Dedekind Domains¶
Noetherian Rings and Modules¶
The Noetherian property arises from the ascending chain condition. The definition and description of the property can be stated as follows:
Definition
Let \(R\) be a unital commutative ring, and \(M\) be an \(R\)-module. Then the following conditions are equivalent:
(1) Let \(\Sigma\) be a subset of \(M\), and \(\Sigma' \subset \Sigma\), then \(\Sigma'\) has a maximal element.
(2) For any ascending chain of submodules \(M_1 \subset M_2 \subset \cdots\), there exists \(n \in \mathbb{N}\) such that \(\forall m \geqslant n, M_m = M_n\).
(3) Every submodule of \(M\) is finitely generated.
\(M\) is called a Noetherian module if any thus every of the above properties hold. \(R\) is called a Noetherian ring if it is Noetherian as an \(R\)-module.
Proof
(2) \(\Rightarrow\) (1) If \(\Sigma'\) has no maximal element, then we can construct an ascending chain \(M_1 \subset M_2 \subset \cdots\) that grows infinitely, which is a contradiction.
(1) \(\Rightarrow\) (3) Let \(N\) be a submodule of \(M\), and \(\Sigma\) be the set of all finitely generated submodules contained in \(N\). Since \(0 \in \Sigma\), \(\Sigma\) is nonempty, and thus has a maximal element \(N_0\). If \(N \neq N_0\), then there exists \(x \in N \setminus N_0\), consider the submodule \(N_0 + Rx\), which is finitely generated and \(N_0 \subsetneq N_0 + Rx \subsetneq N\), which is a contradiction. So \(N = N_0\) is finitely generated.
(3) \(\Rightarrow\) (2) Let \(M_1 \subset M_2 \subset \cdots\) be an ascending chain of submodules of \(M\), let \(N = \bigcup_{i=1}^{\infty} M_i\), which is a submodule of \(M\), so \(N\) is finitely generated. Assume \(N = Rx_1 + \cdots + Rx_r\), and \(x_i \in M_{n_i}\). Let \(n = \operatorname{max}\{n_1, \cdots, n_r\}\), then \(\forall x_i \in N, x_i \in M_n\), so \(N = M_n\), and \(\forall m \geqslant n, M_m = M_n\).
The properties of Noetherian modules are also related to short exact sequences:
Proposition
(1) Let \(0 \to M' \to M \to M/M' \to 0\) be a short exact sequence of \(R\)-modules, then the following conditions are equivalent:
(i) \(M'\) and \(M/M'\) are both Noetherian modules.
(ii) \(M\) is a Noetherian module.
(2) \(\oplus_{i=1}^n M_i\) is a Noetherian module if and only if each \(M_i\) is a Noetherian module.
Hilbert's Basis Theorem
Let \(R\) be a Noetherian ring, then \(R[x_1, x_2, \ldots, x_n]\) is also a Noetherian ring.(The converse is evidently true as well.)
Note that \(\mathcal{O}_{\mathbb{F}}\) is trivially Noetherian as an \(\mathbb{Z}\)-module. Thus we are wodering if \(\mathcal{O}_{\mathbb{F}}\) is a Noetherian ring itself.
Theorem
Assume \(A\) is an integrally closed integral domain and Noetherian. \(\operatorname{char}(\mathbb{F}) = 0\). Consider \([\mathbb{E} : \mathbb{F}] = n\) a field extension, then \(A^{\sharp} = S(\mathbb{E}, A)\) is finitely generated as \(A\) module and is Noetherian.
Proof
(i) Recall that we have already shown that \(A^{\sharp}\) is a submodule of a Noetherian \(A\)-module \(A^n\) (In Lec2), so \(A^{\sharp}\) is finitely generated. So \(A^{\sharp} = A[\alpha_1, \ldots, \alpha_k]\).
(ii) Note that \(A[x_1, \ldots, x_k]\) is Noetherian according to Hilbert's Basis Theorem. And we can map the polynomial ring to \(A^{\sharp}\) by \(x_i \mapsto \alpha_i\). According to its universal property, the map is unique and surjective. So \(A^{\sharp}\) is Noetherian.
Dedekind Domains¶
Definition
A Dedekind domain \(A\) is an integral domain that is Noetherian, integrally closed, with its all non-zero prime ideals being maximal.
Example
Every field is a Dedekind domain. \(\mathbb{Z}\) is also a Dedekind domain. Actually, every PID is a Dedekind domain.
Theorem
Let \(A\) be a Dedekind domain with quotient field \(\mathbb{F}\) of characteristic \(0\). Let \([\mathbb{E} : \mathbb{F}] = n\), then \(A^{\sharp}\) is a Dedekind domain and finitely generated as an \(A\)-module.
Proof
The proposition above implies that \(A^{\sharp}\) is Noetherian. Also we have shown that \(A^{\sharp}\) is integrally closed. So we only need to show that every prime ideal of \(A^{\sharp}\) is maximal.
Consider \(0 \neq \mathfrak{p} \subset A^{\sharp}\) a prime ideal. We are to show that \(\mathfrak{p}\) is maximal, i.e. \(A^{\sharp}/\mathfrak{p}\) is a field. To utilize the property of Dedekind domain \(A\), we shall construct the intermediate field \(A/A \cap \mathfrak{p}\).
TODO
Factorization of Ideals in Dedekind Domain¶
We are to factorize ideals into product of prime ideals in a Dedekind domain. However, we are to ensure that the product is inside:
Definition
For an integral domain \(A\) with quotient field \(\mathbb{F}\). A fractional ideal \(I\) of \(A\) is a \(A\)-submodule of \(\mathbb{F}\) s.t. there exists \(0 \neq d \in A\) such that \(dI \subset A\).
Note that \(I \not \subset A\) in general. If it does, then we may call it an integral ideal of \(A\). Set
For all \(I_1, I_2 \in A^{-1} I(A)\), it is easy to see that \(I_1 I_2, I_1 + I_2 \in A^{-1} I(A)\). Also, \(0, A \in A^{-1} I(A)\). Thus \((A^{-1} I(A) \setminus \{0\}, \cdot)\) is at least a monoid with identity \(A\). Actually it is a group in some cases.
Lemma
Let \(A\) be a Dedekind domain. Then for every prime ideal \(0 \neq \mathfrak{p} \subset A\) is invertible in \(A^{-1} I(A)\). i.e. there exists \(\mathfrak{q} \in A^{-1} I(A)\) such that \(\mathfrak{p} \mathfrak{q} = A\).
Proof
Take \(\mathfrak{q} = \{ x \in \mathbb{F} \mid x \mathfrak{p} \subset A \} \subset \mathbb{F}\). Note that \(\mathfrak{q}\) is an \(A\)-module and for any \(b \in \mathfrak{p}\), \(b \mathfrak{q} \subset A\). Thus \(\mathfrak{q} \in A^{-1} I(A)\). By definition we have \(\mathfrak{p} \mathfrak{q} \subset A\) and \(A \subset \mathfrak{q}\). So \(\mathfrak{p} = \mathfrak{p} A \subset \mathfrak{p} \mathfrak{q} \subset A\).
We are to show that \(\mathfrak{p} \cdot \mathfrak{q} \supset A\). Suppose not, then