History for TorChase

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One can use $\text{Tor}$ to show easily that if $0\to A \to B \to C \to 0$
is exact with $C$ flat then it is pure exact. This requires knowing
that $\text{Tor}$ is "balanced":TorChaseTwo (indeed even well defined) and so we give
a simple diagram chase argument instead. We hope that this chase
gives some indication as to how the chases proceed to prove the
general theory of $\text{Tor}$.
Let $R$ be an associative ring with $1$, and $F$ a left
$R$-module. We say that $F$ is flat if given any exact sequence
of right $R$-modules $0 \to A \to B \to C \to 0$, then $0 \to
{A \otimes F} \to B \otimes F \to C \otimes F \to 0$ is an exact
sequence of Abelian groups. Similarly an exact sequence of right
$R$-modules $0 \to A \to B \to C \to 0$ is said to be pure exact
if $0 \to {A \otimes E} \to B \otimes E \to C \otimes E \to 0$
is an exact sequence of Abelian groups for every left $R$-module $E$.
Suppose $F$ is flat and $0 \to H \to G \to F \to 0$ is exact. We want
to show it is pure exact, so we let $A$ be a right $R$-module. The
tensor product is already right exact so we at least have the
following exact sequence: $A \otimes H {}^{\underrightarrow{\alpha}} A \otimes G \to
A \otimes F \to 0$. By the general theory of $\text{Tor}$, we know that
$\ker(\alpha)=\text{Tor}(A,F)=0$, but we wish to prove this directly.
We consider a partial flat resolution of $A$, that is an exact
sequence $0 \to C \to B \to A \to 0$ with $B$ flat (for instance
the free right $R$-module $R^{(A)}$). We can form the somewhat large commutative
diagram with exact rows and columns:
\[\xymatrix{
& 0 & 0 & 0 &  \\
& A\otimes H \ar[r]^{\alpha} \ar[u] & A \otimes G \ar[r]^{\beta} \ar[u] & A
 \otimes F \ar[r] \ar[u] & 0\\
0\ar[r] & B\otimes H \ar[r]^{\phi} \ar[u]^{c} & B \otimes G \ar[r]^{\gamma}
  \ar[u]^{d} & B \otimes F \ar[r] \ar[u]^{e} & 0\\
 & C\otimes H \ar[r]^{\kappa} \ar[u]^{h} & C \otimes G \ar[r]^{\lambda}
 \ar[u]^{i} & C \otimes F \ar[r] \ar[u]^{j} & 0\\
&&& 0\ar[u]&  \\
}\]
Let $x \in \ker(\alpha)$. Since $x \in A \otimes H$, $x = c(x_1)$
for some $x_1 \in B \otimes H$. Since $0 = \alpha(x) = \alpha c (x_1)
= d\phi(x_1)$ we get $\phi(x_1) \in \ker(d) = \text{im}(i)$, so $\phi(x_1)
= i(x_2)$ for some $x_2 \in C \otimes G$. $0 = \gamma\phi(x_1) =
\gamma i(x_2) = j\lambda(x_2)$, so $\lambda(x_2) \in \ker(j) = 0$,
so $x_2 \in \ker(\lambda) = \text{im}(\kappa)$, so $x_2 = \kappa(x_3)$
for some $x_3 \in C \otimes H$. $\phi(x_1) = i(x_2) = i\kappa(x_3) = \phi h(x_3)$,
so $x_1 - h(x_3) \in \ker(\phi) = 0$, so $x_1=h(x_3)$. Thus $x=c(x_1)=ch(x_3)=0$.
Thus $\ker(\alpha)=0$ and $0 \to A \otimes H \to A \otimes G \to A \otimes F \to 0$
is exact, and thus $0 \to H \to G \to F \to 0$ is pure.