Submodules of a direct sum of ideals in a valuation ring

34 Views Asked by At

Let $K$ be the field of Hahn series in an indeterminate $t$ with exponents in $\mathbb{R}$, coefficients in $\mathbb{F}_2$ and valuation $\nu$. Set $$A:=\{a\in K:\nu(a)\geq 0\},$$ which has unique maximal ideal $$\{a\in K:\nu(a)>0\}.$$ Label this $I_{>0}$. I want to find the submodules of $I_{>0}\oplus I_{>0}$. I have the following useful result:

Let $F$ be a free and finitely generated $A$-module and $M$ be an $A$-submodule of $F$. Then there exists a basis $\{v_i\}$ for $F$ and ideals $L_i$ in $A$ for which $M=\bigoplus_i L_iv_i$.

I also have a similar result for $A$-submodules of finite-dimensional $K$-vector spaces.

So let $M\leq I_{>0}\oplus I_{>0}$. Then there exists a basis $\{v_1,v_2\}$ of $A\oplus A$ and ideals $L_1,L_2$ in $A$ for which $M=L_1v_1\oplus L_2v_2$. If $L_1=L_2=A$, then $M=A\oplus A$, clearly false. But can $L_1=A$ and $L_2<A$, say $L_2=I_{>0}$? Does there exist a basis $\{v_1,v_2\}$ for which $Av_1\oplus I_{>0}v_2\leq I_{>0}\oplus I_{>0}$? I don't know where to go with this.

My overall goal is to classify the submodules of direct sums of finitely many copies of $K$, $A$ and $I_{>0}$, having done the case where only $A$'s or only $K$'s are present.

1

There are 1 best solutions below

0
On

$M$ is a sub $A$-module of $A^n$. For each $r\in \Bbb{R}_{\ge 0}$ let $S_r = t^r A^n \cap M$ and $E_r = t^{-r} S_r / t^{-r} S_{r+\epsilon}$ which is a $k=A/I_{>0}$ vector subspace of $k^n$. Then $r' > r\implies E_r\subset E_{r'}$. Your $A$-module $M$ is given by the $E_r$. Find the $v_j,w_i\in k^n,c_j,d_i\in \Bbb{R}$ such that $E_r= \bigoplus_{c_j < r} v_j k \oplus \bigoplus_{d_i \le r} w_i k$. This defines $M$ as $\bigoplus v_j t^{c_j} I_{>0}\oplus \bigoplus w_i t^{d_i} A$.