research31 min studyLesson 546 in Mathematics

Hom and Ext functors

Compute derived Hom with the correct variance and interpret first Ext through extensions.

On this page

What you’ll understand

  • Compute Ext by resolving the first argument projectively or the second injectively.
  • Track contravariance and covariance separately.
  • Relate a nonzero first Ext group to nonsplit short exact sequences.

Hom records maps, but a map into a quotient need not lift and a map out of a submodule need not extend. Ext measures these missing pieces of Hom exactness. Its two arguments behave differently: the first reverses arrows and the second preserves them. We use left modules throughout this lesson, so Hom is well defined even for a noncommutative ring; without commutativity or extra bimodule data, its natural output is an abelian group.

Two compatible routes to the same invariant

Resolve the first module MM by projective left modules in nonnegative chain degrees. Applying Hom⁡R(−,N)\operatorname{Hom}_R(-,N) reverses the arrows, producing a cochain complex with degree-ii term Hom⁡R(Pi,N)\operatorname{Hom}_R(P_i,N) and differential given by precomposition with the resolution differential. Its cohomology defines Ext. Degree zero recovers ordinary Hom because a map on P0P_0 descends to MM exactly when it vanishes on the image of P1P_1.

Alternatively, resolve NN by injectives in nonnegative cochain degrees and apply Hom⁡R(M,−)\operatorname{Hom}_R(M,-). The resulting cohomology groups are naturally isomorphic to those obtained from the projective calculation. Module categories have both types of resolutions. In a more general abelian category one must specify enough projectives or enough injectives before using the respective construction. The two computations are compatible methods, not a permission to replace a projective resolution with an arbitrary exact complex.

Ext⁡Ri(M,N)=Hi ⁣(Hom⁡R(P∙,N))≅Hi ⁣(Hom⁡R(M,I∙))\operatorname{Ext}_R^i(M,N)=H^i\!\left(\operatorname{Hom}_R(P_\bullet,N)\right)\cong H^i\!\left(\operatorname{Hom}_R(M,I^\bullet)\right)
P resolves M projectively and I resolves N injectively; both formulas use cohomological degree i.↔ Scroll wide equations horizontally.

Variance controls the long exact sequence

A map M′→MM'\to M induces Ext maps from the groups with first argument MM to those with first argument M′M'. A map N→N′N\to N' induces maps in the same direction in the second argument. The comparison theorem makes these maps independent of chosen lifts. Ignoring this reversal can produce a plausible diagram with every first-argument arrow pointing incorrectly, so label the variable being changed before writing any derived map.

For a short exact sequence in the second argument, the first connecting map starts at a Hom group into the quotient and lands in a first Ext group into the submodule. It measures failure to lift that map. With the first argument varying, the Hom arrows reverse, and the corresponding connecting map measures failure to extend. If MM is projective, all its positive Ext groups vanish against every NN. Dually, an injective second argument makes all positive Ext vanish against every first argument.

Hom⁡R(M,B)⟶Hom⁡R(M,C)⟶Ext⁡R1(M,A)⟶Ext⁡R1(M,B)\operatorname{Hom}_R(M,B)\longrightarrow\operatorname{Hom}_R(M,C)\longrightarrow\operatorname{Ext}_R^1(M,A)\longrightarrow\operatorname{Ext}_R^1(M,B)
This is the lifting-obstruction segment associated to 0 to A to B to C to 0 in the second argument.↔ Scroll wide equations horizontally.

First Ext classifies extensions

The group Ext⁡R1(M,N)\operatorname{Ext}_R^1(M,N) also classifies short exact sequences with submodule NN and quotient MM, up to isomorphisms inducing the identities on those endpoints. Its zero class is represented by the split sequence. The group operation is the Baer sum, not the direct sum of middle modules alone. Two abstractly isomorphic middle modules can still represent different extension data when their specified inclusion and quotient maps are retained.

For an integer quotient in the first argument, applying Hom to its length-one free resolution yields multiplication by that integer on the second argument. Hom is its kernel and first Ext its cokernel; higher Ext vanishes. This resembles the earlier Tor calculation, but the meanings and variance differ. In particular, a torsion first argument has no nonzero maps into the integers, yet can have nonzero first Ext into the integers. Confusing Hom with Ext would erase exactly the obstruction the resolution was designed to retain.

Hom and Ext from one multiplication map

Compute Hom⁡Z(Z/8Z,Z/12Z)\operatorname{Hom}_{\mathbb Z}(\mathbb Z/8\mathbb Z,\mathbb Z/12\mathbb Z) and every positive Ext group for the same ordered pair. Identify generators of the kernel and cokernel.

  1. Resolve the first argument by 0→Z→8Z→Z/8Z→00\to\mathbb Z\xrightarrow{8}\mathbb Z\to\mathbb Z/8\mathbb Z\to0. Applying Hom into the second argument gives multiplication by eight on Z/12Z\mathbb Z/12\mathbb Z in cochain degrees zero and one.

  2. The kernel consists of {0,3,6,9}, a cyclic subgroup generated by the residue three. These are exactly the possible images of the residue one under a homomorphism from the integers modulo eight.

  3. The image is {0,4,8}; quotienting the target by this subgroup gives a cyclic cokernel of order four generated by the coset of one. There are no terms above degree one, so all higher Ext groups vanish.

The result

Hom and first Ext are both abstractly Z/4Z\mathbb Z/4\mathbb Z, arising respectively as a kernel and a cokernel. The calculation does not identify their elements or functorial roles.

Explore the model

Interactive lab

Interactive lab · modular

Read the cokernel of an Ext differential

Pause/Reset, then set both modulus to 12 and additive step to 8. Predict the orbit {0,8,4} with period three. Compare the three-element image with the four cosets in its cokernel and the four-element kernel calculated in the example. Reset and set both modulus to 12 and step to 5; predict an invertible map and zero cokernel.

Read the cokernel of an Ext differentialAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOrbit size checks the multiplication differential, while the resolution determines why its cokernel is Ext rather than an unrelated quotient.

Check your understanding

A map u:M′→Mu:M'\to M is given. Which induced map has the correct variance?

Try it yourself

Compute Hom and first Ext from Z/5Z\mathbb Z/5\mathbb Z to Z\mathbb Z. Show that 0→Z→5Z→Z/5Z→00\to\mathbb Z\xrightarrow{5}\mathbb Z\to\mathbb Z/5\mathbb Z\to0 is a nonsplit extension. Compare with first Tor of the same pair.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Apply Hom to the length-one resolution, then separately ask whether the quotient map can have a section.

Compare with the solution

The Hom complex is Z→5Z\mathbb Z\xrightarrow{5}\mathbb Z. Its kernel is zero and its cokernel is Z/5Z\mathbb Z/5\mathbb Z, so Hom vanishes and first Ext does not. The displayed extension is nonsplit because every homomorphism from the finite quotient to the torsion-free middle group is zero, precluding a section. First Tor with the integer second argument is zero because its multiplication-by-five kernel is zero, consistent with flatness of that second argument.

Take it with you

  • Ext reverses its first variable and preserves its second.
  • Projective and injective resolutions compute the same Ext groups.
  • First Ext retains endpoint-labeled extension data and vanishes on split classes.

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