Hom and Ext

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• \(\mathop{\mathrm{Hom}}_R(A, {-})\) is:
  • Covariant
  • Left-exact
  • Is a functor that sends \(f:X\to Y\) to \(f_*: \mathop{\mathrm{Hom}}(A, X) \to \mathop{\mathrm{Hom}}(A, Y)\) given by \(f_*(h) = f\circ h\).
  • Has right-derived functors \(\operatorname{Ext} ^i_R(A, B) \coloneqq R^i \mathop{\mathrm{Hom}}_R(A, {-})(B)\) computed using injective resolutions.
• \(\mathop{\mathrm{Hom}}_R({-}, B)\) is:
  • Contravariant
  • Right-exact
  • Is a functor that sends \(f:X\to Y\) to \(f^*: \mathop{\mathrm{Hom}}(Y, B) \to \mathop{\mathrm{Hom}}(X, B)\) given by \(f^*(h) = h\circ f\).
  • Has left-derived functors \(\operatorname{Ext} ^i_R(A, B) \coloneqq L_i \mathop{\mathrm{Hom}}_R({-}, B)(A)\) computed using projective resolutions.
• For \(N \in ({R}, {S'}){\hbox{-}}\mathsf{biMod}\) and \(M\in ({R}, {S}){\hbox{-}}\mathsf{biMod}\), \(\mathop{\mathrm{Hom}}_R(M, N) \in ({S}, {S'}){\hbox{-}}\mathsf{biMod}\).
  • Mnemonic: the slots of \(\mathop{\mathrm{Hom}}_R\) use up a left \(R{\hbox{-}}\)action. In the first slot, the right \(S{\hbox{-}}\)action on \(M\) becomes a left \(S{\hbox{-}}\)action on Hom. In the second slot, the right \(S'{\hbox{-}}\)action on \(N\) becomes a right \(S'{\hbox{-}}\)action on Hom.

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• \(\operatorname{Ext} ^{>1}(A, B) = 0\) for any \(A\) projective or \(B\) injective.

Write \(F({-}) \coloneqq\mathop{\mathrm{Hom}}_R(A, {-})\). This is left-exact and thus has right-derived functors \(\operatorname{Ext} ^i_R(A, B) \coloneqq R^iF(B)\). To compute this:

• Take an injective resolution: \begin{align*} 1 \to B \xrightarrow{{\varepsilon}} I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} \cdots .\end{align*}

• Remove the augmentation \({\varepsilon}\) and just keep the complex \begin{align*} I^{-}\coloneqq\qty{ 1 \xrightarrow{d^{-1}} I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} \cdots } .\end{align*}

• Apply \(F({-})\) to get a new (and usually not exact) complex \begin{align*} F(I)^{-}\coloneqq\qty{ 1 \xrightarrow{{{\partial}}^{-1}} F(I^0) \xrightarrow{{{\partial}}^0} F(I^1) \xrightarrow{{{\partial}}^1} \cdots } ,\end{align*} where \({{\partial}}^i \coloneqq F(d^i)\).

• Take homology, i.e. kernels mod images: \begin{align*} R^iF(B) \coloneqq{ \ker d^i \over \operatorname{im}d^{i-1}} .\end{align*}

Note that \(R^0 F(B) \cong F(B)\) canonically:

• This is defined as \(\ker {{\partial}}^0 / \operatorname{im}{{\partial}}^{-1} = \ker {{\partial}}^0 / 1 = \ker {{\partial}}^0\).

• Use the fact that \(F({-})\) is left exact and apply it to the augmented complex to obtain \begin{align*} 1 \to F(B) \xrightarrow{F({\varepsilon})} F(I^0) \xrightarrow{{{\partial}}^0} F(I^1) \xrightarrow{{{\partial}}^1} \cdots .\end{align*}

• By exactness, there is an isomorphism \(\ker {{\partial}}^0 \cong F(B)\).

\(\phi: \mathop{\mathrm{Hom}}_{{\mathbb{Z}}}({\mathbb{Z}}, {\mathbb{Z}}/n) \xrightarrow{\sim} {\mathbb{Z}}/n\), where \(\phi(g) \coloneqq g(1)\).

• That this is an isomorphism follows from

• Surjectivity: for each \(\ell \in {\mathbb{Z}}/n\) define a map \begin{align*} \psi_y: {\mathbb{Z}}&\to {\mathbb{Z}}/n \\ 1 &\mapsto [\ell]_n .\end{align*}

• Injectivity: if \(g(1) = [0]_n\), then \begin{align*} g(x) = xg(1) = x[0]_n = [0]_n .\end{align*}

• \({\mathbb{Z}}{\hbox{-}}\)module morphism: \begin{align*} \phi(gf) \coloneqq\phi(g\circ f) \coloneqq(g\circ f)(1) = g(f(1)) = f(1)g(1) = \phi(g)\phi(f) ,\end{align*} where we’ve used the fact that \({\mathbb{Z}}/n\) is commutative.

• \(\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}/m, {\mathbb{Z}}) = 0\).
• \(\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}/m, {\mathbb{Z}}/n) = {\mathbb{Z}}/d\).
• \(\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Q}}, {\mathbb{Q}}) = {\mathbb{Q}}\).

• \(\operatorname{Ext} _{\mathbb{Z}}({\mathbb{Z}}/m, G) \cong G/mG\)
  • Use \(1 \to {\mathbb{Z}}\xrightarrow{\times m} {\mathbb{Z}}\xrightarrow{} {\mathbb{Z}}/m \to 1\) and apply \(\mathop{\mathrm{Hom}}_{\mathbb{Z}}({-}, {\mathbb{Z}})\).
• \(\operatorname{Ext} _{\mathbb{Z}}({\mathbb{Z}}/m, {\mathbb{Z}}/n) = {\mathbb{Z}}/d\).

TFAE in \({\mathsf{R}{\hbox{-}}\mathsf{Mod}}\):

• A SES \(0\to A\to B \to C\to 0\) is split.
• ?