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Can any one suggest me an example of a ring $R$ and an $R$-module $A$ s.t. torsion of module $A$ is not a sub-module?

Torsion of module $A$, i.e. $\operatorname{Tor}(A)$, denotes all torsion elements in module $A$.

*(And we already know that if $R$ is a commutative integral domain then $\operatorname{Tor}(A)$ is a submodule of $A$).

Mat He Mat Cian
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An example for a commutative ring which is not a domain: $R=\{0,a,1-a,1\}$ with $a^2=a,a+a=1+1=0$ and $A=R$. The elements $a,1-a$ are torsion, but $a+(1-a)=1$ isn't.

tomasz
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    This is a right example. Well done. – Sara Tancredi Nov 21 '13 at 05:31
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    From $1+1=0$ we get $-1=1$, so $1-a=1+a$. But more important is that $R=\Bbb Z_2[x]/(x^2+x)$. –  Nov 21 '13 at 07:47
  • Can you explain me please (a) How Tor(A) is not a submodule. (b) What is that 1-a element in R? I didn't see such kind of ring before. (c) What is that A= R.a ?? and why 1-a are torsion???? Nothing is clear.. Please someone explain it to me. – Mat He Mat Cian Nov 21 '13 at 08:19
  • @MatHeMatCian: $a,1-a\in \operatorname{Tor}(A)$, but $1=a+(1-a)$ isn't. What's comlpicated about this? $1-a$ is just one of the elements of the ring. I've defined its structure completely, which part do you not understand? $A=R$ is simply $R$ treated as a module over itself. $a(1-a)=a-a^2=a-a=0$, so $a,1-a$ are both torsion. – tomasz Nov 22 '13 at 02:42
  • @MatHeMatCian: I now see that you might have been confused by the fact that I put two math expressions so close by and separated only by a dot. I fixed it now, hope its clearer now. – tomasz Nov 22 '13 at 02:43
  • @user: How is that relevant? – tomasz Nov 22 '13 at 02:44
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    The way you presented $R$ looks rather like a trick than as an algebraic object obtained by algebraic methods. (Anyway, the comment was addressed to the readers of your answer.) –  Nov 22 '13 at 06:01
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    @user: The "natural" example that first came to mind was that of orthogonal projections in a Hilbert space, and this is just the abstraction. If you insist on having a neat algebraic form, I think saying that it's $({\bf Z}/2{\bf Z})^2$ is neater. – tomasz Nov 23 '13 at 11:45
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Let me propose a more elementary example.

Let $R$ be $(\mathbb Z_6,+,\times)$. Let $A=R$.

Then $A$ is a $R$-module.

Let Tor$(A):=\{a\in A|ra=0 \mbox{ for some nonzero }r\in R\}$.

Note that $\bar{1}$ and $\bar{5}$ are the inverses of each other. So they are not in Tor$(A)$.

Note that $\bar{2}\,\bar{3}=\bar{3}\,\bar{2}=\bar{0}$ and $\bar{3}\,\bar{4}=\bar{0}$. So, $\bar{2}, \bar{3},\bar{4}\in$ Tor$(A)$. Also, $\bar{0}\in$ Tor$(A)$ is trivial.

Now we easily see that $\bar{2}+\bar{3}=\bar{5}\notin$ Tor$(A)$. Hence, Tor$(A)$ is not a $R$-submodule of $A$.

Sam Wong
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actually, the Torsion subset is a submodule for all $R$-modules $M$. Recall that $m\in M$ is called torsion, if there is $r\in R$ which is regular (i.e. not a zero divisor) such that $r.m=0$.

Assume $m$ and $m'$ are torsion with corresponding regular elements $r$ and $r'$. Then $rr'$ is non-zero and still not a zero divisor and we have $rr'.(m+m')=0$. Hence $m+m'$ is torsion. It is trivial to see that if $m$ is torsion then the same is true for $s.m$ for all $s\in R$.

Note also that over a finite ring every module is torsion. To see this, notice that an element of a finite ring is either a unit or a zero-divisor. Hence all regular elements act invertibly on modules and hence cannot kill any non-zero elements.

Aufenthaltsraum
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  • I dispute your claim that over a finite ring every module is torsion. Let $M = R = \Bbb{Z}/(6)$. Then $\text{Tor}(M) = {0, 2, 3, 4}$ yet $2 + 3 = 5 \notin \text{Tor}(M)$. – Daniel Donnelly Mar 11 '16 at 02:48
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    @EnjoysMath The definition ufenthaltsraum uses insists $r\in R$ to be regular and $r.m=0$ and this new definition makes Tor a submodule. – Babai Sep 20 '16 at 11:14
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    This is form wikipedia-"An element m of a module M over a ring R is called a torsion element of the module if there exists a regular element r of the ring (an element that is neither a left nor a right zero divisor) that annihilates m, i.e., r m = 0. In an integral domain (a commutative ring without zero divisors), every non-zero element is regular, so a torsion element of a module over an integral domain is one annihilated by a non-zero element of the integral domain. Some authors use this as the definition of a torsion element but this definition does not work well over more general rings." – Babai Sep 20 '16 at 11:15
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    Only if your ring is commutative! Else this statement is wrong. E.g. http://math.stackexchange.com/questions/1939173/torsion-elements-do-not-form-a-submodule for a proper example! – ctst Jan 17 '17 at 21:05
  • The example above disproves this. The mistake was in your definition of $T(M)$. – DanLewis3264 Mar 18 '20 at 20:39
  • @Dan The example above does not disprove this, it just uses a different definition. It is unclear from the question what definition the OP was using in the first place. (Not that it matters much, but in my experience this definition is more common.) – Anakhand Feb 10 '24 at 19:26