What Is a Torsion Element in Ring Theory?

[Stephen Tashi]

TL;DR Summary

Should the definition of a torsion element be stated in terms of non-zero-divisors? - or should it refer to non-zero elements?

The current Wikipedia article on Torsion element (https://en.wikipedia.org/wiki/Talk:Torsion_(algebra) ) says:

In mathematics, specifically in ring theory, a torsion element is an element of a module that yields zero when multiplied by some non-zero-divisor of the ring.

A ring R can be used to define a module M of the ring over itself. Multiplication of a module element m by a ring element r is the same as multiplication in the ring. If m is not zero and m*r = 0 this makes r a zero divisor - correct? So, by the definition above, M could not have any torsion elements except m=0 (?). Is that a correct line of reasoning?

 

[fresh_42]

TL;DR Summary: Should the definition of a torsion element be stated in terms of non-zero-divisors? - or should it refer to non-zero elements?

The current Wikipedia article on Torsion element (https://en.wikipedia.org/wiki/Talk:Torsion_(algebra) ) says:
A ring R can be used to define a module M of the ring over itself. Multiplication of a module element m by a ring element r is the same as multiplication in the ring. If m is not zero and m*r = 0 this makes r a zero divisor - correct? So, by the definition above, M could not have any torsion elements except m=0 (?). Is that a correct line of reasoning?

It is at least a strange line of reasoning. Where did you get this definition from? Why do you still use M although you set M=R? That is normally a sure sign of a mistake.

I have never heard of "a torsion element". Torsion is used in various places with various meanings.

 

[Stephen Tashi]

Where did you get this definition from?

The definition of "torsion element" came from the Wikipedia article. Were you asking about a definition of something else?

I found an inconclusive discussion of the definition on stackexchange: https://math.stackexchange.com/questions/1174903/what-is-the-standard-definition-of-torsion-element

Why do you still use M although you set M=R?

To make the distinction between a ring versus the ring considered as a module over itself.

 

[fresh_42]

Here is a real source, not a "talk on Wikipedia":
https://ncatlab.org/nlab/search?query=torsion+element
19 hits. Make your choice.

 

[Stephen Tashi]

From https://ncatlab.org/nlab/show/torsion+module, I pick the definition that uses nonzero instead of non-zero-divisor:

Given a ring R, an element m in an R-module M is torsion element if there is a nonzero element r in R such that rm=0.

(It's interesting that ChatGPT gives an equivalent definition.)

 

[fresh_42]

Torsion is when something is turning back to the starting point. A sum that ends up in zero, a product that becomes one.

Your version is:

The torsion (submodule) of a module $T\subseteq M$ is $T=\{m\in M\,|\,\exists r\in R\, : \,rm=0\wedge r \text{ is no zero divisor}\}.$ The elements of $T$ are called torsion elements.

The crucial point is the existence quantifier. Say $t\in T$ is a torsion element. Then the kernel $r\longmapsto r\cdot t$ may not consist of only zero-divisors. There must exist an element $r\in R$ such that $r\cdot t=0$ that is no zero-divisor.

So what is the torsion submodule of $R$? We get $T=\{r'\in R\,|\,\exists r\in R\, : \,r\cdot r'=0\wedge r \text{ is no zero divisor }\}.$ But these two conditions contradict themselves. As soon as $r\cdot r'=0$ both elements $r$ and $r'$ are zero-divisors. Hence, a ring is torsion-free. That does not mean that a ring has no zero-divisors. They simply do not count as torsion elements in this restrictive definition.

The fact that your element is part of the ring is crucial here.

 

[mathwonk]

I always found the use of this word confusing, but as I understand it, in an abelian group a torsion element is simply an element of finite order. More generally, an element x of an R module M, is an R torsion element if it is annihilated by some non zero element of R. This notion is best behaved when R itself has no non trivial zero divisors, i.e. when R is a domain, then the torsion elements of M form a submodule Tor(M). Note that in the ring Z/6 considered as a module over itself, 2 and 3 are torsion elements, but their sum is a unit.

These torsion subgroups are to some extent measured by the derived functors Tor_n, associated to the failure of left exactness of the tensor product, hence the name. These Tor functors occur in topology when computing how homology groups change when the coefficients do.

When the ring is simply the integers, the torsion subgroup of M is the kernel of the natural map from M to MtensorQ, and is isomorphic to the group Tor_1(M,Q/Z). So here is an example where the Tor_1 functor exactly picks out the torsion submodule.

Chapter 17 of Abstract Algebra by Dummit and Foote has a nice short treatment of the topic.

 

[Stephen Tashi]

Your version is:

The torsion (submodule) of a module $T\subseteq M$ is $T=\{m\in M\,|\,\exists r\in R\, : \,rm=0\wedge r \text{ is no zero divisor}\}.$ The elements of $T$ are called torsion elements.

Acutally, I picked the version that only requires that r is nonzero. I understand what you said about the version that requires r not be a zero divisor.

 

[fresh_42]

Acutally, I picked the version that only requires that r is nonzero. I understand what you said about the version that requires r not be a zero divisor.

The idea is that $r\cdot m=\underbrace{m+\ldots+m}_{r\text{ times}}=0,$ i.e. that $m$ is of finite (additive) order $r$ without using an element $r$ that kills other elements anyway (zero divisors).

$r$ cannot be a unit, that would cause $m=0$, and $r$ shouldn't be a zero divisor since then $0=0\cdot m=(r\cdot r')\cdot m= r\cdot (r'\cdot m)$ which says nothing about $m.$ So torsion elements are something between a unit and a zero (divisor).

 

[mathwonk]

I never before heard of the definition that requires the annihilating element of R to be regular. They do that apparently to force the R- torsion elements of M to form an R submodule.

 

[WWGD]

Well, using the definition of torsion elements as ring elements x so that nx=0 for some Natural n( which can be generalized , I believe, to elements r with rx=0 ), I've tried to understand the concept topologically, as with homology classes, as in e.g., the homology of Real Projective Space, which is Z/2. Since homology here is a module, we can somehow meaningfully add classes, in this case we can add [1]+[1]=[2]=[0] , getting torsion. Maybe @mathwonk can elaborplease? can you, olease?

 

[mathwonk]

not quite sure what is being asked for elaboration. the example of topological homology is an example of a module over the ring Z, which is a domain, hence the torsion elements in a homology module (abelian group) do form a submodule.

when you try to prove the torsion elements form a submodule, you have to prove that the sum of two torsion elements is also torsion. If x is annihilated by a, and y is annihilated by b, then (x+y) is annihilated by ab. But in case the ring is not a domain, it could happen that a≠0 and b ≠0 yet ab = 0, so the fact that x+y is annihilated by ab, would not make x+y a torsion element, assuming a torsion element is anything annihilated by a non zero ring element. Redefining torsion elements to be elements that are annihilated by "regular" ring elements would require that a and b are not only non zero, but also are not zero divisors, and thus remove the possibility that ab=0.

so the torsion subset of a module is a submodule, provided we define a torsion element to be one that is annihilated by a regular element. In case the ring is a domain, all non zero elements are regular and it is equivalent to requiring them to be annihilated by a non zero element. In case the ring is not a domain we get a torsion submodule if we define torsion elements to be those that are annihilated by regular elements, but if we use the definition familiar to me, that they are elements annihilated by non - zero elements, then we do not get a submodule.

 

[WWGD]

Just curious , see if you head examples of ab=0 with neither being 0. Maybe Real Projective Space? Also, would this be cup product, cap product?

 

[mathwonk]

check out these great answers on mathoverflow, for useful examples of torsion in topology and geometry (elements of finite order in homology or homotopy or cohomology): (starting with your example of fundamental group of P^2)
https://mathoverflow.net/questions/22583/why-torsion-is-important-in-cohomologyohh yes, cup product raises dimension and homology vanishes in high dimensions so I guess any cohomology element for a finite dimensional manifold has a power which is zero. so I guess in the cohomology ring zero divisors are common.