1
$\begingroup$

Let $S$ be an arbitrary basis of a module $X$ over $R$. Show that any function $f:S\to Y$ of $S$ into any module $Y$ over $R$ extends to a uniquely determined homomorphism $h:X\to Y$ of the module $X$ into the module $Y$

By what I know about basis and modules, I know that any element of $X$ can be written as a linear combination of elements of $S$. I know that $Y$ has a basis, so any element there can be written as a linear combination of this basis. Since the basis representation is unique, I should found a bijection $f$ from basis elements $S$ of $X$ to the basis elements of $Y$, right? Then, I should just extend like this:

$$\overline{f} = h:X\to Y; h\left(\sum_{i\in M}\alpha_i s_i \right) = \sum_{i\in M}\alpha_i f(s_i)$$

$\endgroup$
1
$\begingroup$

$\quad$If a (unitary) $R$-module $X$ has a (nonempty) basis $S$, then yes, any element of $X$ can be written as a $\mathit{unique}$ $R$-linear combination of elements of $S$.
$\quad$Given an $R$-module $Y$ and a function $f:S\rightarrow Y$, it does $\mathit{not}$ follow that $Y$ also has a basis. It does follow that there exists a unique $R$-module homomorphism $h:X\rightarrow Y$ such that $hi=f$, where $i:S\rightarrow X$ is the inclusion function.
$\quad$This is part of a theorem which characterizes $\mathit{free}$ modules, which says that a free (unitary) $R$-module (i.e. one which has a (nonempty) basis) is a free object in the category of (unitary) $R$-modules.
$\quad$To prove it:
(1) define $h$ in the way you described;
(2) verify $hi=f$;
(3) show that $h$ is an $R$-module homomorphism
(4) show that $h$ is unique with the properties in (2) and (3) (suppose there is another $R$-module homomorphism $g:X\rightarrow Y$ such that $gi=f$ and show $g=h$).
$\quad$For a great reference, see Hungerford's $\mathit{Algebra}$, chapter IV, section 2, theorem 2.1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.