Hom functor (${\mathrm{Hom}}_R(-,-)$)

The Hom functor sends two $R$-modules to the abelian group of $R$-linear maps between them. It is contravariant in its first argument and covariant in its second.

The Ext functors are the right derived functors of the Hom functor. They can be computed by applying ${\mathrm{Hom}}_R(-,N)$ to a projective resolution of $M$ and taking cohomology.

For a non-zero-divisor $a\in A$, the free resolution $0\to A\stackrel{a}{\to }A\to A/(a)\to 0$ gives degree-zero kernel and degree-one cokernel, and all higher Ext functors vanish. Directly in an injective resolution of the second variable, multiplication by $a$ is surjective in every term: the map $aA\to I$ taking $ax$ to $xm$ extends to $A\to I$. Thus $0\to {\mathrm{Hom}}_A(A/(a),I^{\bullet })\to I^{\bullet }\stackrel{a}{\to }{I}^{\bullet }\to 0$ is a short exact sequence of complexes and gives the same result. The displayed Hom functor is covariant in $M$; changing it to ${\mathrm{Hom}}_A(M,A/(a))$ changes the variable that is resolved.

A free resolution of $\mathbb{Q}$ has generators $a_n=1/n!$ and independent relations $a_n-(n+1)a_{n+1}$. Thus the Ext functor gives the cokernel of the map on integer sequence productsThe constant sequence one is not in its image. A preimage would require $u_1\equiv {\sum }_{j=1}^{N}j!(\mathrm{mod}(N+1)!)$ for every $N$. For $N\ge 3$, the sum is below $2N!$, tends to infinity, and its distance from $(N+1)!$ also tends to infinity. No fixed integer satisfies all these congruences. Hence ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z})\ne 0$.

Applying the contravariant Hom functor ${\mathrm{Hom}}_R(-,W)$ to $0\to I\to X\to C\to 0$ givesThe connecting map sends $I\to W$ to its pushout of a module extension. Its vanishing means that map extends to $X\to W$. For a hereditary ring, ${\mathrm{Ext}}^2(C,W)=0$, making the indicated restriction on first extension groups surjective. The covariant variable has its corresponding long exact sequence.

An extension group is a value of an Ext functor. In degree one it classifies short exact sequences $0\to N\to E\to M\to 0$ up to isomorphisms fixing the endpoints. The zero class is the split extension; its addition is the Baer sum. Higher degrees can be represented by longer exact extensions, or computed from a projective resolution.

To add two extensions of $M$ by $N$, take their direct sum, form a pullback in a category along the diagonal $M\to M\oplus M$, and form a pushout in a category along addition $N\oplus N\to N$. This produces another extension of $M$ by $N$ and defines addition in ${\mathrm{Ext}}_R^1(M,N)$. In a quiver representation, vertexwise splittings express the two extensions by off-diagonal arrow matrices, and the Baer sum adds those matrices modulo the extension complex of quiver representations coboundaries.

Over a Noetherian ring, a finitely generated module $N$ has a free resolution with every term finitely generated. Hom from a finite free module commutes with localization, and exactness of localization commutes with cohomology. Thus the displayed isomorphism holds for every $i\ge 0$ and arbitrary $M$. The finiteness condition is on the first argument; the second need not be finitely generated.

When $X$ acts as zero, resolve the first module with the Koszul resolution on $(p,X)$. Writing $N=\mathbb{Z}/q$, the resulting groups are $\ker (p:N\to N)$ in degree zero, $(N/pN)\oplus \ker p$ in degree one, and $N/pN$ in degree two, with zero higher groups. This works for composite as well as prime moduli.

For an ideal generated by a finite regular sequence in a commutative ring, the self-Ext functor algebra is exterior on the degree-one dual generators. Contraction chain maps on the Koszul resolution represent those generators, square to zero, and anticommute. The exterior square-zero relations also apply in characteristic two.

Partition the isomorphism classes of simple modules into two sets ${\mathcal{C}}_1,{\mathcal{C}}_2$, with ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ whenever $S,T$ lie in different sets. Every module $M$ of finite composition length then has a unique decomposition $M=U_1\oplus U_2$ into its largest submodules whose Jordan–Hölder factors belong to the respective sets.

For a simple $S\in {\mathcal{C}}_1$ and a finite-length $N$ with factors in ${\mathcal{C}}_2$, induction through the long exact sequence of the Ext functor gives ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $M$: take $0\to N\to M\to S\to 0$, split $N=N_1\oplus N_2$, and suppose $S\in {\mathcal{C}}_1$. The quotient extension $0\to N_2\to M/N_1\to S\to 0$ splits. The inverse image of its $S$-summand complements $N_2$ in $M$ and has only factors in ${\mathcal{C}}_1$. Finally, Hom between modules with factors in different sets is zero, since any nonzero image would have a factor in both sets. Projecting any proposed submodule to the opposite summand therefore proves maximality and uniqueness.