Say $R$ is a ring, not necessarily a domain, and $M$ is an $R$-module. All rings are commutative with 1. An element $m\in M$ is called torsion if $r.m=0$ for some regular element (non-zerodivisor) $r\in R$.

(I learned this definition over non-domains from a lecture of Irena Swanson, and it's noted in the last paragraph of the definition on Wikipedia.)

It's easy to see that torsion elements localize to torsion elements (note $0$ is always torsion), because regular elements never disappear. How about the converse? If an element is locally torsion, is it torsion? That is,

If $f_1,\ldots,f_n$ generate the unit ideal in $R$, and $m\in M_{f_i}$ is torsion over $R_{f_i}$ for each $i$, then is $m$ torsion over $R$?

I started trying to make a high-dimensional variety as a counterexample, but instead wound up proving that for $R$ Noetherian, torsionality is stalk-local 1, so in that case the answer is yes.

So how about a proof or a counterexample for rings in general? This will affect when and whether I think about sections of $O_X$-modules on a non-integral scheme $X$ as "torsion" or not...

1 Proof of stalk-locality in Noetherian case: Say $m\in M$ is non-torsion, meaning $ann(m)$ is contained in the set of zero-divisors of $R$, which equals the union of the associated primes of $R$. By prime avoidance, $ann(m)$ is contained in some associated prime of $R$, say $p=ann(x)$ for $x\in R$. I claim $m$ localizes to a non-torsion element of $M_p$. If $r.m=0$ in $M_p$ for $r\in R$ (WLOG), it means $rs.m=0$ in $M$ for some $s\notin p$. Now, $rs\in ann(m)\subseteq p$, so $r\in p$ and $rx=0$ in $R$. But $x\neq 0$ in $R_p$ since $ann(x)$ is (contained in) $p$, so $r$ is not regular in $R_p$, as required. Hence an element of $M$ which is torsion in every $M_p$ must be torsion in $M$.

  • 1
    $\begingroup$ Here is a reformulation, which is trivial but perhaps useful: Assume $I \subseteq R$ is an ideal, such that $I_{f_i}$ contains a regular element of $R_{f_i}$ for every $i$. Does then $I$ contain a regular element of $R$? $\endgroup$ Commented Oct 20, 2010 at 21:46

3 Answers 3


No, being torsion is not a local property, and I can give a counterexample. [Edit: This took some doing, with my initial answer containing a serious flaw. After completely reworking the construction, this should work now. Apologies for the length of this answer, but I don't see any quick constructions].

The idea is to construct a ring $R$ and an ideal $I$ contained in the zero divisors of $R$, and $f_1,f_2\in R$ satisfying $f_1+f_2=1$ such that $I_{f_i}$ contains a regular element of $R_{f_i}$ for each $i$. This provides a counterexample to the question by taking the module $M=R/I$ and $m=I+1$. Then, ${\rm ann}(m)=I$ consists of zero divisors, so $m$ is not torsion. However, mapping $m$ into $M_{f_i}$ takes ${\rm ann}(m)$ to $I_{f_i}$, which contains regular elements of $R_{f_i}$. So, $m$ is torsion in each $M_{f_i}$.

This does get rather involved, so let's start simple and construct an example showing that being torsion is not a stalk-local property.

Choose a field $k$, set $A=k[X_0,X_1,X_1,X_2,\ldots]$ and let $J$ be the ideal generated by $X_iX_j$ for $i\not=j$ and $X_i(X_i-1)$ for $i\ge1$. Then, $R=A/J$ is the $k$-algebra generated by elements $x_0,x_1,\ldots$ satisfying the relations $x_ix_j=0$ for $i\not=j$ and $x_i(x_i-1)=0$ for $i\ge1$. Let $I\subseteq R$ be the ideal generated by $x_0,x_1,\ldots$. We can see that $x_i\not=0$ by considering the $k$-morphism $A\to k$ taking $X_j$ to 1 (some fixed $j$) and $X_i$ to 0 for $i\not=j$. This takes $J$ to 0, so it defines a $k$-morphism $R\to k$ mapping $x_j$ to 1, so $x_j\not=0$. Then, every $a\in I$ satisfies $ax_j=0$ for large $j$, showing that it is a zero divisor. Also, the $k$-morphism $A\to k$ taking each $X_i$ to zero contains $J$ in its kernel, and defines a morphism $R\to\mathcal{k}$ with kernel $I$, showing that $R/I\cong k$. So $I$ is a maximal ideal. For any prime $\mathfrak{p}$ we either have $\mathfrak{p}\not=I$, in which case the non-empty set $I\setminus\mathfrak{p}$ maps to units (and hence, regular elements) in $R_{\mathfrak{p}}$. Or, we have $\mathfrak{p}=I$ in which case $x_i-1$ maps to a unit and $x_i$ goes to zero in $R_{\mathfrak{p}}$ ($i\ge1$). So, $R_{\mathfrak{p}}\cong k[X]$ with $x_0$ going to the regular element $X$. This shows that $I$ contains regular elements in the localization at any prime, giving the required counterexample for the stalk-local case.

Now, let's move on to the full construction of the counterexample showing that being torsion is not a local property. Simply guessing a set of generators and relations as for the stalk-local case didn't work out so well. Instead, I will start with a simple example of a polynomial ring and then transform it in such a way as to give the properties we are looking for. I find it helpful to first fix the following notation: Start with the base (polynomial) ring $R=\mathbb{Z}[x,y,z]$. A (commutative, unitial) R-algebra is simply a ring with three distinguished elements $x,y,z$, and a morphism of R-algebras is just a ring homomorphism respecting these distinguished elements. For an R-algebra $A$, define $K(A)\subseteq A$ to be the smallest ideal such that, for all $a\in A$, $$ \begin{align} ax\in K(A)&\Rightarrow az\in K(A),\\\\ ay\in K(A)&\Rightarrow a(1-z)\in K(A). \end{align} $$ In particular, $K(A)=0$ implies that $x$ is a regular element in the localization $A_z$ and $y$ is a regular element in $A_{1-z}$. If we can construct such an example where the ideal $Ax+Ay$ consists purely of zero divisors, then that will give the counterexample needed. The idea is to start with $A=\mathbb{Z}[x,y,z]$ and transform it using the following steps.

  • Force the elements of $I=Ax+Ay$ to be zero divisors. So, for each $a\in I$, add an element $b$ to $A$ in as free a way as possible such that $ab=0$. Adding elements to $A$ also has the effect of adding elements to $I$. So, this step needs to be iterated to force these new elements of $I$ to also be zero divisors.
  • Replace $A$ by the quotient $A/K(A)$ to force the condition $K(A)=0$.

The first step above is easy enough. However, we do need to be careful to check that the second step does not undo the first. Suppose that $a\in A$ is a zero divisor, so that $ab=0$ for some non-zero $b$. It is possible that taking the quotient in the second step above takes $b$ to zero, so that $a$ becomes a regular element again. To get around this, we need some stronger condition on $b$ which implies $b\not=0$ and is also stable under taking the quotient. Note that $A(1-b)$ being a proper ideal or, equivalently, $A/(1-b)$ being nontrivial, will imply that $b\not=0$. In turn, this is implied by $K(A/(1-b))$ being a proper ideal. As it turns out, this property of $b$ does remain stable under each of the steps above, and can be used to show that this construction does give the counterexample required. However, note that if $ab=0$ and $K(A/(1-b))$ is proper, then $a=a(1-b)\in A(1-b)$, from which we can deduce that $K(A/(a))$ is a proper ideal. This necessary condition is unchanged by either of the steps above, so we had better check that elements $a\in Ax+Ay$ in our R-algebra do satisfy this from the outset. I'll make the following definition: $A$ satisfies property (P) if $K(A/(a))$ is proper for every $a\in Ax+Ay$. As it turns out, polynomial rings do satisfy this property and, consequently, the construction outlined above works fine.

Now on to the details of the argument.

(1) Let $f\colon A\to B$ be an R-morphism. Then $f(K(A))\subseteq K(B)$. Furthermore,

  • If $I\subseteq A$, $J\subseteq B$ are ideals with $f(I)\subseteq J$ and $K(B/J)$ is proper, then $K(A/I)$ is proper.
  • If $B$ satisfies (P) then so does $A$.

As $f^{-1}(K(B))$ satisfies the defining properties for $K(A)$ (other than minimality), it contains $K(A)$. In particular, if $K(B)$ is proper then $K(A)\subseteq f^{-1}(K(B))$ is proper. Next, if $f(I)\subseteq J$ are ideals, then $f$ induces an R-morphism $A/I\to B/J$ so, if $K(B/J)$ is proper then so is $K(A/I)$.

If $B$ satisfies property (P) and $a\in Ax+Ay$ then $f(a)\in Bx+By$ and $K(B/f(a))$ is proper. So, $K(A/(a))$ is proper and $A$ also satisfies property (P).

(2) If $A$ is a non-trivial ring then the polynomial ring $R\otimes A\cong A[x,y,z]$ satisfies (P).

As $A$ is non-trivial, it must have a maximal ideal $\mathfrak{m}$. Applying (1) to the R-morphism $A[x,y,z]\to(A/\mathfrak{m})[x,y,z]$ reduces to the case where $A=k$ is a field. Then, letting $\bar k$ be the algebraic closure, applying (1) to $k[x,y,z]\to\bar k[x,y,z]$ reduces to the case where $A=k$ is an algebraically closed field.

Now set $B=k[x,y,z]$ and choose $a\in Bx+By$. The idea is to look at the morphism $\theta\colon B/(a)\to k$ taking $x,y,z$ to some $x_0,y_0,z_0\in k$ with $a(x_0,y_0,z_0)=0$. As long as these satisfy $ux_0=0\Rightarrow uz_0=0$ and $uy_0=0\Rightarrow u(1-z_0)=0$ (all $u\in k$) then $K(B/(a))$ will be contained in the kernel of $\theta$, so will be proper. For this to be the case it is enough that both ($x_0\not=0$ or $z_0=0$) and ($y_0\not=0$ or $z_0=1$).

Case 1: We can find $a(x_0,y_0,z_0)=0$ such that $x_0y_0\not=0$. This satisfies the required condition.

Case 2: Whenever $a(x_0,y_0,z_0)=0$ then $x_0y_0=0$. This means that $xy$ is contained in the radical ideal generated by $a$, so $a$ divides $x^ry^r$ some $r\ge1$. Then $a$ is a multiple of $x$ or $y$ and one of $(x_0,y_0,z_0)=(0,1,0)$ or $(1,0,1)$ satisfies the required condition. So, $K(B/(a))$ is proper.

Next, we construct extensions of the R-algebra forcing elements of $Ax+Ay$ to be zero-divisors.

(3) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ with a left-inverse and such that, for every $a\in Ax+Ay$, there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ is proper.

To construct the morphism, set $I=Ax+Ay$ and let $(X_a)_{a\in I}$ be indeterminates over $A$. Let $J$ be the ideal in $A[(X_a)_{a\in I}]$ generated by $(aX_a)_{a\in I}$. Then define $B=R[(X_a)_{a\in I}]/J$ and let $f$ be the canonical homomorphism. Its left inverse is the map taking $X_a$ to 0.

Now, for a fixed $a\in I$, set $b=J+X_a$, so $ab=0$. Consider the morphism $A[(X_c)_{c\in I}]\to A\to A/(a)$ taking each $X_c$ to 0 (for $c\not=a$) and $X_a$ to 1. As its kernel contains $J$, it defines a morphism $g\colon B\to A/(a)$, which takes $b$ to one. Therefore, the ideal $B(1-b)$ maps to 0 and, as $K(A/(a))$ is proper, (1) says that $K(B/(1-b))$ is proper.

(4) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ such that, for every $a\in Bx+By$ there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ is proper.

Set $A_0=A$ and use (3) to construct a sequence of extensions $f_i\colon A_i\to A_{i+1}$ with left inverses such that, for every $a\in A_ix+A_iy$ there is a $b\in A_{i+1}$ with $ab=0$ and $K(A_{i+1}/(1-b))$ is proper. Note that, as each $f_i$ has a left inverse, (1) says that $A_{i+1}$ satisfies (P) whenever $A_i$ does. So, we can keep applying (3) to build up the entire sequence of extensions.

We now take the colimit $B={\rm colim}A_i$ and let $f$ be the induced morphism (i.e, if we consider $A_i\subseteq A_{i+1}$ then $B$ is the union and $f$ is inclusion). As each $A_i\to B$ has a left-inverse, (1) shows that the required properties for $B$ are inherited from the individual $A_i$.

(5) Suppose that $A$ satisfies the following: for every $a\in Ax+Ay$ there is a $b\in A$ with $ab=0$ and $K(A/(1-b))$ is proper. Then, the R-algebra $B=A/K(A)$ satisfies the same property, and also $K(B)=0$.

That $K(B)=0$ follows quickly from the definition of $K$. Suppose $a\in Ax+Ay,b\in A$ are such that $ab=0$ and $K(A/(1-b))$ is proper. Set $C=A/(1-b)$, so that $C/K(C)$ is not trivial. By (1), the canonical morphism $A\to C$ maps $K(A)$ into $K(C)$. So, it induces a morphism $B\to C/K(C)$. This takes $1-b$ to zero, so it induces an R-morphism $B/(1-b)\to C/K(C)$. As $K(C/K(C))=0$, (1) implies that $K(B/(1-b))$ maps to zero, so is proper.

(6) If $B$ is the R-algebra constructed in (5), then $I=Bx+By$ contains only zero-divisors but $x,y\in I$ map to regular elements in $R_z$ and $R_{1-z}$ respectively.

For any $a\in I$ there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ proper. In particular, $(1-b)$ must be a proper ideal, so that $b\not=0$ and $a$ is a zero divisor.

Finally, the property $K(B)=0$ implies that $x$ is regular in $B_z$ and $y$ is regular in $B_{1-z}$.

  • $\begingroup$ I don't have time to further work on this answer right now, so I'll leave the following note. As the z term is an idempotent, it can be removed from most of the construction and only added in at the last step. This should make it a bit simpler, and property (P) can be simplified. $\endgroup$ Commented Dec 13, 2010 at 4:43
  • $\begingroup$ There's still an error here somewhere, because it can't work if z is chosen to be an idempotent. Oh well... $\endgroup$ Commented Dec 13, 2010 at 5:55
  • $\begingroup$ The mistake is in (1). Forcing $z^2=z$ means that property (P) cannot be satisfied $\endgroup$ Commented Dec 13, 2010 at 6:04
  • $\begingroup$ Ok, I see how this can be fixed by dropping the condition $z^2=z$ and using a different property (P) - replacing it by a more intuitive (but still slightly messy) condition. I'll update this, tonight if I get time. $\endgroup$ Commented Dec 13, 2010 at 13:10
  • $\begingroup$ Updated. I think it should all work now. $\endgroup$ Commented Dec 14, 2010 at 1:05

Denote by $Q(R)$ the total ring of fractions of $R$, i.e. the localization of $R$ at the regular elements of $R$. Then $m$ torsion depends only on $M := \langle m \rangle$, namely $M \otimes Q(R) = 0$. This can be tested locally on $Spec(R)$ if the canonical maps $Q(R)_{f_i} \to Q(R_{f_i})$ are isomorphisms. So everything is fine if $R$ is an integral domain. This holds also in other examples, but not in general. See this discussion about the sheaf of meromorphic functions.


This is not an answer, but a (long) comment concerning your statement about torsion sections of sheaves of modules.

If $M$ is a $R$-module and $m \in M$, then $m$ is called torsion if "$\exists r \in R : r \text{ is regular and } rm=0$". Now using the logic of the topos $Sh(X)$, this definition also applies to $R$-modules $M$, where $(X,R)$ is a ringed space. This will be automatically a local notion, and I also claim it is the right notion of torsion (cf. also Mulvey, "Intuitionistic Algebra and Representations of rings" for more examples).

Then we get the following definition: A global section $m \in \Gamma(X,M)$ is torsion iff there is a covering $X = \cup_i U_i$ and sections $r_i \in \Gamma(U_i,R)$, such that $r_i$ is a regular section (i.e. $r_i : R|_{U_i} \to R|_{U_i}$ is injective, or equivalently, every germ of $r_i$ is a regular element in the stalk) and $r_i m|_{U_i}=0$ for every $i$.

  • $\begingroup$ Right. My question is asking whether this agrees with the usual definition. $\endgroup$ Commented Aug 26, 2010 at 18:17

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