Direct Products of Modules .... Bland Proposition 2.1.1 .... ....

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help with some aspects of the proof of Proposition 2.1.1 ...

Proposition 2.1.1 and its proof read as follows:
View attachment 8030

In the statement of the above proposition we read the following:

" ... ... for every \(\displaystyle R\)-module \(\displaystyle N\) and every family \(\displaystyle \{ f_\alpha \ : \ N \rightarrow M_\alpha \}_\Delta\) of \(\displaystyle R\)-linear mappings there is a unique \(\displaystyle R\)-linear mapping \(\displaystyle f \ : \ N \rightarrow \prod_\Delta M_\alpha\) ... ... "The proposition declares the family of mappings \(\displaystyle \{ f_\alpha \ : \ N \rightarrow M_\alpha \}_\Delta\) as \(\displaystyle R\)-linear mappings and also declares that \(\displaystyle f\) (see below for definition of \(\displaystyle f\)!) is an \(\displaystyle R\)-linear mapping ...

... BUT ...

I cannot see where in the proof the fact that they are \(\displaystyle R\)-linear mappings is used ...

Can someone please explain where in the proof the fact that the family of mappings \(\displaystyle \{ f_\alpha \ : \ N \rightarrow M_\alpha \}_\Delta\) and \(\displaystyle f\) are \(\displaystyle R\)-linear mappings is used ... basically ... why do these mappings have to be \(\displaystyle R\)-linear ... ?
Help will be much appreciated ...

Peter======================================================================================The above post mentions but does not define \(\displaystyle f\) ... Bland's definition of \(\displaystyle f\) is as follows:
View attachment 8031Hope that helps ...

Peter***EDIT***

In respect of \(\displaystyle f\) it seems we have to prove \(\displaystyle f\) is an \(\displaystyle R\)-linear mapping ... but then ... where is this done ...

 

[steenis]

Bland defines the map $f:N \longrightarrow \prod_\Delta M_\alpha$ by saying that $f$ is product of the family of mappings $f_\alpha:N \longrightarrow M_\alpha$. YOU have to fill in the proof that $f$ is an R-linear map, by using the fact that $f_\alpha$ is a R-linear maps for all $\alpha \in \Delta$.

The mappings $f_\alpha:N \longrightarrow M_\alpha$ have to be R-maps, because $f$ has to become an R-map and because $\pi_\alpha \circ f = f_\alpha$ for all α∈Δ.

 

[Math Amateur]

Bland defines the map $f:N \longrightarrow \prod_\Delta M_\alpha$ by saying that $f$ is product of the family of mappings $f_\alpha:N \longrightarrow M_\alpha$. YOU have to fill in the proof that $f$ is an R-linear map, by using the fact that $f_\alpha$ is a R-linear maps for all $\alpha \in \Delta$.

The mappings $f_\alpha:N \longrightarrow M_\alpha$ have to be R-maps, because $f$ has to become an R-map and because $\pi_\alpha \circ f = f_\alpha$ for all α∈Δ.

Hi Steenis ... thanks for your help ...

You write:

" ... ... YOU have to fill in the proof that $f$ is an R-linear map, by using the fact that $f_\alpha$ is a R-linear maps for all $\alpha \in \Delta$. ... ... "

Yes ... of course you're right ...

BUT ... need some help ...We need to show \(\displaystyle f\) is an \(\displaystyle R\)-linear map (homomorphism) ...

We are given that the \(\displaystyle f_\alpha\) are \(\displaystyle R\)-linear maps, and we know that the projections \(\displaystyle \pi_\alpha\) are \(\displaystyle R\)-linear maps ...

We also know that \(\displaystyle \pi_\alpha f = f_\alpha\) for each \(\displaystyle \alpha \in \Delta\) ... ... ... ... ... (1)

Now ... we know that if \(\displaystyle f\) is an \(\displaystyle R\)-linear mapping then (1) holds true but ...

... how do you prove that f must necessarily be an \(\displaystyle R\)-linear map ... ...... can you help ... ... ?

Peter

 

[steenis]

To show that $f:N \longrightarrow \prod_\Delta M_\alpha$ is an R-linear map, use the definition of R-linear map. So you have to show that:

$f(x+y) = f(x) + f(y)$ for $x, y \in N$ and
$f(xr) = f(x)r$ for $x \in N$ and $r \in R$ (in case of a right R-module).

$f$ is defined as the product of the family of mappings $f_\alpha:N \longrightarrow M_\alpha$. So $f(x)=(f_\alpha(x))_{\alpha \in \Delta}$ and $f(y)=(f_\alpha(y))_{\alpha \in \Delta}$ for $x, y \in N$ by definition.

Thus $f(x+y) = (f_\alpha(x+y))=(f_\alpha(x)+f_\alpha(y))=(f_\alpha(x))+(f_\alpha(y))=f(x)+f(y)$.

You can do the second part.

 

[steenis]

I am sorry, Peter, did NOT ask me to prove that $f$ is R-linear, you DID ask me why $f$ is necessarily R-linear, given that $\pi_\alpha \circ f = f_\alpha$ in which $\pi_\alpha$ and $f_\alpha$ are R-linear.

I think this is why, take $x, y \in N$ then $\pi_\alpha f(x+y) = \pi_\alpha (f_\alpha(x+y)) = f_\alpha(x+y) = f_\alpha(x) + f_\alpha(y)$.

From this it follows $(f_\alpha(x+y)) = (f_\alpha(x)) + (f_\alpha(y))$ and $f(x+y)=f(x)+f(y)$. (you can do the second rule).