A case where submodules, isomorphic as modules, are isomorphic as submodules

Question

Let $R$ be a discrete valuation ring with maximal ideal $m$ and let $M$ be a finitely generated torsion $R$-module. Let $K$ and $K'$ be two submodules of $M$ which are isomorphic as $R$-modules.

Can I find an automorphism $f$ of $M$ mapping $K$ to $K'$?

I am not so sure about what the answer is. I tried to use the decomposition of $M$ as $$M\cong \bigoplus_{i>0} (R/m^i)^{e_i}, $$ where only finitely many $e_i$'s are non zero.

1. Clearly this holds if $e=\sum_i{e_i}=1$, as then $K$ and $K'$ have to be equal.
2. This also holds if $mM=(0)$, as this reduces to the case of vector spaces.

I cannot see how to pursue the argument. Many thanks in advance for your thoughts on this!

Answer 1

Not true in general. For simplicity, let $R=\mathbb Z_2$, the ring of $2$-adic integers, and $M=\mathbb Z/2\mathbb Z\oplus\mathbb Z/4\mathbb Z$, note that $$K:=(\mathbb Z/2\mathbb Z\times \{0\})\cong (\{0\}\times \{4\mathbb Z, 2+4\mathbb Z\})=:K'$$ as $R$-modules, but there is no automorphism $T$ of $M$ that sends $K$ to $K'$, because the equation $2x=a\in K\setminus\{0\}$ has no solution $x\in M$ while $2x=a\in K'\setminus\{0\}$ does.

The same argument essentially works for general DVR.

Answer 2

The question as posed is in a sense overly-specific: the main point is that if $R$ is a ring, and $M$ an $R$-module, then submodules $N_1,N_2$ of a module $M$ can easily be isomorphic as modules without there having to be an automorphism of the ambient module $M$ which takes one to the other.

For any commutative ring $R$, $\mathrm{End}_R(R) = R$ where if $\theta \colon R \to R$ is an $R$-module homomorphism, we associate to it $\theta(1)$, and conversely if $r_0\in R$ then $\theta_{r_0}(r) :=rr_0$. It follows that $\mathrm{Aut}_R(R)= R^\times$, the group of units in $R$. But then it is clear that any automorphism of $R$ will fix every ideal of $R$, so that if $I_1,I_2$ are distinct ideals, no automorphism of $R$ will map one to the other.

On the other hand, if, for example, $R$ is a domain then any two principal ideals of $R$ are isomorphic to $R$ as $R$-modules, and hence are isomorphic to each other, so that provided $R$ is not a field, it will contain examples of submodules $N_1,N_2$ which are isomorphic as modules but not conjugate under an automorphism of $R$.

The most basic example of the above is probably when $ R=\mathbb Z$ and $N_1=\mathbb Z$, $N_2 =2\mathbb Z$.

The question asks about finitely generated torsion $R$-modules, which means the above does not directly apply, but it is easy enough to adapt: Note that if $\phi\colon M\to M$ is an automorphism, just as an automorphism of $R$ fixed every ideal in $R$, we see that $\phi(IM) = IM$ for any ideal $I$ of $R$, and so in particular, $\phi(\mathfrak m^kM) = \mathfrak m^kM$. On the other hand, as we saw above, as abstract $R$-modules, the ideals $\mathfrak m^k$ are all isomorphic via the maps $\psi_d\colon \mathfrak m^k \to \mathfrak m^{d+k}$ where $\psi_d(x)=x.\pi^d$. Moreover $\psi_d$ induces an isomorphism $\bar{\psi}_d\colon R/\mathfrak m^k \cong \mathfrak m^d/\mathfrak m^{d+k}$.

This gives the obvious generalisation of Just a user's example (to elaborate on his final sentence): if $M= R/\mathfrak m\oplus R/\mathfrak m^k$ where $k>1$, then $\mathfrak m^{k-1}M = \{0\}\oplus \mathfrak m^{k-1}/\mathfrak m^k$ is preserved by any automorphism of $M$, while $\phi(a,b) = (0,a\pi^{k-1})$ restricts to an isomorphism between $N_1 = R/\mathfrak m\oplus 0$ and $N_2 =\mathfrak m^{k-1}M$.