Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $M$.

share|improve this question
1  
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 $R$? –  Martin Brandenburg Oct 20 '10 at 21:46
add comment

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 $B_{1-z}$.

share|improve this answer
    
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. –  George Lowther Dec 13 '10 at 4:43
    
There's still an error here somewhere, because it can't work if z is chosen to be an idempotent. Oh well... –  George Lowther Dec 13 '10 at 5:55
    
The mistake is in (1). Forcing $z^2=z$ means that property (P) cannot be satisfied –  George Lowther Dec 13 '10 at 6:04
    
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. –  George Lowther Dec 13 '10 at 13:10
    
Updated. I think it should all work now. –  George Lowther Dec 14 '10 at 1:05
add comment

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.

share|improve this answer
add comment

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 $i$.

share|improve this answer
    
Right. My question is asking whether this agrees with the usual definition. –  Andrew Critch Aug 26 '10 at 18:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.