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 $\ker {{\partial}}^0 \cong F(B)$.

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

• That this is an isomorphism follows from

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

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

• ${\mathbf{Z}}{\hbox{-}}$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 ${\mathbf{Z}}/n$ is commutative.

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

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

TFAE in ${}_{R}{\mathsf{Mod}}$:

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