research31 min studyLesson 545 in Mathematics

Tensor products and Tor functors

Measure lost tensor exactness using resolutions, torsion kernels, and balanced arguments.

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What you’ll understand

  • Compute Tor using a right-module projective resolution and a left tensor factor.
  • Identify the kernel responsible for the failure of an injection after tensoring.
  • Distinguish objectwise vanishing from the universal flatness criterion.

Tensor product combines generators subject to bilinear relations, so it preserves quotients and is right exact. An injection can become noninjective because the new tensor relations identify elements that were previously distinct. Tor records the higher information lost by taking the tensor product alone. Its definition uses a resolution, but its interpretation is concrete: cycles are relations that become zero after tensoring, and boundaries record the relations already explained by a preceding stage of the resolution.

Respect the sides of the ring action

For a possibly noncommutative ring, take a right module MM and a left module NN. Resolve MM by projective right modules, tensor the chain complex with NN, and take homology. The output is an abelian group unless additional compatible actions are supplied. For a commutative ring, both factors can be treated as modules over the same ring and Tor naturally carries that action. Writing a tensor product of two arbitrary left modules over a noncommutative ring is not justified by notation alone.

The zeroth homology is the ordinary tensor product, while the positive homology groups are the higher Tor groups. Comparison maps and their homotopies make the construction independent of the resolution. One can instead resolve the left factor by projective left modules: the balanced Tor theorem identifies the results. This does not mean tensor product has become exact in both arguments. It means either suitably chosen resolution recovers the same derived failure of exactness.

Tor⁡iR(M,N)=Hi(P∙⊗RN),Tor⁡0R(M,N)≅M⊗RN\operatorname{Tor}_i^R(M,N)=H_i(P_\bullet\otimes_R N),\qquad \operatorname{Tor}_0^R(M,N)\cong M\otimes_R N
Here P is a projective right-module resolution of M; the index is homological and nonnegative.↔ Scroll wide equations horizontally.

A kernel repairs the missing injection

A short exact sequence in either compatible module argument produces a long exact Tor sequence. Its low-degree end contains a map from a first Tor group into the tensor product of the submodule, followed by the map that may have lost injectivity. Exactness identifies the image of that connecting map with the kernel of the failed injection. The connecting map need not itself be injective, so Tor should not automatically be identified with that kernel unless the preceding term is known to vanish.

Over a commutative ring, a module is flat precisely when its first Tor with every module vanishes; equivalently all its positive Tor groups vanish against every module. Vanishing against one test module proves much less. For instance, tensoring with the ring itself creates no positive Tor for any module, including nonflat ones. Universal criteria contain a universal quantifier, and an example involving only a single quotient must be interpreted as a computation rather than a complete certificate of flatness.

Tor⁡1R(M,B)⟶Tor⁡1R(M,C)⟶M⊗RA⟶M⊗RB⟶M⊗RC⟶0\operatorname{Tor}_1^R(M,B)\longrightarrow\operatorname{Tor}_1^R(M,C)\longrightarrow M\otimes_R A\longrightarrow M\otimes_R B\longrightarrow M\otimes_R C\longrightarrow0
This segment comes from a short exact sequence of left modules 0 to A to B to C to 0.↔ Scroll wide equations horizontally.

Finite and infinite resolution behavior

For an integer quotient, the length-one free resolution reduces every Tor calculation to a kernel and a cokernel of one multiplication map. It proves that higher groups vanish without requiring a separate calculation in each degree. Over the dual-number ring, the periodic resolution behaves differently: tensoring with the residue field kills every differential because the nilpotent element acts as zero. Homology then survives in every nonnegative degree, despite the original augmented resolution being exact.

That contrast is an early diagnostic of singular behavior. It also illustrates why counting generators of the original module does not predict resolution length. The construction of a resolution and the calculation after tensoring are distinct stages: exactness before tensoring is a prerequisite, while failure afterward is the intended output. A finite drawing of the first few periodic terms suggests a pattern, but the proof that the pattern continues comes from the uniform kernel-image calculation in every degree.

A cyclic Tor calculation

Compute all Tor⁡iZ(Z/6Z,Z/15Z)\operatorname{Tor}_i^{\mathbb Z}(\mathbb Z/6\mathbb Z,\mathbb Z/15\mathbb Z) using the length-one free resolution of the first factor. Identify actual kernel elements, not only their number.

  1. Tensoring Z→6Z\mathbb Z\xrightarrow{6}\mathbb Z with the second factor yields multiplication by six on Z/15Z\mathbb Z/15\mathbb Z in chain degrees one and zero. There are no higher nonzero chain terms.

  2. The kernel consists of residues rr with 1515 dividing 6r6r. Equivalently, five divides rr, so it is the subgroup {0,5,10}, cyclic of order three. This is the first Tor group.

  3. The image is the subgroup of multiples of three, containing five residues. Its cokernel is cyclic of order three, which is the tensor product. All Tor groups in degrees at least two vanish because the chosen free resolution has length one.

The result

Both degree zero and degree one give Z/3Z\mathbb Z/3\mathbb Z, while every higher group is zero. The two copies arise as a cokernel and a kernel respectively, rather than as the same subset.

Explore the model

Interactive lab

Interactive lab · modular

Enumerate the image in the Tor calculation

Pause/Reset, then set both modulus to 15 and additive step to 6. Predict period five and image {0,3,6,9,12}; use five Step presses to enumerate the orbit. Compare its five image elements with the three kernel elements computed separately. Reset and set both modulus to 15 and step to 4; predict a bijective multiplication map and zero kernel.

Enumerate the image in the Tor calculationAddition 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 forThe orbit gives the image size for cyclic multiplication, allowing a kernel-size check; Tor still requires identifying this map inside a verified resolution.

Check your understanding

For a commutative ring R, which statement correctly characterizes flatness of M?

Try it yourself

For R=k[ε]/(ε2)R=k[\varepsilon]/(\varepsilon^2) and k=R/(ε)k=R/(\varepsilon), tensor the periodic resolution from the previous lesson with kk. Compute every Tor group and explain why the answer rules out a finite projective resolution of kk.

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

Give me a hint

Determine the action of the nilpotent element on each tensor term before taking homology.

Compare with the solution

Each tensor term is R⊗Rk≅kR\otimes_R k\cong k, and every differential becomes multiplication by the zero action of ε\varepsilon, hence is zero. Therefore Tor⁡iR(k,k)≅k\operatorname{Tor}_i^R(k,k)\cong k for every i≥0i\geq0. A projective resolution of finite length would compute zero Tor beyond that length, contradicting these nonzero groups. Thus the displayed periodic resolution cannot be replaced by a finite projective resolution, although each individual term has rank one.

Take it with you

  • Specify right and left module actions before tensoring.
  • Tor measures the failure of exactness after a valid resolution is tensored.
  • Flatness requires vanishing against every compatible module, not a single successful test.

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