Vtyjkyuk

Prove that $mathbbZ_2 otimes_mathbbZ$ is not exact.

I am trying to find an exact sequence such that if we apply the functor above we don't get an exact sequence. It is known that the functor above is exact from the right so I am just trying to prove that the first map does not maintain injectivity. I took the following sequence:

$$0 longrightarrow 3mathbbZ longrightarrow 6mathbbZ longrightarrow 0 longrightarrow 0$$

where $f: 3mathbbZ rightarrow 6mathbbZ$ is given by $f(z) = 2z$.
Clearly this is a short exact sequence. Now if we look at the map:

$$ f_* : mathbbZ_2 otimes_mathbbZ 3mathbbZ longrightarrow mathbbZ_2 otimes_mathbbZ6mathbbZ $$

it is the zero map because the tensor product on the right is null. But the tensor on the left is not null so this means that $ker (f_*) neq 0$ so $f_*$ is not injective.

I want to ask if this solution is correct or I did something stupid. Can someone verify and if there is a better example please provide it?

Thanks in advance.

I don't think your example is an example: your map $f$ is an iso of abelian groups, so $f_*$ is as well. This follows from the fact that $mathbbZ_2 otimes_mathbbZ-$ is a functor, so it respects isos.

Moreover, because tensoring respects isos, and each of your groups $3mathbbZ$ and $6mathbbZ$ are infinite cyclic (hence iso to each other), their tensor products are iso: you cannot have one null and the other not.

The standard example of this is to consider the canonical sequence
$$
0 to mathbbZ to mathbbZ to mathbbZ_2 to 0
$$
which "creates" $mathbbZ_2$ as the quotient $mathbbZ/2mathbbZ$. So, the second arrow is multiplication by $2$ and the third is the quotient map. Apply your functor $mathbbZ_2 otimes_mathbbZ-$ and use the usual isos to get
$$
0 to mathbbZ_2 to mathbbZ_2 to mathbbZ_2 to 0
$$
which will NEVER be short exact, no matter the maps (for counting/quotient reasons).