Exploration 2: Module actions
January 2, 2024.Questions:
- How can we represent algebraic objects as actions?
Recall that -matrices are homomorphisms between free modules . How do we visualize homomorphisms between non-free modules ?
Let’s first focus on endomorphisms. Every endomorphism must satisfy the homomorphism laws for -modules :
- for all
- for all
In other words, is -linear.
-linearity allows us to define module actions. Here, since is -linear, respects addition and scalar multiplication and therefore can be described as acting on . We can express this by extending the -module to a -module, which is an -module together with an action represented by the indeterminate . When the action is scalar multiplication, a -module is just a module over the polynomial ring .
Instead of scalar multiplication by , define to be an application of , and define the action of constant polynomials to be scalar multiplication by . Thus each endomorphism is associated with an -module whose action is to apply some linear combination of , represented by a polynomial in .
By this construction, every endomorphism is associated with a unique -module, and therefore studying these -modules is enough to classify the endomorphisms.
In this section, we explore properties of -modules when is a PID.
When is a PID, it’s equivalent to saying that is a field . So we’ll refer to as in this section.
Importantly, it turns out -modules are torsion, i.e. it has a nontrivial annihilator .
- Since acts as , we can construct a ring homomorphism that maps linear combinations of () to their corresponding endomorphism ().
- The kernel of represents all polynomials that map to the zero map in . Thus , and it is enough to show that has a non-trivial kernel.
- Given that is finitely generated, and that polynomial rings are always infinitely generated (by , is a mapping from an infinitely generated set to a finitely generated set, thus cannot be injective.
- Since injectivity corresponds with having a trivial kernel, has a non-trivial kernel.
By definition, a torsion module is one with a non-trivial annihilator, which we proved true earlier.
Note that this result can be obtained via the Chinese remainder theorem which says when coprime. Observe that two different linear factors will always be coprime in , thus every can be split completely into linear factors and
If is torsion, then it has a nonzero element such that for all . Thus all elements of can be zeroed by , and this doesn’t change when you take any submodule of .
Because we’re working over a PID, we can apply the structure theorem to decompose such a -module into a direct sum of free submodules and cyclic torsion submodules (for an irreducible polynomial ).
But , being torsion, has no nontrivial free submodules, so there is no free part:
In summary, we’ve reduced the problem of classifying endomorphisms to studying the torsion cyclic -submodules of their corresponding -module.
In this section, we derive the notion of eigenvalues in module theory.
Given a -vector space , consider the endomorphism . As before, define a -module as where the action of is to apply . Since is a torsion module over a PID, apply the structure theorem to break it into a direct sum of torsion cyclic modules where . Note that each generates the annihilator for their corresponding cyclic module . In this context where the module is defined over a polynomial ring , we say that is the minimal polynomial for if it is the lowest degree polynomial that annihilates .
Every element in the annihilator of annihilates by definition. Since is the generator of the annihilator of , it must divide every element of this annihilator, and therefore is the least-degree element in the annihilator. Thus is the lowest degree polynomial that annihilates .
We know that the annihilator of is the kernel of the homomorphism . Since the kernel is an ideal of , a PID, the kernel must be generated by a single element . Since is a field, we can choose to be monic, since if it’s not monic we can multiply by the inverse of its leading coefficient to get a monic . This makes the lowest degree monic polynomial in the annihilator of .
In -module , an eigenvalue of an -module endomorphism is one where there exists a nonzero element such that .
An endomorphism is nilpotent if some power of is equal to the zero map.
- If is nilpotent, then for some .
- If , then we have implying , which directly shows that is an eigenvalue.
- Otherwise if , being nilpotent means . So there is some element such that . But then we have Since there exists an element where , this proves thus is an eigenvalue of .
A counterexample for the converse is the identity , which has eigenvalue but is not nilpotent.
It is enough to show that the eigenvalues of are eigenvalues of , since the case has an identical proof. If has an eigenvalue , then for some . Thus in , we have proving that the eigenvalue of is an eigenvalue of .
- If is a root of , then by the factor theorem, is an irreducible factor of . Note that in a UFD like , irreducibles and primes are the same thing.
- Since the annihilator is generated by a prime power, and we know that the prime is , the annihilator is exactly for some . So we’re working with .
- Recall that polynomials in act as endomorphisms on . The endomorphism corresponding to this polynomial should look like , since we defined the action of to treat constant polynomials like as scalar multiplication.
- Since is a root, we have — that is, is nilpotent on the given component , implying that is an eigenvalue of when restricted to the component .
- But since is a direct sum including , and therefore is also an eigenvalue of .
If the decomposition is , then since the roots of each are eigenvalues of , the minimal polynomial must contain a factor for each eigenvalue for each component. This means the minimal polynomial is the LCM of all . Since the decomposition has , we can observe that is a multiple of every , making the minimal polynomial of .
Since the roots of for each correspond to the eigenvalues of (the action of ), it is useful to encapsulate all of the eigenvalues (including repeats) as the characteristic polynomial : a polynomial with a root every time appears as an eigenvalue of . Therefore:
This follows directly from knowing that the roots of each are eigenvalues of . Then the product of each captures all eigenvalues of .
Corollary: The characteristic polynomial of an endomorphism for a finitely generated -module is the product of the minimal polynomials of its cyclic components.
There is another way to compute this characteristic polynomial:
- Since is finitely generated, we can construct its presentation matrix so that . The fact that is equivalent to a quotient of a free module means all endomorphisms of (e.g. ) can be expressed as a -matrix modulo the quotient .
- Now consider the equation , where we treat as a scalar. If this is true for some nonzero , then is an eigenvalue of by definition.
- Rearranging the above to reduces the problem of finding eigenvalues of to finding values of such that zeroes some nonzero .
- Recall that we can define a Smith normal form for every matrix over a Bézout domain , such as . This results in a diagonal matrix where are the invariant factors of (which may differ from the invariant factors of .)
- Thus our equation becomes , which we can simplify to . Since is invertible, is also an arbitrary nonzero element of , so WLOG we may write . Because is diagonal, we may rewrite this as a system of equations: where the are the components of . Note that a nonzero only requires at least one of the to be nonzero. That means we can assume for all but one of the equations.
- When is nonzero, then for the equation to be true, must be equal to zero. This is because , being an integral domain, has no zero divisors. In summary, if we can find values for that make any zero (i.e. the roots of ), then is true for some nonzero , so is true. So the roots of are eigenvalues of .
- This means one can obtain every eigenvalue of by finding the roots of the product . So the characteristic polynomial is , which is exactly the determinant of .
The companion matrix of a polynomial is a matrix constructed so that its characteristic polynomial is exactly . Given a polynomial , its companion matrix is
- For the forward direction: since the companion matrix for every polynomial (of degree ) has an eigenvalue, every polynomial (of degree ) has a root in , therefore is algebraically closed.
- The backward direction is because in an algebraically closed field, all degree polynomials in have a root in , including , whose roots correspond to eigenvalues of .
- Let by the structure theorem. Since the minimal polynomial is exactly and the characteristic polynomial is exactly , they can only coincide when there is only one invariant factor. But that indicates that the decomposition includes only one cyclic submodule , meaning the original module must be cyclic as well.
- The proof in the other direction is trivial: if is cyclic , then both and are equal to .
If a matrix has eigenvalue , then the kernel of is its corresponding eigenspace. If the image of is equal to one of its eigenspaces, the map is a multiplication by a scalar .
In this section, we display the module action of polynomial rings.
We can express such a linear transformation as the action of a polynomial ring on the -module.
Let’s explore the -modules, where is a polynomial ring defined over a field .
For instance, take the -module . Since it’s quotiented by a degree polynomial , the obvious basis is . We use the fact that in the quotient to get the constraint . This means the -matrix (with respect to the basis ) is one where , i.e. we have which corresponds to the presentation matrix
So we have found for the -module .
From the structure theorem, we know that direct sums of -modules describe all -modules. Let’s do this again where is a direct sum.
Take the -module . The obvious basis is . We do the same thing as before, realizing on the left and on the right, and obtaining the constraints on the left and on the right. We have: corresponding to the presentation matrix where blank entries are zeroes.
So we’ve described how to find a presentation matrix for any -module. We could reduce this matrix using the operations we can do on presentation matrices, but there is a kind of standard form for a presentation matrix in the case of -presentation matrices.
The Chinese remainder theorem says when coprime. Consider . Since distinct linear factors are always coprime, we can always split to get a bunch of powers of linear factors, one for each repeated root. For instance, .
For an easy example, consider the -module . To find , first find some basis for (as a vector space). A clever basis is . This conveniently gives us the constraints Converting this into terms of basis vectors, this is which corresponds to the presentation matrix
Such a matrix is known as a Jordan matrix with eigenvalue with multiplicity . It has along the diagonal and ones on the superdiagonal, and this is precisely because we chose the basis in such a way that the coefficients in each constraint are and . Next, here’s a direct sum example. Consider the -module . We’ll end up with the following presentation matrix, which is composed of matrices that are much alike the previous example:
This is a direct sum of Jordan matrices! We call this the Jordan normal form of the presentation matrix, which always exists for -matrices over a PID . This is the standard form for a presentation matrix for such -modules.
In general, given a -module over a PID , you may:
- Express the -module as a direct sum of cyclic modules where is the minimal polynomial of that cyclic component.
- Split each into a direct sum of powers of linear factors (like ) via factoring the , which gives you the eigenvalues and their multiplicities.
- Construct a direct sum of presentation matrices, which is in Jordan normal form.
All this to find the -matrix that defines scalar multiplication by in a -module.
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