(R/I)-Modules .... Dummit and Foote Example (5), 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 aspect of Example (5) of Section 10.1 Basic Definitions and Examples ... ... Example (5) reads as follows:

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I do not fully understand this example and hence need someone to demonstrate (explicitly and completely) why it is necessary for \(\displaystyle am = 0\) for all \(\displaystyle a \in I\) and all \(\displaystyle m \in M\) for us to be able to make \(\displaystyle M\) into an \(\displaystyle (R/I)\)-module. ...
Help will be much appreciated ..

Peter

 

[steenis]

You can prove that $M$ is a (left) $R/I$ module by proving that the map $R/I \times M \longrightarrow M$ satisfies (1) and (2) of the definition on page 337, applied to $R/I$.
But, the most important thing you have to do is to prove that this map is well-defined, i.e., that the multiplication $(r+I)m=rm$ is well-defined.

 

[Math Amateur]

You can prove that $M$ is a (left) $R/I$ module by proving that the map $R/I \times M \longrightarrow M$ satisfies (1) and (2) of the definition on page 337, applied to $R/I$.
But, the most important thing you have to do is to prove that this map is well-defined, i.e., that the multiplication $(r+I)m=rm$ is well-defined.

Thanks steenis ... appreciate your help ...

You write:

" ... ... But, the most important thing you have to do is to prove that this map is well-defined, i.e., that the multiplication $(r+I)m=rm$ is well-defined. ... ... "But how exactly do we prove that the map is well-defined ... ...?

Do we have to show that if \(\displaystyle (r_1 + I)m_1 = (r_2 + I)m_2\) then \(\displaystyle r_1 m_1 = r_2 m_2\) ...

... ... is that right ...Can you help ...

Peter***EDIT***

Another question that you may be able to help with ... I am puzzled about the definition and nature of the "multiplications involved in the equation

\(\displaystyle (r+I)m = rm\) ... ... ... ... ... (1)It seems to me that there are two multiplications involved and I am not sure how they are defined ... indeed suppose the two multiplications are \(\displaystyle \star\) and \(\displaystyle \circ\) ... ... then, (1) becomes \(\displaystyle (r+I) \star m = r \circ m\) But how are \(\displaystyle \star\) and \(\displaystyle \circ\) defined ... where do they come from ... what is their nature ,,,Can you help ... ...

 

[steenis]

Almost right.
You have to prove that if $r_1+I=r_2+I$ then $r_1m=r_2m$.
(Strictly, you have to prove that if $r_1+I=r_2+I$ and $m_1=m_2$ then $r_1m_1=r_2m_2$. But you can see that this is the same.)

 

[Math Amateur]

Almost right.
You have to prove that if $r_1+I=r_2+I$ then $r_1m=r_2m$.
(Strictly, you have to prove that if $r_1+I=r_2+I$ and $m_1=m_2$ then $r_1m_1=r_2m_2$. But you can see that this is the same.)

Thanks Steenis ...

Another question that you may be able to help with ... I am puzzled about the definition and nature of the "multiplications involved in the equation

\(\displaystyle (r+I)m = rm\) ... ... ... ... ... (1)It seems to me that there are two multiplications involved and I am not sure how they are defined ... indeed suppose the two multiplications are \(\displaystyle \star\) and \(\displaystyle \circ\) ... ... then, (1) becomes \(\displaystyle (r+I) \star m = r \circ m\) But how are \(\displaystyle \star\) and \(\displaystyle \circ\) defined ... where do they come from ... what is their nature ,,,Can you help ... ...

Peter

 

[steenis]

Given is that $M$ is a (left) R-module, so it has a multiplication $\phi :R \times M \longrightarrow M$ denoted by $\phi (r,m) = rm$. This map defines the action of the ring $R$ on $M$. $rm$ is the notation of the mutiplication in $M$ as $R$-module. For $r \in R$ and $m \in M$ we know the values of $\phi (r,m)$ in $M$.

Now you want to change $M$ into a (left) $R/I$-module, so you have to define the action of the ring $R/I$ on $M$, by defining a new multiplication $\psi : R/I \times M \longrightarrow M$. You can notate the multiplication by $\psi (r+I,m)=(r+I) \star m$, for $r+I \in R/I$ and $m \in M$. But this is a notation and not a definition of the multiplication in $M$ as $R/I$-module. You still have to define $(r+I) \star m$ and you can do so by defining the value $(r+I) \star m$ in the $R/I$-module $M$ equal to the value $rm$ in the $R$-module $M$. (more difficult $\psi (r+I,m)= \phi (r,m)$).
In the above you have proven that the multiplication $\psi$ is well-defined and has valid vlaues in $M$.

(I wonder why some characters are printed in red.)

 

[Opalg]

(I wonder why some characters are printed in red.)

The backslash in $R\I$ is interpreted by TeX as introducing a command name. Since no command with the name \I has been defined, it gets printed in red as a warning that you have used an undefined command. You presumably meant to use a forward slash, which TeX would have happily printed as $R/I$.

 

[steenis]

Of course, thank you. I have changed it in my post above.

 

[Math Amateur]

Given is that $M$ is a (left) R-module, so it has a multiplication $\phi :R \times M \longrightarrow M$ denoted by $\phi (r,m) = rm$. This map defines the action of the ring $R$ on $M$. $rm$ is the notation of the mutiplication in $M$ as $R$-module. For $r \in R$ and $m \in M$ we know the values of $\phi (r,m)$ in $M$.

Now you want to change $M$ into a (left) $R/I$-module, so you have to define the action of the ring $R/I$ on $M$, by defining a new multiplication $\psi : R/I \times M \longrightarrow M$. You can notate the multiplication by $\psi (r+I,m)=(r+I) \star m$, for $r+I \in R/I$ and $m \in M$. But this is a notation and not a definition of the multiplication in $M$ as $R/I$-module. You still have to define $(r+I) \star m$ and you can do so by defining the value $(r+I) \star m$ in the $R/I$-module $M$ equal to the value $rm$ in the $R$-module $M$. (more difficult $\psi (r+I,m)= \phi (r,m)$).
In the above you have proven that the multiplication $\psi$ is well-defined and has valid vlaues in $M$.

(I wonder why some characters are printed in red.)

Hi steenis,

... your post was most helpful...

I very much appreciate all your assistance with all the above issues ...

Peter