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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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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 '13 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 '13 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 '13 at 20:45
    
I see, thanks.. –  Todd Trimble Jan 23 '13 at 20:46

1 Answer 1

up vote 6 down vote accepted

Let $R$ be a Dedekind domain with field of fractions $K$. A module $M$ is symtrivial if and only if its maximal torsion-free quotient $F$ is a submodule of $K$ and its torsion submodule $T$ is a sum over each prime $p$ such that $F$ is $p$-divisible of a $p$-divisible $p$-torsion module and a $p$-power cyclic module $R/p^n$.

If $F$ and $T$ are modules as described, then we can classify extnesions $0 \to T \to M \to F \to 0$ by the group $Ext^1(F,T)$. If $F\neq 0$, this group is equal to $\prod_p R/p^n / \sum_p R/p^n$.

Proof: First we prove that such a module is symtrivial. To do this, let $a$ and $b$ be two elements and let $I$ be the submodule of $F$ generated by $a$ and $b$ modulo $T$. $I$ is a finitely generated submodule of $M$, hence it is projective, so we can find an injection $I \to M$. Hence $a$ and $b$ can be written as $x_1+t_1$, $x_2+t_2$ with $x_1,x_2 \in I$, $t_1,t_2 \in T$. We need to check:

a. $x_1 \otimes x_2 = x_2\otimes x_1$ for $x_1,x_2 \in I$.

b. $x \otimes t= t\otimes x$ for $x\in I$, $t\in T$

c. $t_1\otimes t_2 = t_2\otimes t_2$ for $t_1,t_2, \in T$.

Proof of a: $I$ is either $0$ or a fractional ideal. The case $I=0$ is trivial, so we may assume that $I$ is a fractional ideal. $I\otimes I$ is $I^2$, another fractional ideal, and the multiplication is clearly commutative.

Proof of b: $t$ is torsion, so assume $n t =0$ for some $n \in R$. For each prime $p$ dividing $n$, we know that $F$ is $p$-divisible, so $x$ is $n$-divisible up to torsion. In other words, we can write $x=ny+t'$ for $t'$ torsion. Then

$$x\otimes t= (ny+t') \otimes t = n (y\otimes t)+(t'\otimes t)= (t'\otimes t)$$

This reduces this case to case c.

Proof of c: Write $T= C+D$ for $C=\sum_p R/p^n$ and $D$ a divisible torsion module. Since $D$ is divisible, the tensor product of $D$ with any torsion module is $0$. Hence $T\otimes T = (C+D)\otimes (C+D)= C\otimes C$. But $C\otimes C = \sum_p R/p^n \otimes R/p^n = \sum_p R/p^n = C$, where the isomorphism is given by the multiplicative structure of $R/p^n$, which is clearly commutative, so $t_1\otimes t_2 = t_2 \otimes t_1$.


Next we prove that any symtrivial module has this form. Given a symtrivial $M$, we write it as an exact sequence $0 \to T \to M \to F \to 0$, with $T$ torsion and $F$ torsion-free. First we show that $F \subseteq K$. This just follows Todd Trimble's argument from the comments - tensoring with $K$ is symmetric monoidal, so preserves symtriviality, so it takes $M$ to a symtrivial vector space, which must be $0$ or $1$-dimensional. The image of the map $M \to M \otimes_R K$ is just $F$, so $F \subseteq K$.

Next we must show that $T_p=0$ if $F$ is not $p$-divisible and is a sum of a cyclic module and a $p$-divisible $p$-module otherwise. To do this we may consider the localization $M_p$, which fits into an exact sequence $0 \to T_p \to F_p \to M_p$. $M_p$ must be symtrivial. $M_p \subset R_p$. $F_p \subseteq K_p$, so $F_p=0,R_p$, or $K_p$. The middle case is the non-$p$-divisible case, and it clearly forces $T_p=0$, since $R_p$ is projective so the exact sequence splits, and we have $(R_p + T_p) \otimes (R_p + T_p) = R_p+T_p+T_p+ T_p\otimes T_p$, with the two $T_p$s switched by the symmetry, so they must both vanish.

If $F_p=0$, $T_p$ is symtrivial. We will demonstrate that if $F_p=K_p$, $T_p$ is symtrivial still. Suppose not. Find an $a,b$ in $T_p$ such that $a\otimes b \neq b \otimes a$ in $T_p \otimes T_p$, but $a \otimes b = b \otimes a$ in $M_p \otimes M_p$. The proof that $a \otimes b = b\otimes a$ must involve finitely many elements of $R_p$,. Consider the submodule $M'$ of $M_p$ generated by $T_p$ and those finitely many elements. The torsion-free quotient of $M'$ is a finitely generated submodule of $K_p$, thus is $R_p$ or $0$, thus is projective, so the submodule splits into a direct sum of torsion and torsion-free parts. But than $T_p\otimes T_p$ is a direct summand of $M'\otimes M'$, so if $a\otimes b \neq b\otimes a$ in $T_p\otimes T_p$, 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.

So $T_p$ is symtrivial. We must show it is a sum of a $p$-divisible $p$-torsion module and a $p$-power cyclic module. Since tensoring with $R/p$ is a symmetric monodical functor, $T_p/pT_p$ is symtrivial, so $T_p/pT_p = 0$ or $R/p$. In the first case, $T_p$ is a $p$-divisible $p$-torsion module, so we are done. So the $p$-adic completion of $T_p$ is, by Nakayama's Lemma, generated by a single element. The $p$-adic completion has the form $\hat{R}_p/p^n$ for some $n$, and $n$ must be finite because otherwise the completion morphism would be a nontrivial morphism to a $\hat{R}_p$, a torsion-free module, which is impossible. The kernel of $T_p \to \hat{T}_p$ is $p$-divisible because its elements are exactly the elements of $T_p$ that are in $p^n T_p$ for each $n$. Since it is $p$-divisible $p$-torsion, it is injective, so the exact sequence $0 \to \cap_n p^n T_p \to T_p \to R/p^n \to 0$ splits. (The last map is a surjection because every map to $R/p^n$ that projects nontrivially onto $R/p$ is surjective.)


Finally, we compute the Ext group. Again write $T= C+D$ where $C= \sum_p R/p^n$ adn $D$ a divisible torsion module. $Ext^1(F,T)= Ext^1(F,C)+Ext^1(F,D)$, but $D$ is injective so $Ext^1(F,D)=0$. and we just need to find $Ext^1(F,C)$. We will do so using the following lemma:

Lemma: Let $X$ be a torsion $R$-module that is the sum of finite $p$-torsion modules for different primes $p$, so $X = \sum_p X_p$. Let $Y$ be a torsion-free $R$-module that is $p$-divisible for each $p$ such that $X_p$ is nontrivial. Then

$$Ext^1(Y,X) = Hom(Y \otimes_R K, \prod_p X_p / \sum_p X_p )$$

Proof: We use the exact sequence $0\to \sum_p X_p \to \prod_p X_p \to \prod_p X_p / \sum_p X_p$. We obtain a long exact sequence, the relevant terms are $Hom(Y, \prod_p X_p) \to Hom(Y, \prod_p X_p / \sum_p X_p) \to Ext^1(Y,X) \to Ext^(Y,\prod_p X_p)$. We can pull out the product, replacing the first term with $\prod_p Hom( Y,X_P)$ and the second term with $Ext^1(Y, X_p)$. These modules must be $p$-divisible and must be boundedly $p$-torsion, so they vanish, and hence we get an isomorphism $Hom(Y, \prod_p X_p / \sum_p X_p) =Ext^1(Y,X)$. But $\prod_p X_p/ \sum_p X_p$ is divisible, and torsion-free, hence it is a $K$-vector space, so the Homs uniquely factor thoruhg $Y \otimes_R K$.

In our case setting $X=C$, $Y=F$, $Y \otimes_R K$ is a one-dimensional vector space if $F$ is nontrivial, so the Homs are just $\prod_p R/p^n / \sum_p R/p^n$

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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 '13 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 '13 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 '13 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 '13 at 0:55
    
@Will: Is each submodule of $K$ symtrivial ? –  tj_ Jan 27 '13 at 19:09

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