The definition of right-exact-ness of $-\otimes_R M$
The meaning of "$- \otimes M $ is a right exact functor" is, whenever
$$ 0\to B\to A\to C\to 0$$ is a short exact sequence (in the category of $R$-modules), the induced sequence
Assume a commutative ring $R$ and a $R$-module $M$.
$$ B\otimes_R M\to A\otimes_R M\to C\otimes_R M\to 0$$ is also a short exact sequence.
The description above is basically paraphrasing the definition of right-exact-ness word-for-word.
What about the left $0$?
The question is what happens to the $0$ at the left of the induced sequence. Or, what is the same, whether the induced map $B\otimes_R M\to A\otimes_R M$ is injective or not.
It depends.
- It can happen that that induced map is not injective. For example, consider the exact sequence
$$0\to 2\Bbb Z\stackrel{i}\to\Bbb Z\stackrel p\to\Bbb Z/\Bbb 2Z\to0$$
where $i$ is the inclusion map (and hence $p$ is the quotient map). After we apply $-\otimes_\Bbb Z\Bbb Z/2\Bbb Z$,
the sequence becomes
$$0
\to2\Bbb Z\otimes_\Bbb Z\Bbb Z/2\Bbb Z\stackrel{i\otimes_\Bbb Z\Bbb Z/2\Bbb Z}
\to\Bbb Z\otimes_\Bbb Z\Bbb Z/2\Bbb Z\stackrel {p\otimes_\Bbb Z\Bbb Z/2\Bbb Z}
\to\Bbb Z/\Bbb 2Z\otimes_\Bbb Z\Bbb Z/2\Bbb Z
\to0$$
The module $2\Bbb Z\otimes_\Bbb Z\Bbb Z/2\Bbb Z$ is isomorphic to $\Bbb Z\otimes_\Bbb Z\Bbb Z/2\Bbb Z=\Bbb Z\Bbb Z/2\Bbb Z$ since $2\Bbb Z$ is isomporhic to $\Bbb Z$. It is generated by $2\otimes_\Bbb Z\overline1\not=0$.
$$(i\otimes_\Bbb Z\Bbb Z/2\Bbb Z)(2\otimes_\Bbb Z\overline1)
=2\otimes_\Bbb Z\overline1
=(2\cdot1)\otimes_\Bbb Z\overline1
=1\otimes_\Bbb Z(2\cdot\overline1)
=1\otimes_\Bbb Z 0=0,
$$
which shows the map $i\otimes_\Bbb Z\Bbb Z/2\Bbb Z$ is not injective.
- It can happen that induced map is still injective.
If that is true for any injective map $B\to A$, then $-\otimes_R M$ is an exact functor. We call $M$ a flat $R$-module.
For example, every free $R$-module is also a flat $R$-module.
We can, of course, always write a sequence of any maps. What concerns us is whether we can call it an exact sequence.
Given a short exact sequence $$0 \to \mathfrak{a} \to A \to A / \mathfrak{a} \to 0 \,,$$ we can, in general, only write a "right-exact" short sequence,
$$\mathfrak{a} \otimes M \to A \otimes M \to (A / \mathfrak{a}) \otimes M \to 0.$$ Here "right-exact" means although the sequence is "short" and exact, there is no zero on the left.
Only with more conditions on $\mathfrak{a}, A, M$, etc., we may have a "full-exact" short sequence $$0 \to \mathfrak{a} \otimes M \to A \otimes M \to (A / \mathfrak{a}) \otimes M \to 0 \,$$ with the zero on the left.
