Section 5.3 Computing Ext and Tor
Subsection The Constructions
“”―
Given \(R\)-modules \(M\) and \(N\text{,}\) we have two possible ways to compute \(\operatorname{Tor}_{i}^{R}(M, N)\) from the definition.
Construction 6.31. Find a projective resolution
\begin{equation*}
\cdots \longrightarrow P_{2} \longrightarrow P_{1} \longrightarrow P_{0} \longrightarrow M \longrightarrow 0
\end{equation*}
of \(M\text{.}\) Applying \(\otimes_{R} N\) to the resolution (not counting \(M\) ), we get a complex
\begin{equation*}
\cdots \longrightarrow P_{2} \otimes_{R} N \longrightarrow P_{1} \otimes_{R} N \longrightarrow P_{0} \otimes_{R} N \longrightarrow 0
\end{equation*}
Its homology is \(\operatorname{Tor}_{*}^{R}(M, N)\) :
\begin{equation*}
\operatorname{Tor}_{i}^{R}(M, N)=\mathrm{H}_{i}\left(\cdots \longrightarrow P_{2} \otimes_{R} N \longrightarrow P_{1} \otimes_{R} N \longrightarrow P_{0} \otimes_{R} N \longrightarrow 0\right)
\end{equation*}
Alternatively, we can find a free resolution of \(N\text{,}\) say
\begin{equation*}
\cdots \longrightarrow Q_{2} \longrightarrow Q_{1} \longrightarrow Q_{0} \longrightarrow N \longrightarrow 0
\end{equation*}
apply \(M \otimes_{R} \quad\text{,}\)
\begin{equation*}
\cdots \longrightarrow M \otimes_{R} Q_{2} \longrightarrow M \otimes_{R} Q_{1} \longrightarrow M \otimes_{R} Q_{0} \longrightarrow 0
\end{equation*}
and compute the homology of the resulting complex:
\begin{equation*}
\operatorname{Tor}_{i}^{R}(M, N)=\mathrm{H}_{i}\left(\cdots \longrightarrow M \otimes_{R} Q_{2} \longrightarrow M \otimes_{R} Q_{1} \longrightarrow M \otimes_{R} Q_{0} \longrightarrow 0\right)
\end{equation*}
Similarly, we have two possible ways to compute \(\operatorname{Ext}_{R}^{i}(M, N)\text{.}\)
Construction 6.32. Find a projective resolution
\begin{equation*}
\cdots \longrightarrow P_{2} \longrightarrow P_{1} \longrightarrow P_{0} \longrightarrow M \longrightarrow 0
\end{equation*}
of \(M\text{.}\) Applying the contravariant functor \(\operatorname{Hom}_{R}(, N)\) to the resolution gives us a cocomplex rather than a complex:
\begin{equation*}
0 \longrightarrow \operatorname{Hom}_{R}\left(P_{0}, N\right) \longrightarrow \operatorname{Hom}_{R}\left(P_{1}, N\right) \longrightarrow \operatorname{Hom}_{R}\left(P_{2}, N\right) \longrightarrow \cdots
\end{equation*}
Its homology is \(\operatorname{Ext}_{R}^{*}(M, N)\) :
\begin{equation*}
\operatorname{Ext}_{R}^{i}(M, N)=\mathrm{H}^{i}\left(0 \longrightarrow \operatorname{Hom}_{R}\left(P_{0}, N\right) \longrightarrow \operatorname{Hom}_{R}\left(P_{1}, N\right) \longrightarrow \operatorname{Hom}_{R}\left(P_{2}, N\right) \longrightarrow \cdots\right)
\end{equation*}
Alternatively, we can find an injective resolution of \(N\text{,}\) say
\begin{equation*}
0 \longrightarrow N \longrightarrow E^{0} \longrightarrow E^{1} \longrightarrow E^{2} \longrightarrow \cdots
\end{equation*}
apply the covariant functor \(\operatorname{Hom}_{R}(M, \quad)\text{,}\) which yields the cocomplex
\begin{equation*}
0 \longrightarrow \operatorname{Hom}_{R}\left(M, E^{0}\right) \longrightarrow \operatorname{Hom}_{R}\left(M, E^{1}\right) \longrightarrow \operatorname{Hom}_{R}\left(M, E^{2}\right) \longrightarrow \cdots
\end{equation*}
and compute the cohomology of the resulting cocomplex: \(\operatorname{Ext}_{R}^{i}(M, N)=\mathrm{H}^{i}\left(0 \longrightarrow \operatorname{Hom}_{R}\left(M, E^{0}\right) \longrightarrow \operatorname{Hom}_{R}\left(M, E^{1}\right) \longrightarrow \operatorname{Hom}_{R}\left(M, E^{2}\right) \longrightarrow \cdots\right)\text{.}\)
It helps to keep a few simple ideas in mind:
If \(P\) is a projective \(R\)-module, then
\begin{equation*}
\operatorname{Tor}_{i}^{R}(M, P)=\operatorname{Tor}_{i}^{R}(P, M)=0
\end{equation*}
and
\begin{equation*}
\operatorname{Ext}_{R}^{i}(P, M)=0
\end{equation*}
for all \(i>0\) and all \(R\)-modules \(M\text{,}\) since \(0 \rightarrow P \rightarrow 0\) is a projective resolution for \(M\text{.}\) This is a special case of Remark 6.3.
If \(E\) is an injective \(R\)-module,
\begin{equation*}
\operatorname{Ext}_{R}^{i}(M, E)=0
\end{equation*}
for all \(i>0\) and all \(R\)-modules \(M\text{.}\) Free resolutions are often easier to compute explicitly, and the best path towards finding \(\operatorname{Ext}_{R}^{n}(M, N)\text{.}\)
Relating one of our modules to other, easier modules via a short exact sequence can often simplify complicated computations.
Subsection Examples
“”―
Let’s compute some examples.
Example 5.38.
Let’s compute \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(2), \mathbb{Z} /(3))\text{.}\) Injective resolutions are not so easy to find, so we start from a projective resolution for \(\mathbb{Z} /(2)\) :
\begin{equation*}
0 \longrightarrow \mathbb{Z} \stackrel{2}{\longrightarrow} \mathbb{Z} \longrightarrow \mathbb{Z} /(2) \longrightarrow 0
\end{equation*}
Notice that \(\operatorname{pdim}_{\mathbb{Z}}(\mathbb{Z} /(2)) \neq 0\text{,}\) since \(\mathbb{Z} /(2)\) is not a projective \(\mathbb{Z}\)-module. We found a free resolution of length 1 for \(\mathbb{Z} /(2)\text{,}\) so it must be that \(\operatorname{pdim}_{\mathbb{Z}}(\mathbb{Z} /(2))=1\text{.}\) This immediately tells us that \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(2), \mathbb{Z} /(3))=0\) for all \(i \leqslant 2\text{.}\) Now we apply \(\operatorname{Hom}_{\mathbb{Z}}(, \mathbb{Z} /(3))\) to our free resolutions for \(\mathbb{Z} /(2)\text{,}\) and obtain
\begin{equation*}
0 \longrightarrow \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Z} /(3)) \stackrel{2^{*}}{\longrightarrow} \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Z} /(3)) \longrightarrow 0
\end{equation*}
By Exercise 2.4, \(\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Z} /(3)) \cong \mathbb{Z} /(3)\text{,}\) via the isomorphism \(f \mapsto f(1)\text{.}\) Since \(2^{*}\) was the map \(f \mapsto(2 \cdot \quad) \circ f=2 f()\text{,}\) we can simplify our complex to
\begin{equation*}
0 \longrightarrow \mathbb{Z} /(3) \stackrel{2}{\longrightarrow} \mathbb{Z} /(3) \longrightarrow 0
\end{equation*}
Notice that multiplication by 2 is an isomorphism on \(\mathbb{Z} /(3)\text{,}\) so the complex above is exact, and \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(2), \mathbb{Z} /(3))=0\) for all \(i\text{.}\)
Example 5.39.
Given an integer \(n>1\text{,}\)
\begin{equation*}
0 \longrightarrow \mathbb{Z} \stackrel{n}{\longrightarrow} \mathbb{Z} \stackrel{\pi}{\longrightarrow} \mathbb{Z} /(n) \longrightarrow 0
\end{equation*}
with \(\pi\) the canonical projection is a free resolution for \(\mathbb{Z} /(n)\) over \(\mathbb{Z}\text{.}\) Notice that since \(\mathbb{Z} /(n)\) is not a free \(\mathbb{Z}\)-module, there is no shorter free resolution for \(\mathbb{Z} / \mathfrak{n}\text{.}\) Now we can use this resolution to compute \(\operatorname{Tor}_{i}^{\mathbb{Z}}(\mathbb{Z} /(n), M)\) and \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(n), M)\) for any \(\mathbb{Z}\)-module \(M\text{.}\) For Tor,
\begin{equation*}
\operatorname{Tor}_{i}^{\mathbb{Z}}(\mathbb{Z} /(n), M)=\mathrm{H}_{i}\left(0 \longrightarrow \mathbb{Z} \otimes_{\mathbb{Z}} M \stackrel{n \otimes 1}{\longrightarrow} \mathbb{Z} \otimes_{\mathbb{Z}} M \longrightarrow 0\right)
\end{equation*}
By Lemma 2.54, \(\mathbb{Z} \otimes_{\mathbb{Z}} M \cong M\text{,}\) via the map \(k \otimes m \mapsto k m\text{,}\) and the map \(n \otimes 1_{M}\) corresponds to multiplication by \(n\) on \(M\text{.}\) Therefore,
\begin{equation*}
\operatorname{Tor}_{i}^{\mathbb{Z}}(\mathbb{Z} /(n), M)=\mathrm{H}_{i}(0 \longrightarrow M \stackrel{n}{\longrightarrow} M \longrightarrow 0),
\end{equation*}
So
\begin{equation*}
\operatorname{Tor}_{i}^{\mathbb{Z}}(\mathbb{Z} /(n), M)= \begin{cases}M / n M & \text { for } i=0 \\ \left(0:_{M} n\right) & \text { for } i=1 \\ 0 & \text { otherwise }\end{cases}
\end{equation*}
Notice that \(\operatorname{Tor}_{0}^{\mathbb{Z}}(\mathbb{Z} /(n), M)=M / n M=\mathbb{Z} / n \mathbb{Z} \otimes_{\mathbb{Z}} M\text{,}\) as we already knew from Proposition 5.4.
Similarly, we can compute all the Ext modules from \(\mathbb{Z} /(n)\) :
\begin{equation*}
\operatorname{Ext}_{i}^{\mathbb{Z}}(\mathbb{Z} /(n), M)=\mathrm{H}_{i}\left(0 \longrightarrow \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \stackrel{n^{*}}{\longrightarrow} \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \longrightarrow 0\right)
\end{equation*}
By Exercise 2.4, \(\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \cong M\text{,}\) via the map \(f \mapsto f(1)\text{,}\) and \(n^{*}=\operatorname{Hom}_{\mathbb{Z}}(n, M)\) corresponds to multiplication by \(n\) on \(M\text{.}\) So
\begin{equation*}
\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(n), M)=\mathrm{H}^{i}(0 \longrightarrow M \stackrel{n}{\longrightarrow} M \longrightarrow 0)
\end{equation*}
We conclude that
\begin{equation*}
\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(n), M)= \begin{cases}M / n M & \text { for } i=1 \\ \left(0:_{M} n\right) & \text { for } i=0 \\ 0 & \text { otherwise }\end{cases}
\end{equation*}
Notice that \(\operatorname{Ext}_{\mathbb{Z}}^{1}(\mathbb{Z} /(n), M)=\left(0:_{M} n\right)=\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z} /(n), M)\text{,}\) as we already knew from Proposition 5.4.
Example 5.40.
Alternatively, we can compute \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(n), M)\) and \(\operatorname{Tor}_{i}^{\mathbb{Z}}(\mathbb{Z} /(n), M)\) by looking at some long exact sequences. The long exact sequence for Tor induced by the short exact sequence
\begin{equation*}
0 \longrightarrow \mathbb{Z} \stackrel{n}{\longrightarrow} \mathbb{Z} \longrightarrow \mathbb{Z} /(n) \longrightarrow 0
\end{equation*}
is
\begin{equation*}
\begin{aligned}
& \cdots \longrightarrow \operatorname{Tor}_{n+1}^{\mathbb{Z}}(\mathbb{Z} /(n), M) \longrightarrow \operatorname{Tor}_{n}^{\mathbb{Z}}(\mathbb{Z}, M) \longrightarrow \operatorname{Tor}_{n}^{\mathbb{Z}}(\mathbb{Z}, M) \longrightarrow \operatorname{Tor}_{n}^{\mathbb{Z}}(\mathbb{Z} /(n), M) \longrightarrow \cdots \\
& \cdots \longrightarrow \operatorname{Tor}_{1}^{\mathbb{Z}}(\mathbb{Z} /(n), M) \longrightarrow \mathbb{Z} \otimes_{\mathbb{Z}} M \longrightarrow \mathbb{Z} \otimes_{\mathbb{Z}} M \longrightarrow
\end{aligned}
\end{equation*}
Since \(\mathbb{Z}\) is a projective \(\mathbb{Z}\)-module and thus flat, \(\operatorname{Tor}_{i}^{\mathbb{Z}}(\mathbb{Z}, M)=0\) for all \(i>0\text{.}\) As a consequence, the long exact sequence above forces \(\operatorname{Tor}_{2}^{\mathbb{Z}}(\mathbb{Z} /(n), M)=0\text{.}\) So our long exact sequence really gets reduced to
\begin{equation*}
0 \longrightarrow \operatorname{Tor}_{1}^{\mathbb{Z}}(\mathbb{Z} /(n), M) \longrightarrow \mathbb{Z} \otimes_{\mathbb{Z}} M \longrightarrow \mathbb{Z} \otimes_{\mathbb{Z}} M \longrightarrow \mathbb{Z} /(n) \otimes_{\mathbb{Z}} M \longrightarrow 0
\end{equation*}
Now \(\mathbb{Z} \otimes_{\mathbb{Z}} M \cong M\) via \(k \otimes m \mapsto k m\text{,}\) and this isomorphism turns \(n \otimes 1_{M}\) into multiplication by \(n\) on \(M\text{,}\) same as above. So \(\operatorname{Tor}_{1}^{\mathbb{Z}}(\mathbb{Z} /(n), M)\) is the kernel of multiplication by \(n\) on \(M\text{,}\) or \(\left(0:_{M} n\right)\text{.}\) If we want to compute \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(n), M)\text{,}\) we should now look at the long exact sequence
\begin{equation*}
\begin{array}{r}
0 \longrightarrow \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z} /(n), M) \longrightarrow \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \stackrel{n^{*}}{\longrightarrow} \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \longrightarrow \operatorname{Ext}_{\mathbb{Z}}^{n}(\mathbb{Z}, M) \longrightarrow \operatorname{Ext}_{\mathbb{Z}}^{1}(\mathbb{Z} /(n), M) \longrightarrow \\
\cdots \operatorname{Ext}_{\mathbb{Z}}^{n}(\mathbb{Z}, M) \longrightarrow \operatorname{Ext}_{\mathbb{Z}}^{n+1}(\mathbb{Z} /(n), M) \longrightarrow \cdots
\end{array}
\end{equation*}
Again, \(\mathbb{Z}\) is a free \(\mathbb{Z}\)-module, so \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z}, M)=0\) for all \(i>0\text{.}\) Then \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} /(n), M)=0\) for all \(i>1\text{,}\) and our long exact sequence is actually just
\begin{equation*}
0 \longrightarrow \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z} /(n), M) \longrightarrow \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \stackrel{n^{*}}{\longrightarrow} \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \longrightarrow \operatorname{Ext}_{\mathbb{Z}}^{1}(\mathbb{Z} /(n), M) \longrightarrow 0
\end{equation*}
So \(\operatorname{Ext}_{\mathbb{Z}}^{1}(\mathbb{Z} /(n), M)\) is the cokernel of \(n^{*}\text{.}\) As before, notice that \(\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, M) \cong M\) via the map \(f \mapsto f(1)\text{,}\) and \(n^{*}\) corresponds to multiplication by \(n\) on \(M\text{.}\) We conclude that \(\operatorname{Ext}_{\mathbb{Z}}^{1}(\mathbb{Z} /(n), M) \cong M / n M\text{.}\)
Exercise 5.41.
Let \(k\) be a field and \(R=k[x] /\left(x^{3}\right)\text{.}\)
- Compute \(\operatorname{Tor}_{i}^{R}(k, k)\) for all \(i\text{.}\)
- Show that \(\operatorname{Ext}_{R}^{i}(k, k) \neq 0\) for all \(i\text{.}\)
Exercise 5.42.
Let \(k\) be a field, \(R=k[x, y]\text{,}\) and \(\quad=(x, y)\text{.}\)
- Show that\begin{equation*} 0 \longrightarrow R \stackrel{\left(\begin{array}{c} y \\ x \end{array}\right)}{\longrightarrow} R^{2} \stackrel{x \quad y)}{\longrightarrow} R \longrightarrow 0 \end{equation*}is a free resolution for \(k=R /\text{.}\)
- Compute \(\operatorname{Tor}_{i}^{R}(k, k)\) for all \(i\text{.}\)
- Show that\begin{equation*} \operatorname{Tor}_{1}(, k) \cong \operatorname{Tor}_{2}(k, k) \end{equation*}
