Extension of A by B given a short exact sequence

Question

The following is taken from: Module Theory An approach to linear algebra Electronic edition by T.S.Blyth

Backgroun:

Exercise 3.11

A short exact sequence of the form $$(f,E,g)\equiv 0\to A\xrightarrow{f}E\xrightarrow{g}B\to 0$$ is called an extension of $A$ by $B$.
Given any $R$-modules $A$ and $B$, show that there exists at least one extension of $A$ by $B$.

Questions:

Is the author asking the reader to show what the module $E$ is supposed to be in order to make the above sequence of mappings into a short exact sequence? If so, it seems that the module $E$ would have to be $E=A\times B$ where the mapping $f:A\to A\times B$ is defined as $a\mapsto (a,0)$ and $g:A\times B\to B$ is defined as $(a,b)\mapsto b.$

Is my understanding of the question correct?

Thank you in advance.

Answer 1

Yes, the direct sum $E=A\oplus B$ is always an $R$-module extension of $A$ by $B$, with the maps $f$ and $g$ you have given. To complete the proof you need to show that the resulting sequence is exact at each point and that all maps are $R$-module homomorphisms. This is not difficult. In fact it is more or less obvious.