The Steinitz Realization Problem (2024)

Theorem 1.1.

Every element of $\operatorname{Cl}(K)$ occurs as the Steinitz class of a degree $n$ extension of $K$ with squarefree discriminant.

We give a sketch of the proof. A rank n ring over $\mathcal{O}_{K}$ is a ring that is locally free rank of $n$ as an $\mathcal{O}_{K}$-module. By work of Wood [wood], a binary $n$-ic form

gives rise to a rank $n$ ring $R_{f}$. If $\mathfrak{a}$ is an integral ideal of $\mathcal{O}_{K}$ such that $f_{n-1}\in\mathfrak{a}$ and $f_{n}\in\mathfrak{a}^{2}$, we construct a rank $n$ ring $R_{f}(\mathbf{a})$ and an inclusion $R_{f}\xhookrightarrow{}R_{f}(\mathbf{a})$. When $f$ is irreducible, both rings are orders in a number fields. The Steinitz class of $R_{f}(\mathfrak{a})$ is $[\mathfrak{a}^{-1}]$ and $\operatorname{Disc}(R_{f}(\mathbf{a}))=\mathfrak{a}^{-2}\operatorname{Disc}(R_$.

Theorem 1.2.

By work of Wood [wood], a binary $n$-ic form $f_{0}x^{n}+\dots+f_{n}y^{n}\in\operatorname{Sym}^{n}(\mathcal{O}_{K}\oplus$ gives rise to a rank $n$ ring $R_{f}$ as follows. As an $\mathcal{O}_{K}$-module, put $R_{f}\coloneqq\mathcal{O}_{K}^{n}$ and label the copies of $\mathcal{O}_{K}$ so we write:

Let $1=\zeta_{1},\dots,\zeta_{n-1}$ and $\zeta_{n}=-f_{n}$. For $1\leq i\leq j\leq n-1$ and $1\leq k\leq n$, define the multiplication maps $(\mathcal{O}_{K})_{i}\otimes(\mathcal{O}_{K})_{j}\rightarrow(\mathcal{O}_{K})_$ by:

1. (1)

   $x\otimes y\rightarrow-f_{i+j-k}\zeta_{k}xy$ if $\max(i+j-n,1)\leq k\leq i$;

2. (2)

   $x\otimes y\rightarrow f_{i+j-k}\zeta_{k}xy$ if $j<k\leq\min(i+j,n)$;

3. (3)

   and $x\otimes y\rightarrow 0$ otherwise.

For any nonzero integral ideal $\mathfrak{a}$ of $K$, define $R_{f}(\mathfrak{a})$ to be the locally free $\mathcal{O}_{K}$-module given by

There is a natural inclusion of $\mathcal{O}_{K}$-modules given by $R_{f}\subseteq R_{f}(\mathfrak{a})$.

Lemma 2.1.

$R_{f}(\mathfrak{a})$ has a $\mathcal{O}_{K}$-algebra structure extending that of $R_{f}$ if and only if $f_{n-1}\in\mathfrak{a}$ and $f_{n}\in\mathfrak{a}^{2}$. When $f_{n-1}\in\mathfrak{a}$ and $f_{n}\in\mathfrak{a}^{2}$, there is a unique $\mathcal{O}_{K}$-algebra structure on $R_{f}(\mathfrak{a})$ extending that of $R_{f}$, the Steinitz class of $R_{f}(\mathfrak{a})$ is $[\mathfrak{a}^{-1}]$ and $\mathfrak{a}^{-2}(\operatorname{Disc}(R_{f}))=(\operatorname{Disc}(R_{f}($.

Proof.

The first statement follows directly from the multiplication table. The second is immediate from the construction.∎

The proof given in this section is essentially that given by Kedlaya [ked]. The main differences are that we work over a number field and impose a few additional congruence conditions needed for our application. We repeat much of Kedlaya’s construction verbatim, and encourage the reader to refer to Kedlaya’s paper for more detail. Fix a class $[\mathfrak{a}]\in\operatorname{Cl}(K)$, and pick a representative prime ideal $\mathfrak{a}$ coprime to $(n!)$. For $a_{1},a_{2},\dots,a_{n-1},b\in K$, put:

The integral in the definition of $P_{a,b}(x)$ should be interpreted formally; $Q_{a}(x)$ is a polynomial. The discriminant of $P_{a,b}$, up to sign, is the product of $P_{a,b}(x)$ evaluated at the roots of $Q_{a}(x)$. So, we obtain the following equality of ideals:

Suppose $a_{1}\in\mathfrak{a}$ and $b\in\mathfrak{a}^{2}$. If we write $P_{a,b}(x)=x^{n}+f_{1}x^{n-1}+\dots+f_{n}$, then $f_{n}=b\in\mathfrak{a}^{2}$ and

Furthermore, $\mathfrak{a}^{2}$ divides $(n^{n}P_{a,0}(a_{1}/n)+n^{n}b)$ and hence divides $\Delta_{a,b}$. Therefore, we wish to show that there exists a choice of $a_{1},\dots,a_{n-1},b$ with $a_{1}\in\mathfrak{a}$ and $b\in\mathfrak{a}^{2}$ such that $P_{a,b}(x)$ has integral coefficients, is irreducible, and the ideal $\mathfrak{a}^{-2}\Delta_{a,b}$ is squarefree. We will do this by fixing an appropriate choice of $a_{1},\dots,a_{n-1}$ and then applying a squarefree sieve to choose $b$.

We will first show that there exist $a_{1},\dots,a_{n-1}\in\mathcal{O}_{K}$ such that:

1. (1)

   $a_{1}\in\mathfrak{a}$ and $a_{i}\equiv 1\pmod{\mathfrak{a}}$ for $i=2,\dots,n-1$;

2. (2)

   $P_{a,0}$ has integral coefficients;

3. (3)

   for every prime ideal $\mathfrak{p}\neq\mathfrak{a}$, there exists $b$ such that $\Delta_{a,b}$ is not divisible by $\mathfrak{p}^{2}$;

4. (4)

   there exists $b_{2}\in\mathfrak{a}^{2}$ such that $\Delta_{a,b_{2}}$ is not divisible by $\mathfrak{a}^{3}$;

5. (5)

   there exists a prime ideal $\mathfrak{p}_{1}$ and $b_{1}\in\mathcal{O}_{K}$ such that $P_{a,b_{1}}$ is irreducible modulo $\mathfrak{p}_{1}$;

6. (6)

   and the polynomials $(n^{n}P_{a,0}(a_{1}/n)+n^{n}b),(P_{a,0}(a_{2})+b),\dots,(P_{a,0}(a_{n-1})+b)$ are pairwise coprime.

Lemma 3.1 (Kedlaya, Lemma 2.3 [ked]).

Let $\gamma\in\mathcal{O}_{K}$ be be such that $(\gamma)$ is a principal prime ideal coprime to $(n(n-1))$. Then there exist infinitely many primes $\mathfrak{p}_{1}$ modulo which the polynomial $R(x)=x^{n}-\gamma^{n-1}x+\gamma$ is irreducible and its derivative $R^{\prime}(x)=nx^{n-1}-\gamma^{n-1}$ splits into linear factors.

Proof.

The polynomial $R^{\prime}(x)$ has splitting field $L=K(\zeta_{n-1},n^{1/(n-1)})$, in which $(\gamma)$ does not ramify because $(\gamma)$ does not divide $(n(n-1))$. Thus $R$ is an Eisenstein polynomial with respect to any prime above $(\gamma)$ in $L$; in particular, $R$ is irreducible over $L$. By the Chebotarev density theorem, there exist infinitely many totally split prime ideals of $L$ modulo which $R$ is irreducible; the norm to $K$ of any such prime ideal is a prime ideal of the desired form.∎

Lemma 3.2 (Kedlaya, Lemma 2.4 [ked]).

Let $\gamma$, $\mathfrak{p}_{1}$, and $R(x)$ be as in Lemma3.1 and ensure that $\mathfrak{p}_{1}\nmid\mathfrak{a}(n!)$. Apply Lemma3.2 to obtain $a^{\prime}_{1},\dots,a^{\prime}_{n-1}$ so that

have distinct values. Choose a prime ideal $\mathfrak{p}_{2}\nmid\mathfrak{a}(n!)\mathfrak{p}_{1}$ such the reductions have well-defined and distinct values modulo $\mathfrak{p}_{2}$. Now use the Chinese remainder theorem to choose $a_{1},\dots,a_{n-1}\in\mathcal{O}_{K}$ subject to the following congruence conditions:

• •

  $a_{1}$ is coprime to $(n)$ and is contained in $\mathfrak{a}$ and the ideal generated by $(n-1)!$;

• •

  the integers $a_{2},\dots,a_{n-1}$ are contained in $(n!)$ and $a_{i}\equiv-1$ modulo $\mathfrak{a}$ for $i=2,\dots,n-1$;

• •

  the quantities $a_{1}/n,\dots,a_{n-1}$ are congruent to the roots of $R^{\prime}(x)$ modulo $\mathfrak{p}_{1}$;

• •

  and $a_{i}\equiv a^{\prime}_{i}$ modulo $\mathfrak{p}_{2}$.

We’ll now show that this choice of $a_{1},\dots,a_{n-1}$ satisfies conditions $(1)-(6)$.

1. (1)

   Follows directly from the choice of $a_{i}$.

2. (2)

   Observe that:

   $$Q_{a}(x)\equiv(nx-a_{1})x^{n-2}=nx^{n-1}-a_{1}x^{n-2}\pmod{(n!)},$$

   so $P_{a,0}$ has integer coefficients.

3. (3)

   Suppose that $\mathfrak{p}\neq\mathfrak{a}$ is a prime ideal dividing $(n!)$. Then for $b\in\mathcal{O}_{K}$, we have $P_{a,0}(a_{i})+b\equiv b\pmod{\mathfrak{p}}$ for $i=2,\dots,n-1$ because $a_{i}\in\mathfrak{p}$. If $\mathfrak{p}\mid(n)$, then

   $$n^{n}P_{a,0}(a_{1}/n)+n^{n}b\equiv a_{1}^{n}\pmod{\mathfrak{p}}$$

   and if $\mathfrak{p}\nmid(n)$, then

   $$n^{n}P_{a,0}(a_{1}/n)+n^{n}b\equiv n^{n}b\pmod{\mathfrak{p}}$$

   Consequently, $\Delta_{a,1}$ is not divisible by $\mathfrak{p}$. Now suppose that $\mathfrak{p}$ is a prime ideal coprime to $\mathfrak{a}(n!)$, so $\lvert\mathcal{O}_{K}/\mathfrak{p}\rvert>n$. There are at least $\lvert\mathcal{O}_{K}/\mathfrak{p}\rvert-n$ choices of $b\in\mathcal{O}_{K}/\mathfrak{p}$ for which $\Delta_{a,b}$ is indivisible by $p$, because each linear factor rules out exactly one choice of $b$.

4. (4)

   Choose $b\in\mathfrak{a}^{2}$ so that $P_{a,0}(a_{1}/n)+b\not\equiv 0\pmod{\mathfrak{a}^{3}}$. To conclude it suffices to show that $P_{a,0}(a_{i})\not\equiv 0\pmod{\mathfrak{a}}$ for all $i=2,\dots,n-1$. Because $a_{i}\equiv-1$ modulo $\mathfrak{a}$ for $i=2,\dots,n-1$ and $P_{a,0}(0,-a_{1},\dots,-a_{n-1})$ is a nonzero polynomial with positive coefficients in $a_{2},\dots,a_{n-1}$, we can ensure this by choosing $\mathfrak{a}$ to have sufficiently large norm.

5. (5)

   Ensured by the fact that the quantities $a_{1}/n,\dots,a_{n-1}$ are congruent to the roots of $R^{\prime}(x)$ modulo $\mathfrak{p}_{1}$.

6. (6)

   Ensured by the fact that $a_{i}\equiv a^{\prime}_{i}$ modulo $\mathfrak{p}_{2}$.

Lemma 3.3.

Proof.