Torsion Elements and the Torsion Submodule, Tor(M) .... Dummit and Foote, Section 10.1 ....

[Math Amateur]

I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ...

I am currently studying Chapter 10: Introduction to Module Theory ... ...

I need some help with an Exercise 8(b) of Section 10.1 ...

Exercise 8 of Section 10.1 reads as follows:

https://www.physicsforums.com/attachments/8312Can someone please show me an example as requested in Part (b) of the above exercise ... and demonstrate that the given example is not a submodule ...

Peter

 

[GJA]

Hi Peter,

Try thinking about $\mathbb{Z}_{6}$ and see what you can come up with.

 

[Math Amateur]

Hi Peter,

Try thinking about $\mathbb{Z}_{6}$ and see what you can come up with.

Thanks GJA ...

OK ... then consider the ring \(\displaystyle R = \mathbb{Z}_{6} \equiv \mathbb{Z} / 6 \mathbb{Z} = \{ \overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5} \}
\) ...

... and consider \(\displaystyle M = R\) as a left module over itself ...Now \(\displaystyle \overline{2}\) is a nonzero element of \(\displaystyle R\) ...

So, now \(\displaystyle \overline{3} \in M\) is torsion since there exists a nonzero element, namely \(\displaystyle \overline{2} \in R \) such that \(\displaystyle \overline{2} \cdot \overline{3} = \overline{0}
\)
That is \(\displaystyle \overline{3} \in \text{Tor} (M)\) ...Similarly we have \(\displaystyle \overline{3}\) is a nonzero element of \(\displaystyle R\) ...

So, now \(\displaystyle \overline{2} \in M\) is torsion since there exists a nonzero element, namely \(\displaystyle \overline{3} \in R\) such that \(\displaystyle \overline{3} \cdot \overline{2} = \overline{0}\)

That is \(\displaystyle \overline{2} \in \text{Tor} (M)\) ...
But ... \(\displaystyle \overline{2} + \overline{3} = \overline{5} \notin \text{Tor} (M)\) ... ...

Therefore, \(\displaystyle \text{Tor} (M)\) does not satisfy additive closure ...

So ... \(\displaystyle \text{Tor} (M)\) is not a submodule of M ...Is the above correct? ... ...Thanks again for the suggestion ...

Peter

 

[GJA]

Hi Peter,

Nicely done! Quick note on where the idea for considering $\mathbb{Z}_{6}$ came from: Trying to use a field is out because each non-zero is invertible (so things like $\mathbb{R}, \mathbb{Q},$ and $\mathbb{Z}_{p}$ are out). Good places to look for counterexamples when multiplication is involved with fields off the table are then $\mathbb{Z}_{n}$ ($n$ non-prime) and matrices.

 

[Math Amateur]

Hi Peter,

Nicely done! Quick note on where the idea for considering $\mathbb{Z}_{6}$ came from: Trying to use a field is out because each non-zero is invertible (so things like $\mathbb{R}, \mathbb{Q},$ and $\mathbb{Z}_{p}$ are out). Good places to look for counterexamples when multiplication is involved with fields off the table are then $\mathbb{Z}_{n}$ ($n$ non-prime) and matrices.

Thanks GJA ...

Appreciate your help ...

Peter