What does it mean for the tensor functor to be right exact?

Question

If I am applying the right exact functor $- \otimes M $ to the short exact sequence $$0 \to \mathfrak{a} \to A \to A / \mathfrak{a} \to 0 \,,$$ do we have to write the new sequence as $$\mathfrak{a} \otimes M \to A \otimes M \to (A / \mathfrak{a}) \otimes M \to 0 \,,$$ (without the zero on the left ) or we can still write it as $$0 \to \mathfrak{a} \otimes M \to A \otimes M \to (A / \mathfrak{a}) \otimes M \to 0 \,,$$ (with the zero on the left)?

If the second case is the right, can anyone tell me what is exactly the meaning of tensor is a right exact functor?

Answer 1

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.

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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.