I just started learning about modules. What happens if a free module is generated using elements of a finite set $X$ as a basis, but by multiplying each basis with a different ring? For example, let $X = \{a,b\}$, and consider, $$ M = a\mathbb{Z}_3 + b\mathbb{Z}_5. $$ Is this a module? A sum of modules? Another object?
If I take, $$ N = (b-a)\mathbb{Z}_3. $$ Can I say what M / N is? Would it be isomorphic to $a(\mathbb{Z}_3 + \mathbb{Z}_5)$?
Thank you very, very much.
Edit 1
From what the others say, who know much more than me, the setup does not make sense with $\mathbb{Z}_3$, $\mathbb{Z}_5$. What if I consider the following version instead?
$$ M = a\mathbb{R}[x] + b\mathbb{R}[y]. $$ Is this a module? A sum of modules? Another object?
If I take, $$ N = (b-a), $$ Can I say what M / N is? Would it be isomorphic to $a(\mathbb{R}[x] + \mathbb{R}[y])$?
Edit 2
I just found this that seems relevant: modules over direct sum of different rings
For the first $+$ to make sense, you would have to be adding them in some category, and one that makes sense immediately in this case is the category of abelian groups (or $\mathbb Z$-modules, if you prefer.) The action of $\mathbb Z$ on any particular element is just "repeated addition."
But this does not really allow $a,b$ to be "a basis" since $(a\cdot 1_{\mathbb Z_3})3=0$, for example. But what you're describing sounds an awful lot like the abelian group $\mathbb Z_3\times\mathbb Z_5$ otherwise.
Obviously, what you're describing isn't a $\mathbb Z_3$ or $\mathbb Z_5$ module (for example, $(a\cdot 1_{\mathbb Z_3})5=a\cdot 2\cdot 1_{\mathbb Z_3}$ which ought to be zero if it is a $\mathbb Z_5$ module, but isn't.)
You can however, make it into a $\mathbb Z_{15}$ module, but the "basis" still doesn't work since, for example, $(a\cdot 1_{\mathbb Z_3})3=0$ but $3\neq 0$ in $\mathbb Z_{15}$.