MHB How Can One Show That a Given Example is Not a Submodule?

[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 $$R = \mathbb{Z}_{6} \equiv \mathbb{Z} / 6 \mathbb{Z} = \{ \overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5} \}
$$ ...

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

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

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

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

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

So ... $$\text{Tor} (M)$$ 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