§ Tensor is right exact

Consider an exact sequence
0AfBgC0 0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0
We wish to consider the operation of tensoring with some ring RR. For a given ring morphism h:PQh: P \rightarrow Q this induces a new morphism Rh:RARBR \otimes h: R \otimes A \rightarrow R \otimes B defined by h(ra)rh(a)h(r \otimes a) \equiv r \otimes h(a). So we wish to contemplate the sequence:
RARfRBRgRC 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 Z\mathbb Z:

§ A detailed example

02ZiZπZ/2Z0 0 \rightarrow 2\mathbb Z \xrightarrow{i} \mathbb Z \xrightarrow{\pi} \mathbb Z / \mathbb 2Z \rightarrow 0
Where ii is for inclusion, π\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: Z,2Z,Z/2Z\mathbb Z, \mathbb 2Z, \mathbb Z/\mathbb 2Z. By analogy with fields, tensoring with the base ring Z\mathbb Z is unlikely to produce anything of interest. 2Z\mathbb 2Z maybe more interesting, but see that the map 1Z22Z1 \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), Z/2Z\mathbb Z / \mathbb 2Z. So let's tensor by this element:
Z/2Z2ZiZ/2ZZπZ/2ZZ/2Z \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 Z/2Z2Z\mathbb Z/2\mathbb Z \otimes 2 \mathbb Z has elements of the form (0,)=0(0, *) = 0, We might imagine that the full ring collapses since 12k)=2(1k)=2k=01 \otimes 2k) = 2(1 \otimes k) = 2 \otimes k = 0 (since 2=02 = 0 in Z/2Z\mathbb Z/2\mathbb Z). But this in fact incorrect! Think of the element 121 \otimes 2. We cannot factorize this as 2(11)2(1 \otimes 1) since 1∉2Z1 \not \in 2 \mathbb Z. So we have the element 12Z/2Z/×2Z1 \otimes 2 \in \mathbb Z/2\mathbb Z / \times 2 \mathbb Z.
  • See that Z/2ZZZ/2Z\mathbb Z/2\mathbb Z \otimes \mathbb Z \simeq \mathbb Z/2\mathbb Z: Factorize (k,l)=l(k,1)=(kl,1)Z/2Z(k, l) = l(k, 1) = (kl, 1) \simeq \mathbb Z/2 \mathbb Z.
  • Similarly, see that Z/2ZZ/2ZZ/2Z\mathbb Z/2\mathbb Z \otimes \mathbb Z/2\mathbb Z \simeq \mathbb Z/2\mathbb Z. Elements 00,01,1000 \otimes 0, 0 \otimes 1, 1 \otimes 0 \simeq 0 and 1111 \otimes 1 \simeq 1.
  • In general, Let's investigate elements abZ/nZZ/mmZa \otimes b \in \mathbb Z/n\mathbb Z \otimes \mathbb Z/m\mathbb mZ . We can write this as ab11ab 1 \otimes 1. The 111 \otimes 1 gives us a "machine" to reduce the number by nn and by mm. So if we first reduce by nn, we are left with rr (for remained) for some ab=αn+rab = \alpha n + r. We can then reduce rr by mm to get ab=αn+βm+rab = \alpha n + \beta m + r'. So if r=0r' = 0, then we get ab=αn+βmab = \alpha n + \beta m. But see that all elements of the form αn+βm\alpha n + \beta m is divisible by gcd(n,m)gcd(n, m). Hence, all multiples of gcd(n,m)gcd(n, m) are sent to zero, and the rest of the action follows from this. So we effectively map into Z/gcd(m,n)Z\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 Z×Z/2ZZ/2Z\mathbb Z \times \mathbb Z/2\mathbb Z \simeq \mathbb Z/2 \mathbb Z. We should think of Z\mathbb Z as Z/Z\mathbb Z/\mathbb \infty Z, since we have "no torsion" or "torsion at infinity". So we get the tensor product should have gcd(2,)=2gcd(2, \infty) = 2.
  • Now see that the first two components of the tensor give us a map from Z/2Z2ZiZ/2ZZ\mathbb Z/2\mathbb Z \otimes \mathbb 2Z \xrightarrow{i} \mathbb Z/2\mathbb Z \otimes \mathbb Z which sends:
x2kx2kZ/2ZZ=2(xk)=(2xk)=0k=0 \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 2Z2\mathbb Z is "freed" when injecting into Z\mathbb Z. This latent energy explodes on contant with Z/2Z\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 Z/2ZZπZ/2ZZ/2Z \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)(x,ymod2)(x, y) \mapsto (x, y \mod 2). This clearly gives us all the elements: For example, we get 000 \otimes 0 as the preimage of 0×2k0 \times 2k and we get 111 \otimes 1 as the preimage of (predictably) 1(2k+1)1 \otimes (2k+1). Hence, the map is surjective.
So finally, we have the exact sequence:
Z/2Z2ZiZ/2ZZπZ/2ZZ/2Z0 \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)(0 \rightarrow \dots) since ii' is no longer injective. It fails injectivity as badly as possible, since i(x)=0i'(x) = 0. Thus, tensoring is RIGHT EXACT. It takes right exact sequences to right exact sequences!

§ The general proof

Given the sequence:
AiBπC0 A \xrightarrow{i} B \xrightarrow{\pi} C \rightarrow 0
We need to show that the following sequence is exact:
RAiRBπRC0 R \otimes A \xrightarrow{i'} R \otimes B \xrightarrow{\pi'} R \otimes C \rightarrow 0
  • First, to see that π\pi' is surjective, consider the basis element rcRCr \otimes c \in R \otimes C. Since π\pi is surjective, there is some element bBb \in B such that π(b)=c\pi(b) = c. So the element rbBr \otimes b \in B maps to rcr \otimes c by π\pi'; π(rb)=rπ(b)=rc\pi'(r \otimes b) = r \otimes \pi(b) = r \otimes c(by definition of π\pi, and choice of bb). This proves that BπRC0B \xrightarrow{\pi'} R \otimes C \rightarrow 0is exact.
  • Next, we need to show that im(i)=ker(π)im(i') = ker(\pi').
  • To show that im(i)ker(π)im(i') \subseteq ker(\pi'), consider an arbitrary rar \otimes a. Now compute:
π(i(ra))=π(ri(a))=rπ(i(a))By exactness of AiBπCπ(i(a))=0:=r0=0 \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(ra)im(i)i'(r \otimes a) \in im(i') is in the kernel of π\pi'. Next, let's show ker(π)im(i)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(π)im(i') = ker(\pi') iff coker(i)=RCcoker(i') = R \otimes C. This follows because:
coker(i)=(RB)/im(i)Since im(i)=ker(π)=(RB)/ker(π)Isomorphism theorem: =im(π)π is surjective: =RC \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)=RCcoker(i) = R \otimes C, then I have that im(i)=ker(π)im(i') = ker(\pi'). So let's prove this:
coker(i)=(RB)/im(i)=(RB)/i(RA)Definition of i=(RB)/(Ri(A)) \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 (RB)/(Ri(A))R(B/i(A))(R \otimes B)/( R \otimes i(A)) \simeq R \otimes (B/i(A)) (informally, "take RR common"). Define the quotient map q:BB/i(A)q: B \rightarrow B/i(A). This is a legal quotient map because i(A)=im(i)ker(π)i(A) = im(i) \simeq ker(\pi) is a submodule of BB.
q:BB/i(A)f:RBR(B/i(A))f(rb)=rq(b)rbRBf=Rqrq(b)RB/i(A) \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)ker(f). It contains all those elements such that rq(b)=0r \otimes q(b) = 0. But this is only possible if q(b)=0q(b) = 0. This means that bi(A)=im(i)=ker(π)b \in i(A) = im(i) = ker(\pi). Also see that for every element r(b+i(A))R(B/i(A))r \otimes (b + i(A)) \in R \otimes (B/i(A)), there is an inverse element rbRBr \otimes b \in R \otimes B. So, the map ff is surjective . Hence, im(f)R(B/i(A))im(f) \simeq R \otimes (B/i(A)). Combining the two facts, we get:
domain(f)/ker(f)im(f)(RB)/(R(B/i(A)))R(B/i(A))coker(i)=(RB)/(R(B/i(A)))R(B/i(A))=RC \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)RCcoker(i) \simeq R \otimes C.