MHB A problem about submodules in "Advanced Modern Algebra" of Rotman

[steenis]

Please, can someone help me with this?

Let $M$ be a left $R$-module over a ring $R$.
Let $J$ be a left ideal in $R$ generated by $r$: $J=Rr=<r>$.
Now $JM=\{am \ | \ a \in J \ and \ m \in M\}$
Prove that $JM$ is a submodule of $M$.

This is an example in Rotman's book "Advanced Modern Algebra" 2nd edition 2010, page 404.

 

[I like Serena]

Please, can someone help me with this?

Let $M$ be a left $R$-module over a ring $R$.
Let $J$ be a left ideal in $R$ generated by $r$: $J=Rr=<r>$.
Now $JM=\{am \ | \ a \in J \ and \ m \in M\}$
Prove that $JM$ is a submodule of $M$.

This is an example in Rotman's book "Advanced Modern Algebra" 2nd edition 2010, page 404.

Hi steenis,

From wiki:

Suppose M is a left R-module and N is a subgroup of M. Then N is a submodule (or R-submodule, to be more explicit) if, for any n in N and any r in R, the product r ⋅ n is in N.
​

So don't we need to verify that $JM$ is a subgroup of $M$, and that for any $a_1m_1$ in $JM$ and any $\tilde r$ in $R$, the product $\tilde r a_1 m_1$ is in $JM$, knowing that $a_1 = r_1r$?

 

[steenis]

Yes, I thought so too. But how do I prove that if $x,y \in JM$ then $x+y \in JM$ ?.
That is $x=r_1rm$ and $y=r_2rn$, then $r_1rm+r_2rn$ has to be in $JM$.
I feel stupid but I do not see it.

 

[steenis]

It is true if $r \in Z(R)$, the center of $R$. Howerver Rotman explicitly stated "If $r$ is an element of $R$ not in the center of $R$ $...$".

 

[Euge]

The set $JM$ should consist of finite sums of the form $am$ where $a\in J$ and $m\in M$, otherwise closure under addition fails. For example, let $R$ be the ring of $2\times 2$ real matrices, and let $M = R$, viewed as a left $R$-module. If $r = \begin{bmatrix}1&0\\0&0\end{bmatrix}$, then $J = \left\{\begin{bmatrix}a&0\\c&0\end{bmatrix}:a,c\in \Bbb R\right\}$, and so $JM = \left\{\begin{bmatrix}ax&ay\\cx&cy\end{bmatrix}: a,c,x,y\in \Bbb R\right\}$. In particular, elements of $JM$ have zero determinant. This set is not a submodule of $M$, for both $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\0&1\end{bmatrix}$ are in $JM$, but the sum, $\begin{bmatrix}1&0\\0&1\end{bmatrix}$, is not a member of $JM$ since it has nonzero determinant.

I actually tried sending an email to him about Example 6.18(v) but received an email (from his daughter, I believe) stating that he passed away 18 months ago. :(

 

[steenis]

That is sad news, I did not know that. I saw on the internet that he passed away on 16 october 2016, that is 83 years old.

Thank you for looking at my problem, but I do not understand your example, because if $R$ is the ring of 2x2 real matrices and $r$ is the identity, then $J=R$. Can you explain your example once more?

 

[Euge]

That was a typo; the second row was meant to be a row of zeros. I'm fixing some of the other typos.

 

[Euge]

The typos are now fixed.

 

[steenis]

Now I understand, thank you.

Thank you for trying to contact Rotman.

Strangely, this example still appears in the next edition of his book (B-1.26(vi)).

 

[Euge]

Funny you mentioned that -- looking back at the section, I see that he used nearly the same 'example' as my counterexample. (Giggle)