8 250415

If $(e_i{)}_{1⩽i⩽r}$ is an $F$-basis of $V$ and $(I_i{)}_{1⩽i⩽r}$ is a family of fractional ideals, then $M={\oplus }_{i=1}^r{I}_i{e}_i$ is a fractional module in $V$.

Lemma

Let $I_1,I_2$ be fractional ideals.

• $I_1+I_2$, which is the $R$-submodule of $F$ generated by $I_1\cup I_2$
• $I_1\cap I_2$
• $I_1{I}_2$, which is the $R$-submodule of $F$ generated by all $x_1{x}_2$ with $x_1\in I_1$, $x_2\in I_2$

are all fractional ideals.

Proposition

• For any fractional ideal $I$, there is a natural identification $I^{\ast }=\{x\in F:xI\subseteq R\}$.
• Let $V$ be an $r$-dimensional $F$-space with a basis $(e_i{)}_{1⩽i⩽r}$, and let $(I_i{)}_{1⩽i⩽r}$ be a family of fractional ideals. Set $M={\oplus }_{i=1}^r{I}_i{e}_i$. Then
  • $M^{\ast }={\oplus }_{i=1}^r{I}_i^{\ast }{e}_i^{\ast }$, where $e_i^{\ast }\in V^{\ast }$.
  • the fractional module $M$ is a finitely generated projective $R$-module iff for all $1⩽i⩽r$, we have $I_i^{\ast }{I}_i=R$. In particular, a fractional ideal $I$ is finitely generated projective $R$-module iff $I^{\ast }I=R$.

\begin{proof} See ^fw9ghu. \end{proof}

Definition

A Dedekind Domain is a integral domain $R$ all ideals of which are finitely generated projective.

A principal ideal domain is a integral domain $R$ all ideals of which are free.

Local Ring

Definition

We say a ring $R$ (possibly non-commutative) is local if the set of non-units of $R$ form an ideal of $R$.

Theorem

TFAE for a ring $R$.

• $R$ is a local ring
• the set of nonunits of $R$ coincides with $J(R)$, that is, $J(R)$ is a unique maximal left (or right) ideal of $R$
• $R/J(R)$ is a division ring.

Proposition

Let $R$ be a commutative ring. Assume $R$ is local with unique maximal ideal $I$. Then every finitely generated projective $R$-module is free.

\begin{proof} See ^spgbvh. \end{proof}

Corollary

Local Dedekind domain = Local PID

\begin{proof} See ^wzfm6a. \end{proof}

Proposition

Let $R$ be an integral domain. Let $S$ be a nonempty multiplicatively closed subset of $R\setminus \{0\}$.

• If $R$ is a PID, so is $S^{-1}R$.
• If $R$ is a Dedekind domain, so is $S^{-1}R$.

\begin{proof} See ^88mhms. \end{proof}

Invertible ideals

Definition

A finitely generated fractional ideal is called invertible.

Lemma

An integral domain $R$ is a Dedekind domain iff all fractional ideals are invertible.

\begin{proof} See ^jn8801. \end{proof}

For Dedekind domain, the dual $I^{\ast }$ of a fractional ideal $I$ is denoted by $I^{-1}$.

Lemma

The set of fractional ideals of a Dedekind domain $R$ is an abelian group. The identity element is $R$, the inverse of a fractional ideal $I$ is $I^{-1}$.

Definition

A fractional ideal $I$ of $R$ is called integral ideal if $I\subseteq R$.

Let $I$ and $J$ be fractional ideal of $R$, we say $J$ divides $I$, that is, $J\mid I$, if there is an integral ideal such that $I=JK$.

Lemma

Let $I$ be fractional ideals in a Dedekind domain $R$. TFAE:

• $I\subseteq J$;
• $J\mid I$.

\begin{proof} See ^8e8e1e \end{proof}

Corollary

Every nonzero prime ideal of a Dedekind domain $R$ is maximal.

\begin{proof} See ^ib3ruw \end{proof}

Corollary

A Dedekind domain in integral closed.

\begin{proof} See ^3gnir0. \end{proof}

Theorem

Let $R$ be a Dedekind domain. Every nonzero proper ideal of $R$ can be written in a unique way as a product of prime ideals.

\begin{proof} See ^zio9cz. \end{proof}

Lemma

Let $R$ be a Dedekind domain. For any fractional ideal $I$, there exists integral ideals $J$ and $K$ such that $I=J{K}^{-1}$.

\begin{proof} See ^814a8z. \end{proof}

Definition

For every fractional ideal $I$, we write $I={\prod }_p{p}^{{\nu }_p(I)}$ with ${\nu }_p(I)\in \mathbb{Z}$ and $p$ runs over all prime ideals. We call ${\nu }_p(I)$ the valuation of $I$ at $p$.

Discrete Valuation

Definition

Let $k$ be a field. An (exponential) valuation of $k$ is a real-valued function $\nu$ defined on $k^{\ast }=k\setminus \{0\}$ satisfying

• $\nu (ab)=\nu (a)+\nu (b)$ and
• $\nu (a+b)⩾\min \{\nu (a),\nu (b)\}$.

Note that $\nu :K^{\times }\to (R,+)$ is a group homomorphism. The image $\nu (k^{\times })$ is called the value group of $\nu$, and we set $\nu (0)=\infty$. We call $(k,\nu )$ valuation field.

Facts.

• $\nu (1)=\nu (-1)=0$, $\nu (a)=\nu (-a)$, $\nu (a^{-1})=-\nu (a)$
• $\nu (a)>\nu (b)$ yields $\nu (a+b)=\nu (b)$
  • Since $\nu (a+b)⩾\min \{\nu (a),\nu (b)\}=\nu (b)$, it suffices to show $\nu (a+b)>\nu (b)$ is impossible. Otherwise assume $\nu (a+b)>\nu (b)$ holds, then $\nu (b)=\nu (a+b+(-a))⩾\min \{\nu (a+b),\nu (a)\}>\nu (b)$, which is impossible. Hence $\nu (a+b)=\nu (b)$.

Definition

Let $(k,\nu )$ be a valuation field. Then $R=\{a\in k:\nu (a)⩾0\}$ is a subring of $k$, which is called valuation ring of $\nu$.

Let $R^{-1}=\{a^{-1}:a\in R\setminus \{0\}\}=\{a\in k:\nu (a)<0\}$. Then $k=R\cup R^{-1}$ and so $k$ is a fractional field of $R$.

The set $P=\{a\in l:\nu (a)>0\}$ is an ideal of $R$, which is called the valuation ideal of $\nu$.

The group of units of $R$ is $\{a\in k:\nu (a)=0\}$.

Note that $R$ is a local ring with $P=J(R)=\{\text{nonunits of }R\}$, and $R/P$ is called residue field of $\nu$.

Let $\lambda \in R$ with $\lambda >0$. Define ${\nu }^{\prime }(a)=\lambda \nu (a)$ for all $a\in k^{\times }$. Then $\nu$ is also a valuation. We say $\nu$ is equivalent to ${\nu }^{\prime }$, and write $\nu \sim {\nu }^{\prime }$.

Definition

A valuation $\nu$ of $k$ is called discrete if its value group is isomorphic to $\mathbb{Z}$. In this situation, $R$ is called a discrete valuation ring.

In the case $\nu (k^{\times })=\mathbb{Z}$, $\nu$ is said to be normalized.

Theorem

Let $(k,\nu )$ be a discrete normalized valuation field. Let $R$ be its valuation ring, and let $P$ be its valuation ideal.

• Let $\pi \in R$ with $\nu (\pi )=1$. Then any $\alpha \in k$ is expressed as $\alpha ={\pi }^{i}u$ where $i=\nu (\alpha )$, $u\in R\setminus P$.
• Any nonzero ideal of $R$ is of the form $p^i=({\pi }^i)$ for some $i⩾0$. In particular, $R$ is a $PID$.
• $R$ is integral closed.

\begin{proof} See ^bb0319. \end{proof}

Corollary

Two discrete valuations $\nu$ and ${\nu }^{\prime }$ of $k$ are equivalent iff they have the same valuation ideal.

\begin{proof} See ^ox907z. \end{proof}

Theorem

Discrete valuation ring is a local PID.

\begin{proof} See ^ve57xd. \end{proof}