TITLE: $M\times N$ Doesn’t Have a Module Structure
QUESTION [6 upvotes]: In Keith Conrad's notes (page 4) is written:

For two $R-$modules $M$ and $N$ , $M\oplus N$ and $M\times N$ are the
  same sets, but $M\oplus N$ is an $R-$module and $M\times N$ doesn’t
  have a module structure.

Out of context this phrase seems strange. I think of $\oplus_{i\in I}M_i$ and $\prod_{i\in I}M_i$ as a sets of collections $\{(m_i)_{i\in I}\}$ with term-wise operations (a difference only in case infinite $I$). Or as a universal objects in category of $R-$modules. In both cases $\prod_{i\in I}M_i$ is a module (as I think).
I get the impression that we "make believe" that $M\times N$ doesn’t have a module structure.
It seems to me that Keith Conrad considers $"\oplus"$ (and further $"\otimes"$) be a module (object of category $R-$mod), but $"\times"$ be only a set of variables for multi-linear functions. Why?
Further written:

... addition on $R$ is linear function $R\oplus R\to R$, but addition
  on R is not a bilinear function $R\times R\to R$, as we saw above.
  Multiplication as a function $R\times R\to R$ is bilinear, but as a
  function $R\oplus R\to R$ it is not linear.

It strange question but why $"\oplus"$ has a natural addition and $"\times"$ has a natural multiplication?
Sorry for such vague questions, I hope you understand what confuses me.

REPLY [2 votes]: The author is wrong, he means something else, which is explained right afterwards. I suggest just to skip this sentence. If $M,N$ are modules and $|-|$ denotes the underlying set of a module, then a bilinear map is not a homomorphism on $M \times N$, but rather a (bilinear) map on $|M| \times |N|$. This is probably why the author doesn't want $M \times N$ to be a module. But, of course, it is a module, with underlying set $|M \times N| = |M| \times |N|$. Never forget about forgetful functors! 
[Converted my comment into an answer.]