S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 5: Multilinear Algebra

20 Determinants, and Modules over a Ring

· Denis Auroux

Original Professor Auroux’s handwritten notes (PDF), behind a password for now:

Recall

  • pure tensors , and
  • , : the same for symmetric, resp. alternating, multilinear maps. Skew-symmetrization: This is zero whenever for some , and so, by multilinearity, whenever are linearly dependent. Thus whenever !

Volume and determinant

If , then (if is a basis of , it is spanned by ).

A choice of isomorphism is determined by a volume form , , i.e. a nondegenerate alternating multilinear map (Think of it as: the signed volume of the parallelepiped with edge vectors is naturally , which becomes a scalar given .)

Example 20.1. In a real inner product space with orthonormal basis , the natural volume form is , so (the volume of the unit cube is ). (Warning: reordering the basis gives ... orientation!)

Using the basis to identify , with for each : the determinant of an matrix!

Recall that the determinant of a matrix is is the only quantity which is

  • multilinear in the columns of the matrix,
  • alternating (swapping two columns gives ),
  • .

Even though the notion of determinant/volume of vectors requires a choice of volume form (an isomorphism ), the notion of determinant of a linear operator requires no such choice!

  • Usual definition: given , define , in any basis, using (the usual proof is painfully explicit), so under change of basis .
  • Better definition: the exterior power is a functor, so induces a linear operator (explicitly, ). But , and any linear operator on a 1-dimensional vector space is a scalar multiple of . So define such that (This expresses the fact that scales the volume of parallelepipeds in by a factor of , without having to choose to measure those volumes.)

Using this definition of the determinant via , independence of the choice of basis is manifest, and so is the fact that !

Linear algebra over rings: modules

(Artin §14.1–14.2.)

Let be a commutative ring (with ), i.e. relax the field axioms to not require multiplicative inverses. Main examples: , , , .

Definition 20.2 (Module). A module over a ring is a set with two operations:

  • , addition, such that is an abelian group;
  • , scalar multiplication, such that , , , , .

Example 20.3.

  • with componentwise operations is the free module of rank over .
  • Any abelian group is a -module (). Check the details.

Definition 20.4 (Spanning sets, independence, bases).

  • spans (or is a generating set) if every element of is a (finite) linear combination , , . Equivalently: the map , , is surjective. is finitely generated if it has a finite spanning set.
  • The elements of are (linearly) independent if is injective, i.e. , , implies for all .
  • The elements of form a basis if is an isomorphism. In this case, say is a free module.

General fact about modules: nothing is true!

  • A basis need not exist!

    Example: as a -module: for all , so can’t be injective!

  • Even if is free (admits a basis):

    • a linearly independent set may not be a subset of a basis.

      Example: as a -module: there is no basis containing .

    • a spanning set need not contain a subset which is a basis.

      Example: as a -module: spans (since ), but aren’t independent (), and neither subset nor spans all of .

  • A submodule of a finitely generated module need not be finitely generated.

    Example: , polynomials in infinitely many variables. as an -module is generated by the element . is a submodule, but not finitely generated (any finite subset only involves finitely many ’s, and can’t span the other ’s).

    (By contrast, this holds for modules over Noetherian rings, including , and many others.)

Definition 20.5 (Module homomorphisms). For modules over , a module homomorphism is a map such that and .

Observe: is itself an -module: , .

For free modules, things work as expected: ( is determined by the images of the basis vectors of ). But we can have nonzero modules such that !

Example 20.6. , with multiplication . Then , because satisfies , so it must map to such that , which implies .

Remark.

  • is a module over itself (the free module of rank 1). A submodule of is called an ideal: this is a subset such that

    • is an abelian subgroup of ;
    • : multiplication by any element of takes to itself.

    Example: the ideals in are , and the ideals in are , i.e. generated by a single element. This is very special. ( and are “principal ideal domains”. This has to do with Euclidean division algorithms: .)

  • The quotient of an -module by a submodule is an -module.

    Example: as a -module; as a -module (the example above).

    (The quotient of itself by a submodule, i.e. an ideal, is in fact not just an -module but also a ring in its own right.)

Finitely generated abelian groups

The study of modules is a vast subject, which we won’t study further, with one exception: we’re returning to group theory, but we start with a short account of the classification of finitely generated abelian groups (-modules).

Notes not reached in this lecture, covered at the start of Lecture 21

Theorem 20.7. Any finitely generated abelian group is isomorphic to a product of cyclic groups

(And using iff , we can rearrange the finite factors, e.g. to arrange that all are powers of primes.)

The proof involves the study of “linear algebra over ”! One of the ingredients is:

Proposition 20.8. If is a finitely generated -module, then there exist and such that . (Equivalently: there is an exact sequence .)

(The quotient by a -submodule is the same as the quotient of an abelian group by a subgroup.)

This relies on:

Lemma 20.9. Any submodule ( subgroup) of is finitely generated (in fact, free of rank ).

Proof. By induction on . True for : the subgroups of are and , .

Assume the result holds for , and consider a submodule . The map restricts to a homomorphism , where

  • is a submodule of , hence finitely generated (free) by induction;
  • is a subgroup of , hence free (of rank 0 or 1).

And if and are finitely generated (resp. free), then so is ! The proof is just as in midterm problem 4: let be generators (resp. a basis) of , and generators of . Then for all there exist such that , so , so hence : the generate. (Basis: left as an exercise; we won’t need it anyway.) ∎

Proof of the proposition. If is finitely generated, with generators , then is surjective, and is a subgroup/submodule of , hence finitely generated by the lemma. Let be generators of ; then and now we have an exact sequence , with , inducing an isomorphism . ∎