Skip to main content
MathOverflow

Return to Answer

added 7 characters in body
Source Link
Johannes Hahn
Johannes Hahn
  • 9.7k
  • 2
  • 33
  • 66

The similarity has nothing to do with boolean algebras, but with orders in general. Filters can be defined for every partial order: A subset $\Phi$ of a poset $\Lambda$ is a filter if

  • $\Phi\neq\emptyset$
  • $\forall a,b\in\Phi \exists c\in\Phi: c\leq a \wedge c\leq b$.
  • $\forall a\in \Phi\forall b\in\Lambda: a\leq b \implies b\in\Phi$

If $\Lambda$ has a greatest element $\infty$, then the first condition can be replaced by $\infty\in\Phi$. If $\Lambda$ has meets, then the second condition can be replaced by $\forall a,b\in\Phi: a\wedge b \in \Phi$.

What you observe is simply that the order on $\Lambda:=K^\times / R^\times$ is defined in such a way that $\nu(R)$ is a filter in $\Lambda\sqcup\{\infty\}$ (where we define $\nu(0) := \infty$ as usual), namely the filter of all elements which are greater or equal $0=\nu(1)$.

In fact you can do this more generally find that $\nu^{-1}(\Phi)$ is an $R$-submodule of $K$ for every filter $\Phi\subseteq\Lambda\sqcup\{\infty\}$.

At least for some rings, say UFD rings $R$ and their quotient fields $K$, the reverse is also true and we get a bijection $$\begin{array}{rcl} \{L\subseteq K \;R\text{-submodule}\} & \overset{\cong}{\leftrightarrow} & \{\Phi \subseteq \Lambda\sqcup\{\infty\} \;\text{filter}\} \\ L &\mapsto& \nu(L) \\ \nu^{-1}(\Phi) &\leftarrow& \Phi \end{array}$$

If the filter is also a submonoid of $(\Lambda,+)$, then $\nu^{-1}(\Phi)$ is also multiplicatively closed, i.e. a $R$-algebra.

Now what a submonoid that is a filter must contain $0$ and therefore all larger elements, i.e. they contain all of $\Lambda_0$$\Lambda_{\geq 0}$. They are also additively closed. In other words, they are a positive cone for an extension of the partial order on $\Lambda$. The maximal submonoid-filters therefore correspond to maximal extensions of the partial order, i.e. total orders. And $K^\times / R^\times$ is totally ordered iff $R$ is a valuation ring.

The similarity has nothing to do with boolean algebras, but with orders in general. Filters can be defined for every partial order: A subset $\Phi$ of a poset $\Lambda$ is a filter if

  • $\Phi\neq\emptyset$
  • $\forall a,b\in\Phi \exists c\in\Phi: c\leq a \wedge c\leq b$.
  • $\forall a\in \Phi\forall b\in\Lambda: a\leq b \implies b\in\Phi$

If $\Lambda$ has a greatest element $\infty$, then the first condition can be replaced by $\infty\in\Phi$. If $\Lambda$ has meets, then the second condition can be replaced by $\forall a,b\in\Phi: a\wedge b \in \Phi$.

What you observe is simply that the order on $\Lambda:=K^\times / R^\times$ is defined in such a way that $\nu(R)$ is a filter in $\Lambda\sqcup\{\infty\}$ (where we define $\nu(0) := \infty$ as usual), namely the filter of all elements which are greater or equal $0=\nu(1)$.

In fact you can do this more generally find that $\nu^{-1}(\Phi)$ is an $R$-submodule of $K$ for every filter $\Phi\subseteq\Lambda\sqcup\{\infty\}$.

At least for some rings, say UFD rings $R$ and their quotient fields $K$, the reverse is also true and we get a bijection $$\begin{array}{rcl} \{L\subseteq K \;R\text{-submodule}\} & \overset{\cong}{\leftrightarrow} & \{\Phi \subseteq \Lambda\sqcup\{\infty\} \;\text{filter}\} \\ L &\mapsto& \nu(L) \\ \nu^{-1}(\Phi) &\leftarrow& \Phi \end{array}$$

If the filter is also a submonoid of $(\Lambda,+)$, then $\nu^{-1}(\Phi)$ is also multiplicatively closed, i.e. a $R$-algebra.

Now what a submonoid that is a filter must contain $0$ and therefore all larger elements, i.e. they contain all of $\Lambda_0$. They are also additively closed. In other words, they are a positive cone for an extension of the partial order on $\Lambda$. The maximal submonoid-filters therefore correspond to maximal extensions of the partial order, i.e. total orders. And $K^\times / R^\times$ is totally ordered iff $R$ is a valuation ring.

The similarity has nothing to do with boolean algebras, but with orders in general. Filters can be defined for every partial order: A subset $\Phi$ of a poset $\Lambda$ is a filter if

  • $\Phi\neq\emptyset$
  • $\forall a,b\in\Phi \exists c\in\Phi: c\leq a \wedge c\leq b$.
  • $\forall a\in \Phi\forall b\in\Lambda: a\leq b \implies b\in\Phi$

If $\Lambda$ has a greatest element $\infty$, then the first condition can be replaced by $\infty\in\Phi$. If $\Lambda$ has meets, then the second condition can be replaced by $\forall a,b\in\Phi: a\wedge b \in \Phi$.

What you observe is simply that the order on $\Lambda:=K^\times / R^\times$ is defined in such a way that $\nu(R)$ is a filter in $\Lambda\sqcup\{\infty\}$ (where we define $\nu(0) := \infty$ as usual), namely the filter of all elements which are greater or equal $0=\nu(1)$.

In fact you can do this more generally find that $\nu^{-1}(\Phi)$ is an $R$-submodule of $K$ for every filter $\Phi\subseteq\Lambda\sqcup\{\infty\}$.

At least for some rings, say UFD rings $R$ and their quotient fields $K$, the reverse is also true and we get a bijection $$\begin{array}{rcl} \{L\subseteq K \;R\text{-submodule}\} & \overset{\cong}{\leftrightarrow} & \{\Phi \subseteq \Lambda\sqcup\{\infty\} \;\text{filter}\} \\ L &\mapsto& \nu(L) \\ \nu^{-1}(\Phi) &\leftarrow& \Phi \end{array}$$

If the filter is also a submonoid of $(\Lambda,+)$, then $\nu^{-1}(\Phi)$ is also multiplicatively closed, i.e. a $R$-algebra.

Now what a submonoid that is a filter must contain $0$ and therefore all larger elements, i.e. they contain all of $\Lambda_{\geq 0}$. They are also additively closed. In other words, they are a positive cone for an extension of the partial order on $\Lambda$. The maximal submonoid-filters therefore correspond to maximal extensions of the partial order, i.e. total orders. And $K^\times / R^\times$ is totally ordered iff $R$ is a valuation ring.

Source Link
Johannes Hahn
Johannes Hahn
  • 9.7k
  • 2
  • 33
  • 66

The similarity has nothing to do with boolean algebras, but with orders in general. Filters can be defined for every partial order: A subset $\Phi$ of a poset $\Lambda$ is a filter if

  • $\Phi\neq\emptyset$
  • $\forall a,b\in\Phi \exists c\in\Phi: c\leq a \wedge c\leq b$.
  • $\forall a\in \Phi\forall b\in\Lambda: a\leq b \implies b\in\Phi$

If $\Lambda$ has a greatest element $\infty$, then the first condition can be replaced by $\infty\in\Phi$. If $\Lambda$ has meets, then the second condition can be replaced by $\forall a,b\in\Phi: a\wedge b \in \Phi$.

What you observe is simply that the order on $\Lambda:=K^\times / R^\times$ is defined in such a way that $\nu(R)$ is a filter in $\Lambda\sqcup\{\infty\}$ (where we define $\nu(0) := \infty$ as usual), namely the filter of all elements which are greater or equal $0=\nu(1)$.

In fact you can do this more generally find that $\nu^{-1}(\Phi)$ is an $R$-submodule of $K$ for every filter $\Phi\subseteq\Lambda\sqcup\{\infty\}$.

At least for some rings, say UFD rings $R$ and their quotient fields $K$, the reverse is also true and we get a bijection $$\begin{array}{rcl} \{L\subseteq K \;R\text{-submodule}\} & \overset{\cong}{\leftrightarrow} & \{\Phi \subseteq \Lambda\sqcup\{\infty\} \;\text{filter}\} \\ L &\mapsto& \nu(L) \\ \nu^{-1}(\Phi) &\leftarrow& \Phi \end{array}$$

If the filter is also a submonoid of $(\Lambda,+)$, then $\nu^{-1}(\Phi)$ is also multiplicatively closed, i.e. a $R$-algebra.

Now what a submonoid that is a filter must contain $0$ and therefore all larger elements, i.e. they contain all of $\Lambda_0$. They are also additively closed. In other words, they are a positive cone for an extension of the partial order on $\Lambda$. The maximal submonoid-filters therefore correspond to maximal extensions of the partial order, i.e. total orders. And $K^\times / R^\times$ is totally ordered iff $R$ is a valuation ring.