{m}$, contradiction. Hence $t^2\in {\mathfrak{m}}_{\mathfrak{m}}\setminus {\mathfrak{m}}_{\mathfrak{m}}^2$. If ${\mathfrak{m}}_{\mathfrak{m}}$ were principal, say ${\mathfrak{m}}_{\mathfrak{m}}=a{R}_{\mathfrak{m}}$, then $t^2=au$ for some $u\in R_{\mathfrak{m}}$. Because $t^2\mathrm{\notin }{\mathfrak{m}}_{\mathfrak{m}}^2=a{\mathfrak{m}}_{\mathfrak{m}}$, the element $u$ is a unit, so ${\mathfrak{m}}_{\mathfrak{m}}=t^2{R}_{\mathfrak{m}}$. But then $t^3\in {\mathfrak{m}}_{\mathfrak{m}}$ implies $t=t^3\mathrm{/}{t}^2\in R_{\mathfrak{m}}$, impossible: if $t=r\mathrm{/}s$ with $r\in R$ and $s\mathrm{\notin }\mathfrak{m}$, then $r=ts$ would have a linear $t$-term while elements of $R=k[t^2,t^3]$ have no such term. Therefore ${\mathfrak{m}}_{\mathfrak{m}}$ is not principal.

Since ${\mathfrak{m}}_{\mathfrak{m}}$ is not principal, [L1] shows that $\mathfrak{m}$ is not invertible.