## Classification of symtrivial modules over a PID

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Let us call a module $M$ over a commutative ring $R$ symtrivial if the symmetry $M \otimes M \to M \otimes M, a \otimes b \mapsto b \otimes a$ equals the identity (the same notion applies to arbitrary symmetric monoidal categories). Equivalently, every bilinear map on $M \times M$ is symmetric. Obviously $R$, and more generally every invertible $R$-module, is symtrivial, but there are many more symtrivial modules, which follows from the following closure properties.

• A directed colimit of symtrivial modules is symtrivial.
• A direct sum $\oplus_{i \in I} M_i$ is symtrivial iff all $M_i$ are symtrivial and $M_i \otimes M_j = 0$ for $i \neq j$.
• Every quotient of a symtrivial module is symtrivial.
• Every symmetric monoidal functor, possibly non-unital, preserves symtrivial modules. For example, localizations of symtrivial modules are symtrivial.
• If $A$ is a commutative $R$-algebra, then $A$, regarded as an $R$-module, is symtrivial iff $R \to A$ is an epimorphism in the category of commutative rings.
• If $M_{\mathfrak{p}}$ is symtrivial for all $\mathfrak{p} \in \mathrm{Spec}(R)$, then $M$ is symtrivial.

From Nakayama one can deduce that a finitely generated $R$-module is symtrivial iff all $M_{\mathfrak{p}}$ are cyclic $R_{\mathfrak{p}}$-modules.

Question. Is there a classification of symtrivial $\mathbb{Z}$-modules?

Almost equivalent one may ask for a classification of symtrivial $R$-modules, where $R$ is a PID (or even a Dedekind domain?). There is a classification of epimorphisms of commutative rings $R \to A$, see Torsten Schöneberg's answer here. Either $A=R/rR$ for some $0 \neq r \in R$, or

$A \cong A_{n,\widehat{P}} := (\widehat{P} \setminus P)^{-1} R[(x_p)_{p \in P}] / (x_p (1-p x_p),p^{n(p)} (1-p x_p))_{p \in P},$

where $\widehat{P}$ is a set of primes in $R$, $P \subseteq \widehat{P}$ is a subset and $n : P \to \mathbb{N}^+$ is a function. These are in particular symtrivial $R$-modules.

Every locally cyclic $R$-module (:= every finitely generated submodule is cyclic) is symtrivial; these are precisely the submodules of $Q(R)$ and of $Q(R)/R$. Since $Q(R) \otimes Q(R)/R=0$, also $Q(R) \oplus Q(R)/R$ is symtrivial.

We can continue this way and use the closure properties to optain lots of examples of symtrivial modules. Nevertheless, I wonder if any classification is available (similar to the classification of epimorphisms, where in fact every epi can be optained by a canonical order of closure properties).

PS: I invented the notion of symtrivial objects for myself and also wonder if anybody else has worked with them or if this notion is already known under a different name. Any information is appreciated. The background is that for a given cocomplete symmetric monoidal category $C$ one can show that $\mathsf{gr}_{\mathbb{N}}(C)$ is the universal cocomplete symmetric monoidal category over $C$ together with a symtrivial object (namely $1_C[-1]$). Thus, in the context of graded objects symtrivial objects pop out quite naturally, and my question is equivalent to the classification of cocontinuous symmetric monoidal $R$-linear functors $\mathsf{gr}_{\mathbb{N}}(R) \to \mathsf{Mod}(R)$.

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1
Quick guess is that they are the abelian groups of rank less than $2$, i.e., abelian groups such that the $\mathbb{Q}$-vector space $\mathbb{Q} \otimes_{\mathbb{Z}} A$ is of dimension $0$ or $1$. If so, would that be a satisfactory classification? – Todd Trimble Jan 23 at 20:28

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Notice that $F: \mathbb{Q} \otimes_{\mathbb{Z}} -$ is a strong symmetric monoidal functor, so if $\sigma \neq 1: F(A) \otimes F(A) \to F(A) \otimes F(A)$, then $\sigma \neq 1: A \otimes A \to A \otimes A$. This means that if $A$ is of rank 2 of greater, then $A$ cannot be symtrivial. Thus being rank $0$ or $1$ is a necessary condition for symtriviality. I think it's also sufficient. – Todd Trimble Jan 23 at 20:34

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If $A$ is a commutative ring with $\mathbb{Z}$-rank $\leq 1$, then $\mathbb{Z} \to A$ doesn't have to be an epimorphism and therefore $A$ is not symtrivial over $\mathbb{Z}$. For example, $A = \mathbb{Z}/2 \times \mathbb{Z}/2$. – Martin Brandenburg Jan 23 at 20:45
I see, thanks.. – Todd Trimble Jan 23 at 20:46

## 1 Answer

We can describe a symtrivial module over a Dedekind domain $R$ with field of fractions $K$ as follows:

(1) Torsion-free symtrivial modules are submodules of $K$.

(2) The torsion part of a symtrivial module is symtrivial

(3) A torsion module is symtrivial if and only if it is a direct sum of symtrivial $p$-power-torsion modules

(4) A $p$-power-torsion module $M$ is symtrivial if and only if it satisfies either $M/pM=0$ or $M/pM=R/pR$,.

(5) If a $p$-power-torsion module $M$ satisfies $M/pM=R/pR$, then it is the direct sum of a $p$-divisible $p$-torsion module and a cyclic module. (Obviously if $M/pM=0$ then it is a $p$-divisible $p$-torsion module.)

(6) If $0 \longrightarrow T \longrightarrow M \longrightarrow F \longrightarrow 0$ is an exact sequence, with $T$ torsion, $M$ symtrivial, and $F$ torsion-free, then for each prime $p$, either $F$ is $p$-divisible or $T$ has no $p$-power-torsion.

(7) If $T$ is torsion and symtrivial, $F$ is torsion-free and symtrivial, and for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-power-torsion, then $T \oplus F$ is symtrivial.

If none of my proofs are mistaken, the remaining questions are

(Q1) For $T$ torsion symtrivial, $F$ torsion-free symtrivial, such that for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-torsion, what is $\mathrm{Ext}^1(F,T)$?

(Q2) Do all elements of $\mathrm{Ext}^1(F,T)$ give symtrivial modules?

Proofs:

(1) This is immediate, by Todd Trimble's argument.

(2) Indeed, take a module $M$ with torsion submodule $T$ such that $a \otimes b \neq b \otimes a$ in $T\otimes T$, but $a\otimes b = b \otimes a$ in $M\otimes M$. The equality $a \otimes b = b\otimes a$ must be the consequence of finitely many relations, involving finitely many elements. Consider the submodule $M'$ generated by the whole torsion module and those finitely many elements. The torsion-free quotient of $M'$ is a finitely generated submodule of $K$, thus is a fractional ideal, thus projective, so the submodule splits into a direct sum of torsion and torsion-free parts. But than $T\otimes T$ is a direct summand of $M'\otimes M'$, so if $a\otimes b \neq b\otimes a$ in $T\otimes T$, they do not equal each other in $M'\otimes M'$ - but a complete set of relations implying that they do are relations of $M' \otimes M'$, a contradiction.

(3) This is immediate from things noted in the original question.

(4) Only if follows from the fact that quotients of symtrivial modules are symtrivial. For if, first note that if $M/pM=0$, then $M$ is $p$-divisible and $p$-torsion, so its tensor product with itself is trivial. If $M/pM=R/pR$, let $x$ be a lift of $1 \in R/pR = M/pM$ to $M$. For any $a \otimes b$, choose $n \geq 0$ such that $p^n a = p^n b = p^n x = 0$. Choose $u,v \in \mathbb{Z}$ with $a \equiv ux \pmod {p^n}$ and $b \equiv vx \pmod {p^n}$. Then $a \otimes b = ux \otimes b = ux \otimes vx = vx \otimes ux = b \otimes ux = b \otimes a$.

(5) Let $\pi$ be a uniformizer of $R_p$. This defines a surjection $p^{n-1}M/p^nM \to p^n M/p^{n+1}M$, so either $p^nM/p^{n+1}M$ is eventually $0$ or we get a nontrivial morphism $M \to \lim_n M/p^n M = \lim_n R/p^nR=\hat{R}_p$ which is torsion-free, so it's eventually $0$. Then we have a surjection $M \to M/p^n M= R/p^nR$ whose kernel is $p$-divisible. Since $p$-divisible $p$-torsion groups are divisible, they are injective, so the exact sequence splits and we get a direct sum.

(6) Localize the exact sequence at $p$. Then $F_p$ is a submodule of $K_p$, so either $R_p$ or $K_p$. In the second case $F$ is $p$-divisible so take the first case. $R_p$ is projective so the exact sequence splits into a direct sum, so $R_p \otimes T_p = 0$ by one of the notes in the question, but $R_p \otimes T_p=T_p$ so $T_p=0$.

(7) It suffices to prove that $T \otimes F=0$, by one of the notes in the question. But we can divide $T$ into $p$-power-torsion modules, and a $p$-divisble module tensor a $p$-power-torsion module is trivial.

Answers to Q1 and Q2:

Q1: We disambiguate: "strongly locally cyclic" means that its localization at each prime is cyclic. "weakly locally cyclic" means that each finitely generated submodule is cyclic. $K_p/R_p$ is in the second class but not the first.

We can write $T$ as the direct sum of divisible torsion module $D$ and a strongly locally cyclic module $C$. Divisible modules are injective, so $Ext^1(F,T)=Ext^1(F,C)$.

$Ext^1(F,C)=\left(\prod_p C_p\right)/C$

Coose any nontrivial homomorphism $R \to F$, then the morphism is injective and the kernel is weakly locally cyclic, say $L$. We have a long exact sequence

$Hom(L,C) \to Hom(F,C) \to Hom(R,C) \to Ext^1(L,C) \to Ext^1(F,C) \to Ext^1(R,C)$

$Hom(R,C)=C$, $Ext^1(R,C) = 0$

$Hom(F,C) = 0$ because $F$ is $p$-divisible for all $p$ such that $C_p$ is nontrivial, so the image of any homomorphism inside $C_p$ is trivial, so the whole homomorphism is trivial.

$Ext^1(L,C) = \prod_p Ext^1(L_p,C_p)$ because $L$ and $C$ are torsion. Whenever $C_p$ is nonzero, $L_p=K_p/R_p$ so $Ext^1(L_p,C_p)=Ext^1(K_p,R_p,R/p^a) = R/p^a=C_p$. So we have a short exact sequence

$0 \to C \to \prod_p C_p \to Ext^1(F,C) \to 0$

If we're interested in classifying extensions abstractly, we just have to mod out by the action of the automorphism group.

Q2: All are symtrivial.

Indeed, let $M$ be any module whose torsion and non-torsion parts satisfy the given conditions, and choose $a \otimes b \in M \otimes M$. Choose $x \in M$ such that $a \in \langle x,t \rangle$, $b\in \langle x,t\rangle$. then $a \otimes b= (k_1 x+ t_1) \otimes (k_2 x + t_2) = k_1k_2 x\otimes x + k_1 x\otimes t_2 + k_2 t_1 \otimes x + t_1 \otimes t_2$. Then $t_1 \otimes t_2 = t_2 \otimes t_1$, so the only thing that remains to check is if $t_1\otimes x=x\otimes t_1$. Suppose $n t_1=0$, then choose $y$ such that $ny \equiv x$ mod $T$, so

$x\otimes t_1=(x-ny)\otimes t_1 =t_1 \otimes (x-ny) = t_1 \otimes x$.

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Thanks a lot for the new version, it is very clear. I still have some problems with (5). I see that $R/p^n R \to M/p^n M$ is surjective, but why is it injective? – Martin Brandenburg Jan 25 at 14:44
The kernel is a submodule of $R/p^nR$, which is a quotient of a local ring of a Dedekind domain, hence a DVR, so all ideals are powers of the maximal ideal, $p$. – Will Sawin Jan 25 at 19:17
(5) Yes, but how can we exclude $R/p^k R \cong M/p^n M$ for $k<n$? (Q2) Why is there some $y$ such that $ny \equiv x$ mod $T$? This is only clear when $M/T = K$. – Martin Brandenburg Jan 25 at 23:41
(5) Because then $p^{n-1}M/p^nM= p^{n-1}(R/p^kR)=0$. (Q2) Because for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-torsion – Will Sawin Jan 26 at 0:55
@Will: Is each submodule of $K$ symtrivial ?  – TJ Jan 27 at 19:09
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