$0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$

We wish to consider the operation of tensoring with some ring $R$.
For a given ring morphism $h: P \rightarrow Q$ this induces a new
morphism $R \otimes h: R \otimes A \rightarrow R \otimes B$ defined
by $h(r \otimes a) \equiv r \otimes h(a)$.
So we wish to contemplate the sequence:
$R \otimes A \xrightarrow{R \otimes f} R \otimes B \xrightarrow{R \otimes g} R \otimes C$

To see if it is left exact, right exact, or both. Consider the classic sequence of
modules over $\mathbb Z$:
$0 \rightarrow 2\mathbb Z \xrightarrow{i} \mathbb Z \xrightarrow{\pi} \mathbb Z / \mathbb 2Z \rightarrow 0$

Where $i$ is for inclusion, $\pi$ is for projection. This is an exact sequence, since it's of the form
kernel-ring-quotient. We have three natural choices to tensor with: $\mathbb Z, \mathbb 2Z, \mathbb Z/\mathbb 2Z$.
By analogy with fields, tensoring with the base ring $\mathbb Z$ is unlikely to produce anything of interest.
$\mathbb 2Z$ maybe more interesting, but see that the map $1 \in \mathbb Z \mapsto 2 \in 2 \mathbb Z$ gives us an
isomorphism between the two rings. That leaves us with the final and most interesting element (the one with torsion),
$\mathbb Z / \mathbb 2Z$. So let's tensor by this element:
$\mathbb Z/2\mathbb Z \otimes 2\mathbb Z \xrightarrow{i'}
\mathbb Z/2\mathbb Z \otimes \mathbb Z \xrightarrow{\pi'}
\mathbb Z/2\mathbb Z \otimes \mathbb Z / \mathbb 2Z$

- See that $\mathbb Z/2\mathbb Z \otimes 2 \mathbb Z$ has elements of the form $(0, *) = 0$, We might imagine that the full ring collapses since $1 \otimes 2k) = 2(1 \otimes k) = 2 \otimes k = 0$ (since $2 = 0$ in $\mathbb Z/2\mathbb Z$). But this in fact incorrect! Think of the element $1 \otimes 2$. We
*cannot*factorize this as $2(1 \otimes 1)$ since $1 \not \in 2 \mathbb Z$. So we have the element $1 \otimes 2 \in \mathbb Z/2\mathbb Z / \times 2 \mathbb Z$. - See that $\mathbb Z/2\mathbb Z \otimes \mathbb Z \simeq \mathbb Z/2\mathbb Z$: Factorize $(k, l) = l(k, 1) = (kl, 1) \simeq \mathbb Z/2 \mathbb Z$.
- Similarly, see that $\mathbb Z/2\mathbb Z \otimes \mathbb Z/2\mathbb Z \simeq \mathbb Z/2\mathbb Z$. Elements $0 \otimes 0, 0 \otimes 1, 1 \otimes 0 \simeq 0$ and $1 \otimes 1 \simeq 1$.
- In general, Let's investigate elements $a \otimes b \in \mathbb Z/n\mathbb Z \otimes \mathbb Z/m\mathbb mZ$ . We can write this as $ab 1 \otimes 1$. The $1 \otimes 1$ gives us a "machine" to reduce the number by $n$ and by $m$. So if we first reduce by $n$, we are left with $r$ (for remained) for some $ab = \alpha n + r$. We can then reduce $r$ by $m$ to get $ab = \alpha n + \beta m + r'$. So if $r' = 0$, then we get $ab = \alpha n + \beta m$. But see that all elements of the form $\alpha n + \beta m$ is divisible by $gcd(n, m)$. Hence, all multiples of $gcd(n, m)$ are sent to zero, and the rest of the action follows from this. So we effectively map into $\mathbb Z/ gcd(m, n) \mathbb Z$
- In fact, we can use the above along with (1) write finitely generated abelian groups as direct sum of cyclic groups, (2) tensor distributes over direct sum. This lets us decompose tensor products of all finitely generated abelian groups into cyclics.
- This gives us another heuristic argument for why $\mathbb Z \times \mathbb Z/2\mathbb Z \simeq \mathbb Z/2 \mathbb Z$. We should think of $\mathbb Z$ as $\mathbb Z/\mathbb \infty Z$, since we have "no torsion" or "torsion at infinity". So we get the tensor product should have $gcd(2, \infty) = 2$.
- Now see that the first two components of the tensor give us a map from $\mathbb Z/2\mathbb Z \otimes \mathbb 2Z \xrightarrow{i} \mathbb Z/2\mathbb Z \otimes \mathbb Z$ which sends:

$\begin{aligned}
&x \otimes 2k \mapsto x \otimes 2k \in \mathbb Z/2\mathbb Z \otimes \mathbb Z \\
&= 2 (x \otimes k) \\
&= (2x \otimes k) \\
&=0 \otimes k = 0
\end{aligned}$

- This map is not injective, since this map kills everything! Intuitively, the "doubling" that is latent in $2\mathbb Z$ is "freed" when injecting into $\mathbb Z$. This latent energy explodes on contant with $\mathbb Z/2 \mathbb Z$ giving zero. So, the sequence is no longer left-exact, since the map is not injective!

- So the induced map is identically zero! Great, let's continue, and inspect the tail end $\mathbb Z/2\mathbb Z \otimes \mathbb Z \xrightarrow{\pi} \mathbb Z/2\mathbb Z \otimes \mathbb Z / \mathbb 2Z$. Here, we sent the element $(x, y) \mapsto (x, y \mod 2)$. This clearly gives us all the elements: For example, we get $0 \otimes 0$ as the preimage of $0 \times 2k$ and we get $1 \otimes 1$ as the preimage of (predictably) $1 \otimes (2k+1)$. Hence, the map is surjective.

$\mathbb Z/2\mathbb Z \otimes 2\mathbb Z \xrightarrow{i'}
\mathbb Z/2\mathbb Z \otimes \mathbb Z \xrightarrow{\pi'}
\mathbb Z/2\mathbb Z \otimes \mathbb Z / \mathbb 2Z \rightarrow 0$

We do NOT have the initial $(0 \rightarrow \dots)$ since $i'$ is no longer injective.
It fails injectivity as badly as possible, since $i'(x) = 0$. Thus, tensoring is
RIGHT EXACT. It takes right exact sequences to right exact sequences!
$A \xrightarrow{i} B \xrightarrow{\pi} C \rightarrow 0$

We need to show that the following sequence is exact:
$R \otimes A \xrightarrow{i'} R \otimes B \xrightarrow{\pi'} R \otimes C \rightarrow 0$

- First, to see that $\pi'$ is surjective, consider the basis element $r \otimes c \in R \otimes C$. Since $\pi$ is surjective, there is some element $b \in B$ such that $\pi(b) = c$. So the element $r \otimes b \in B$ maps to $r \otimes c$ by $\pi'$; $\pi'(r \otimes b) = r \otimes \pi(b) = r \otimes c$ (by definition of $\pi$, and choice of $b$). This proves that $B \xrightarrow{\pi'} R \otimes C \rightarrow 0$is exact.

- Next, we need to show that $im(i') = ker(\pi')$.
- To show that $im(i') \subseteq ker(\pi')$, consider an arbitrary $r \otimes a$. Now compute:

$\begin{aligned}
&\pi'(i'(r \otimes a)) \\
&= \pi'(r \otimes i(a)) \\
&= r \otimes \pi(i(a))
& \text{By exactness of $A \xrightarrow{i} B \xrightarrow{\pi} C$, $\pi(i(a)) = 0$:} \\
&= r \otimes 0 \\
&= 0
\end{aligned}$

So we have that any element in $i'(r \otimes a) \in im(i')$ is in the kernel of $\pi'$.
Next, let's show $ker(\pi') \subseteq im(i')$. This is the "hard part" of the proof. So let's
try a different route. I claim that if $im(i') = ker(\pi')$ iff $coker(i') = R \otimes C$. This
follows because:
$\begin{aligned}
&coker(i) = (R \otimes B)/ im(i') \\
& \text{Since } im(i') = ker(\pi')
&= (R \otimes B)/ker(\pi') \\
& \text{Isomorphism theorem: } \\
&= im(\pi') \\
& \text{$\pi'$ is surjective: } \\
&= R \otimes C
\end{aligned}$

Since each line was an equality, if I show that $coker(i) = R \otimes C$, then I have that $im(i') = ker(\pi')$.
So let's prove this:
$\begin{aligned}
&coker(i) = (R \otimes B)/ im(i') \\
&= (R \otimes B)/i'(R \otimes A) \\
& \text{Definition of $i'$: } \\
&= (R \otimes B)/(R \otimes i(A)) \\
\end{aligned}$

I claim that the $(R \otimes B)/( R \otimes i(A)) \simeq R \otimes (B/i(A))$ (informally, "take $R$ common").
Define the quotient map $q: B \rightarrow B/i(A)$. This is a legal quotient map because $i(A) = im(i) \simeq ker(\pi)$
is a submodule of $B$.
$\begin{aligned}
q : B \rightarrow B/i(A) \\
f: R \otimes B \rightarrow \rightarrow R \otimes (B / i(A)) \\
f(r \otimes b) = r \otimes q(b) \\
r \otimes b \in R \otimes B \xrightarrow{f = R \otimes q } r \otimes q(b) \in R \otimes B/i(A)
\end{aligned}$

Let's now study $ker(f)$. It contains all those elements such that $r \otimes q(b) = 0$.
But this is only possible if $q(b) = 0$. This means that $b \in i(A) = im(i) = ker(\pi)$.
Also see that for every element $r \otimes (b + i(A)) \in R \otimes (B/i(A))$, there is an inverse
element $r \otimes b \in R \otimes B$. So, the map $f$ is $\begin{aligned}
&domain(f)/ker(f) \simeq im(f) \\
&(R \otimes B)/(R \otimes (B/i(A))) \simeq R \otimes (B/i(A))
&coker(i) = (R \otimes B)/(R \otimes (B/i(A))) \simeq R \otimes (B/i(A)) = R \otimes C
\end{aligned}$

Hence, $coker(i) \simeq R \otimes C$.