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Exploration 2: Module actions

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Questions:


Recall that RR-matrices are homomorphisms between free modules RnR^n. How do we visualize homomorphisms between non-free modules MM?

Let’s first focus on endomorphisms. Every endomorphism T:MMT:M\to M must satisfy the homomorphism laws for RR-modules MM:

In other words, TT is RR-linear.

RR-linearity allows us to define module actions. Here, since TT is RR-linear, TT respects addition and scalar multiplication and therefore can be described as acting on MM. We can express this by extending the RR-module MM to a R[t]R[t]-module, which is an RR-module together with an action represented by the indeterminate tt. When the action is scalar multiplication, a R[t]R[t]-module is just a module over the polynomial ring R[t]R[t].

Instead of scalar multiplication by tt, define tt to be an application of TT, and define the action of constant polynomials rRr\in R to be scalar multiplication by rr. Thus each endomorphism T:MMT:M\to M is associated with an R[t]R[t]-module MM' whose action is to apply some linear combination of TT, represented by a polynomial in R[t]R[t].

By this construction, every endomorphism is associated with a unique R[t]R[t]-module, and therefore studying these R[t]R[t]-modules is enough to classify the endomorphisms.

In this section, we explore properties of R[t]R[t]-modules when R[t]R[t] is a PID.

When R[t]R[t] is a PID, it’s equivalent to saying that RR is a field FF. So we’ll refer to R[t]R[t] as F[t]F[t] in this section.

Importantly, it turns out F[t]F[t]-modules are torsion, i.e. it has a nontrivial annihilator AnnF[t](M)\Ann_{F[t]}(M).

Theorem: Given a F[t]F[t]-module MM where tt acts as an endomorphism MMM\to M, there is some polynomial fF[t]f\in F[t] whose corresponding endomorphism End(M)\in\End(M) is the zero map. In other words, AnnF[t](M)\Ann_{F[t]}(M) is nontrivial.
  • Since tt acts as TEnd(M)T\in\End(M), we can construct a ring homomorphism φ:F[t]End(M)\varphi:F[t]\to\End(M) that maps linear combinations of tt (iaiti\sum_i a_it^i) to their corresponding endomorphism (iaiTi\sum_i a_iT^i).
  • The kernel of φ\varphi represents all polynomials F[t]\in F[t] that map to the zero map in End(M)\End(M). Thus AnnF[t](M)=ker φ\Ann_{F[t]}(M)=\ker\varphi, and it is enough to show that φ\varphi has a non-trivial kernel.
  • Given that End(M)\End(M) is finitely generated, and that polynomial rings F[t]F[t] are always infinitely generated (by {1,t,t2,}\{1,t,t^2,\ldots\}, φ\varphi 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, φ\varphi has a non-trivial kernel.

Corollary: Every finitely generated F[t]F[t]-module MM (where tt acts as an endomorphism MMM\to M) is torsion.

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 F[t]/(fg)F[t]/(f)F[t]/(g)F[t]/(f\cdot g)\iso F[t]/(f)\oplus F[t]/(g) when f,gf,g coprime. Observe that two different linear factors tαt-\alpha will always be coprime in F[t]F[t], thus every fF[t]f\in F[t] can be split completely into linear factors and

Theorem: Torsion RR-modules MM have no nontrivial free submodules.

If MM is torsion, then it has a nonzero element aRa\in R such that am=0am=0 for all mMm\in M. Thus all elements of MM can be zeroed by aa, and this doesn’t change when you take any submodule of MM.

Because we’re working over a PID, we can apply the structure theorem to decompose such a F[t]F[t]-module MM into a direct sum of free submodules F[t]\iso F[t] and cyclic torsion submodules F[t]/(fk)\iso F[t]/(f^k) (for an irreducible polynomial ff).

M(F[t])nfreeF[t]/(f1k1)F[t]/(f2k2)F[t]/(fnkn)torsionM\iso\underbrace{\left(F[t]\right)^n}_{\text{free}}\oplus\underbrace{F[t]/(f_1^{k_1})\oplus F[t]/(f_2^{k_2})\oplus\ldots\oplus F[t]/(f_n^{k_n})}_{\text{torsion}}

But MM, being torsion, has no nontrivial free submodules, so there is no free part:

MF[t]/(f1k1)F[t]/(f2k2)F[t]/(fnkn)M\iso F[t]/(f_1^{k_1})\oplus F[t]/(f_2^{k_2})\oplus\ldots\oplus F[t]/(f_n^{k_n})

In summary, we’ve reduced the problem of classifying endomorphisms to studying the torsion cyclic F[t]F[t]-submodules F[t]/(fk)F[t]/(f^k) of their corresponding F[t]F[t]-module.

In this section, we derive the notion of eigenvalues in module theory.

Given a FF-vector space VV, consider the endomorphism T:VVT:V\to V. As before, define a F[t]F[t]-module VTV_T as VV where the action of tt is to apply TT. Since VTV_T is a torsion module over a PID, apply the structure theorem to break it into a direct sum of torsion cyclic modules iF[t]/(fi)\bigoplus_i F[t]/(f_i) where f1f2fnf_1\mid f_2\mid\ldots\mid f_n. Note that each fif_i generates the annihilator for their corresponding cyclic module F[t]/(fi)F[t]/(f_i). In this context where the module MM is defined over a polynomial ring F[t]F[t], we say that fif_i is the minimal polynomial for MM if it is the lowest degree polynomial that annihilates MM.

Theorem: ff is the minimal polynomial for a cyclic module F[t]/(f)F[t]/(f).

Every element in the annihilator of F[t]/(f)F[t]/(f) annihilates F[t]/(f)F[t]/(f) by definition. Since ff is the generator of the annihilator of F[t]/(f)F[t]/(f), it must divide every element of this annihilator, and therefore is the least-degree element in the annihilator. Thus ff is the lowest degree polynomial that annihilates F[t]/(f)F[t]/(f).

Theorem: Every F[t]F[t]-module MM (where tt acts as an endomorphism MMM\to M) has a minimal polynomial fF[t]f\in F[t] that generates the annihilator of MM.

We know that the annihilator of MM is the kernel of the homomorphism φ:F[t]End(M)\varphi:F[t]\to\End(M). Since the kernel is an ideal of F[t]F[t], a PID, the kernel must be generated by a single element fF[t]f\in F[t]. Since FF is a field, we can choose ff to be monic, since if it’s not monic we can multiply ff by the inverse of its leading coefficient to get a monic ff. This makes ff the lowest degree monic polynomial in the annihilator of MM.

In RR-module MM, an eigenvalue λ\lambda of an RR-module endomorphism T:MMT:M\to M is one where there exists a nonzero element mMm\in M such that Tm=λmTm=\lambda m.

An endomorphism T:MMT:M\to M is nilpotent if some power of TT is equal to the zero map.

Lemma: An RR-module endomorphism T:MMT:M\to M has an eigenvalue λ\lambda if TλIT-\lambda I is nilpotent.
  • If TλIT-\lambda I is nilpotent, then (TλI)k=0(T-\lambda I)^k=0 for some k1k\ge 1.
  • If k=1k=1, then we have TλI=0T-\lambda I=0 implying T=λIT=\lambda I, which directly shows that λ\lambda is an eigenvalue.
  • Otherwise if k2k\ge 2, being nilpotent means (TλI)k1char "338=0(T-\lambda I)^{k-1}\ne 0. So there is some element mMm\in M such that n=(TλI)k1mchar "338=0n=(T-\lambda I)^{k-1}m\ne 0. But then we have (TλI)n=(TλI)[(TλI)k1m]=(TλI)km=0(T-\lambda I)n=(T-\lambda I)\left[(T-\lambda I)^{k-1}m\right]=(T-\lambda I)^km=0 Since there exists an element nMn\in M where (TλI)n=0(T-\lambda I)n=0, this proves Tn=λnTn=\lambda n thus λ\lambda is an eigenvalue of TT.

A counterexample for the converse is the identity II, which has eigenvalue λ=1\lambda=1 but is not nilpotent.

Lemma: The eigenvalues of an endomorphism of a direct sum T:MNMNT:M\oplus N\to M\oplus N are exactly the eigenvalues of TMT_M and TNT_N combined (where TMT_M denotes TT restricted to MM.)

It is enough to show that the eigenvalues of TMT_M are eigenvalues of TT, since the NN case has an identical proof. If TMT_M has an eigenvalue λ\lambda, then TMm=λmT_Mm=\lambda m for some mMm\in M. Thus in MNM\oplus N, we have T(m,0)=(TMm,TN0)=(λm,0)=λ(m,0)T(m,0)=(T_Mm,T_N0)=(\lambda m,0)=\lambda(m,0) proving that the eigenvalue λ\lambda of MM is an eigenvalue of MNM\oplus N.

Theorem: For a torsion component F[t]/(f)F[t]/(f) of a F[t]F[t]-module MM, the roots λ\lambda of its minimal polynomial ff correspond to eigenvalues of TT (the action of tt).
  • If λ\lambda is a root of ff, then by the factor theorem, (tλ)(t-\lambda) is an irreducible factor of ff. Note that in a UFD like F[t]F[t], irreducibles and primes are the same thing.
  • Since the annihilator is generated by a prime power, and we know that the prime is (tλ)(t-\lambda), the annihilator is exactly (tλ)k(t-\lambda)^k for some k1k\ge 1. So we’re working with F[t]/(tλ)kF[t]/(t-\lambda)^k.
  • Recall that polynomials in F[t]F[t] act as endomorphisms on MM. The endomorphism corresponding to this polynomial (tλ)k(t-\lambda)^k should look like (TλI)k(T-\lambda I)^k, since we defined the action of F[t]F[t] to treat constant polynomials like λ\lambda as scalar multiplication.
  • Since λ\lambda is a root, we have (TλI)k=0(T-\lambda I)^k=0 — that is, TλIT-\lambda I is nilpotent on the given component F[t]/(tλ)kF[t]/(t-\lambda)^k, implying that λ\lambda is an eigenvalue of TT when restricted to the component F[t]/(tλ)kF[t]/(t-\lambda)^k.
  • But since MM is a direct sum including F[t]/(tλ)kF[t]/(t-\lambda)^k, and therefore λ\lambda is also an eigenvalue of TT.

Theorem: The minimal polynomial for a F[t]F[t]-module MM is exactly the largest invariant factor fnf_n of MM.

If the decomposition is MiF[t]/(fi)M\iso\bigoplus_iF[t]/(f_i), then since the roots of each fif_i are eigenvalues of MM, the minimal polynomial must contain a factor (tλ)(t-\lambda) for each eigenvalue λ\lambda for each component. This means the minimal polynomial is the LCM of all fif_i. Since the decomposition has f1f2fnf_1\mid f_2\mid\ldots\mid f_n, we can observe that fnf_n is a multiple of every fif_i, making fnf_n the minimal polynomial of MM.

Since the roots of ff for each F[t]/(f)F[t]/(f) correspond to the eigenvalues of T:MMT:M\to M (the action of tt), it is useful to encapsulate all of the eigenvalues (including repeats) as the characteristic polynomial χTF[t]\chi_T\in F[t]: a polynomial with a root λ\lambda every time λ\lambda appears as an eigenvalue of TT. Therefore:

Theorem: The characteristic polynomial for a F[t]F[t]-module MM is exactly the product of all invariant factors fif_i of MM.

This follows directly from knowing that the roots of each fif_i are eigenvalues of MM. Then the product of each fif_i captures all eigenvalues of MM.

Corollary: The characteristic polynomial of an endomorphism T:MMT:M\to M for a finitely generated F[t]F[t]-module MM is the product of the minimal polynomials fif_i of its cyclic components.

There is another way to compute this characteristic polynomial:

Theorem: Over a finitely generated F[t]F[t]-module MM, the characteristic polynomial of an endomorphism T:MMT:M\to M is exactly det(TtI)\det(T-tI).
  • Since MM is finitely generated, we can construct its presentation matrix AA so that MF[t]n/im AM\iso F[t]^n/\im A. The fact that MM is equivalent to a quotient of a free module F[t]nF[t]^n means all endomorphisms of MM (e.g. TT) can be expressed as a F[t]F[t]-matrix modulo the quotient im A\im A.
  • Now consider the equation Tm=tmTm=tm, where we treat tF[t]t\in F[t] as a scalar. If this is true for some nonzero mMm\in M, then tt is an eigenvalue of TT by definition.
  • Rearranging the above to (TtI)m=0(T-tI)m=0 reduces the problem of finding eigenvalues of TT to finding values of tt such that TtIT-tI zeroes some nonzero mMm\in M.
  • Recall that we can define a Smith normal form PDQ1PDQ^{-1} for every matrix over a Bézout domain F[t]F[t], such as TtIT-tI. This results in a diagonal matrix D=[f1f2fn]D=\left[\begin{matrix}f_1&&&\\&f_2&&\\&&\ddots&\\&&&f_n\end{matrix}\right] where fif_i are the invariant factors of TtIT-tI (which may differ from the invariant factors of MM.)
  • Thus our equation becomes (PDQ1)m=0(PDQ^{-1})m=0, which we can simplify to D(Q1m)=0D(Q^{-1}m)=0. Since QQ is invertible, Q1mQ^{-1}m is also an arbitrary nonzero element of MM, so WLOG we may write Dm=0Dm=0. Because DD is diagonal, we may rewrite this as a system of equations: f1m1=0f2m2=0fnmn=0\begin{aligned} f_1m_1&=0\\ f_2m_2&=0\\ \vdots\\ f_nm_n&=0 \end{aligned} where the mim_i are the components of mm. Note that a nonzero mm only requires at least one of the mim_i to be nonzero. That means we can assume mi=0m_i=0 for all but one of the equations.
  • When mim_i is nonzero, then for the equation fimi=0f_im_i=0 to be true, fif_i must be equal to zero. This is because F[t]F[t], being an integral domain, has no zero divisors. In summary, if we can find values λ\lambda for tt that make any fif_i zero (i.e. the roots of fif_i), then Dm=0Dm=0 is true for some nonzero mMm\in M, so (TλI)m=0(T-\lambda I)m=0 is true. So the roots of fif_i are eigenvalues of TT.
  • This means one can obtain every eigenvalue of TT by finding the roots of the product ifi\prod_i f_i. So the characteristic polynomial χT\chi_T is ifi\prod_i f_i, which is exactly the determinant of TtIT-tI.

The companion matrix of a polynomial fF[t]f\in F[t] is a matrix constructed so that its characteristic polynomial is exactly ff. Given a polynomial tn+an1tn1++a1t+a0t^n+a_{n-1}t^{n-1}+\ldots+a_1t+a_0, its companion matrix is tn+an1tn1++a1t+a0    [000a0100a1010a20001an1]t^n+a_{n-1}t^{n-1}+\ldots+a_1t+a_0\iff\left[\begin{matrix} 0&0&\cdots&0&-a_0\\ 1&0&\cdots&0&-a_1\\ 0&1&\cdots&0&-a_2\\ \vdots&\vdots&\ddots&0&\vdots\\ 0&0&\cdots&1&-a_{n-1} \end{matrix}\right]

Corollary: All FF-matrices MM have at least one eigenvalue iff FF is an algebraically closed field.
  • For the forward direction: since the companion matrix for every polynomial F[x]\in F[x] (of degree 1\ge 1) has an eigenvalue, every polynomial F[x]\in F[x] (of degree 1\ge 1) has a root in FF, therefore FF is algebraically closed.
  • The backward direction is because in an algebraically closed field, all degree 1\ge 1 polynomials in F[t]F[t] have a root in FF, including det(MtI)\det(M-tI), whose roots correspond to eigenvalues of MM.

Theorem: Given a F[t]F[t]-module MM where tt acts as an endomorphism MMM\to M, MM is cyclic iff the minimal and characteristic polynomials coincide.
  • Let MiF[t]/(fi)M\iso\bigoplus_iF[t]/(f_i) by the structure theorem. Since the minimal polynomial is exactly fnf_n and the characteristic polynomial is exactly ifi\prod_i f_i, they can only coincide when there is only one invariant factor. But that indicates that the decomposition includes only one cyclic submodule F[t]/(fi)F[t]/(f_i), meaning the original module MM must be cyclic as well.
  • The proof in the other direction is trivial: if MM is cyclic MF[t]/(f)M\iso F[t]/(f), then both fnf_n and ifi\prod_i f_i are equal to ff.

If a matrix TT has eigenvalue λ\lambda, then the kernel of TλIT-\lambda I is its corresponding eigenspace. If the image of TT is equal to one of its eigenspaces, the map is a multiplication by a scalar λ\lambda.

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 F[x]F[x] on the FF-module.


Let’s explore the F[t]F[t]-modules, where F[t]F[t] is a polynomial ring defined over a field FF.

For instance, take the F[t]F[t]-module F[t]/(t2)3F[t]/(t-2)^3. Since it’s quotiented by a degree 33 polynomial (t2)3=t36t2+12t8(t-2)^3=t^3-6t^2+12t-8, the obvious basis is {1,t,t2}\{1,t,t^2\}. We use the fact that 0=t36t2+12t80=t^3-6t^2+12t-8 in the quotient to get the constraint T(t2)=t3=6t212t+8T(t^2)=t^3=6t^2-12t+8. This means the F[t]F[t]-matrix TT (with respect to the basis {1,t,t2}\{1,t,t^2\}) is one where T(t2)=6t212t+8T(t^2)=6t^2-12t+8, i.e. we have T[100]=[010],T[010]=[001],T[001]=[8126]T\left[\begin{matrix}1\\0\\0\end{matrix}\right] =\left[\begin{matrix}0\\1\\0\end{matrix}\right],\quad T\left[\begin{matrix}0\\1\\0\end{matrix}\right] =\left[\begin{matrix}0\\0\\1\end{matrix}\right],\quad T\left[\begin{matrix}0\\0\\1\end{matrix}\right] =\left[\begin{matrix}8\\-12\\6\end{matrix}\right] which corresponds to the presentation matrix [0081012016]\left[\begin{matrix}0&0&8\\1&0&-12\\0&1&6\end{matrix}\right]

So we have found TT for the F[t]F[t]-module F[t]/(t2)3F[t]/(t-2)^3.

From the structure theorem, we know that direct sums of F[t]F[t]-modules describe all F[t]F[t]-modules. Let’s do this again where MM is a direct sum.

Take the F[t]F[t]-module MF[t]/(t2)F[t]/(t2)2M\iso F[t]/(t-2)\oplus F[t]/(t-2)^2. The obvious basis is {(1,0),(0,1),(0,t)}\{(1,0),(0,1),(0,t)\}. We do the same thing as before, realizing 0=t20=t-2 on the left and 0=(t2)2=t24t+40=(t-2)^2=t^2-4t+4 on the right, and obtaining the constraints T(1)=t=2T(1)=t=2 on the left and T(t)=t2=4t4T(t)=t^2=4t-4 on the right. We have: T[1  ]=[2  ],T[ 10]=[ 01],T[ 01]=[ 44]T\left[\begin{matrix}1\\~\\~\end{matrix}\right] =\left[\begin{matrix}2\\~\\~\end{matrix}\right],\quad T\left[\begin{matrix}~\\1\\0\end{matrix}\right] =\left[\begin{matrix}~\\0\\1\end{matrix}\right],\quad T\left[\begin{matrix}~\\0\\1\end{matrix}\right] =\left[\begin{matrix}~\\-4\\4\end{matrix}\right] corresponding to the presentation matrix [20414]=[2][0414]\left[\begin{matrix}2&&\\&0&-4\\&1&4\end{matrix}\right] =\left[\begin{matrix}2\end{matrix}\right] \oplus\left[\begin{matrix}0&-4\\1&4\end{matrix}\right] where blank entries are zeroes.

So we’ve described how to find a presentation matrix for any F[t]F[t]-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 F[t]F[t]-presentation matrices.

The Chinese remainder theorem says F[t]/(fg)F[t]/(f)F[t]/(g)F[t]/(f\cdot g)\iso F[t]/(f)\oplus F[t]/(g) when f,gf,g coprime. Consider F[t]/(f)F[t]/(f). Since distinct linear factors are always coprime, we can always split ff to get a bunch of powers of linear factors, one for each repeated root. For instance, F[t]/(tα)2F[t]/(tα)3F[t]/(t-\alpha)^2\oplus F[t]/(t-\alpha)^3.

For an easy example, consider the F[t]F[t]-module MF[t]/(tα)4M\iso F[t]/(t-\alpha)^4. To find TT, first find some basis for MM (as a vector space). A clever basis is {1,(tα),(tα)2,(tα)3}\{1,(t-\alpha),(t-\alpha)^2,(t-\alpha)^3\}. This conveniently gives us the constraints T(1)=t(1)=(tα)4+α(tα)3=α(tα)3T(tα)=t(tα)=(tα)3+α(tα)2T(tα)2=t(tα)2=(tα)2+α(tα)T(tα)3=t(tα)3=(tα)+α(1)\begin{aligned} T(1)&=t(1)&&=(t-\alpha)^4+\alpha(t-\alpha)^3=\alpha(t-\alpha)^3\\ T(t-\alpha)&=t(t-\alpha)&&=(t-\alpha)^3+\alpha(t-\alpha)^2\\ T(t-\alpha)^2&=t(t-\alpha)^2&&=(t-\alpha)^2+\alpha(t-\alpha)\\ T(t-\alpha)^3&=t(t-\alpha)^3&&=(t-\alpha)+\alpha(1)\\ \end{aligned} Converting this into terms of basis vectors, this is T[1000]=[α000],T[0100]=[1α00],T[0010]=[01α0],T[0001]=[001α]T\left[\begin{matrix}1\\0\\0\\0\end{matrix}\right] =\left[\begin{matrix}\alpha\\0\\0\\0\end{matrix}\right],\quad T\left[\begin{matrix}0\\1\\0\\0\end{matrix}\right] =\left[\begin{matrix}1\\\alpha\\0\\0\end{matrix}\right],\quad T\left[\begin{matrix}0\\0\\1\\0\end{matrix}\right] =\left[\begin{matrix}0\\1\\\alpha\\0\end{matrix}\right],\quad T\left[\begin{matrix}0\\0\\0\\1\end{matrix}\right] =\left[\begin{matrix}0\\0\\1\\\alpha\end{matrix}\right] which corresponds to the presentation matrix [α1α1α1α]\left[\begin{matrix}\alpha&1&&\\&\alpha&1&\\&&\alpha&1\\&&&\alpha\end{matrix}\right]

Such a matrix is known as a Jordan matrix with eigenvalue α\alpha with multiplicity 44. It has α\alpha 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 11 and α\alpha. Next, here’s a direct sum example. Consider the F[t]F[t]-module MF[t]/(tα)4F[t]/(tα)3F[t]/(tβ)3M\iso F[t]/(t-\alpha)^4\oplus F[t]/(t-\alpha)^3\oplus F[t]/(t-\beta)^3. We’ll end up with the following presentation matrix, which is composed of matrices that are much alike the previous example: [α1α1α1α][α1α1α][β1β1β]\left[\begin{matrix}\alpha&1&&\\&\alpha&1&\\&&\alpha&1\\&&&\alpha\end{matrix}\right] \oplus\left[\begin{matrix}\alpha&1&\\&\alpha&1\\&&\alpha\end{matrix}\right] \oplus\left[\begin{matrix}\beta&1&\\&\beta&1\\&&\beta\end{matrix}\right]

This is a direct sum of Jordan matrices! We call this the Jordan normal form of the presentation matrix, which always exists for RR-matrices over a PID RR. This is the standard form for a presentation matrix for such RR-modules.

In general, given a RR-module over a PID RR, you may:

All this to find the F[t]F[t]-matrix TT that defines scalar multiplication by tt in a F[t]F[t]-module.

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