Exploration 2: Module actions

January 2, 2024.

Questions:

How can we represent algebraic objects as actions?

Recall that $R$ -matrices are homomorphisms between free modules $R^n$ . How do we visualize homomorphisms between non-free modules $M$ ?

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

$T(m+n)=T(m)+T(n)$ for all $m,n\in M$
$T(km)=kT(m)$ for all $k\in R,m\in M$

In other words, $T$ is $R$ -linear.

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

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

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

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

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

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

Theorem: Given a

F[t]

-module

M

where

t

acts as an endomorphism

M\to M

, there is some polynomial

f\in F[t]

whose corresponding endomorphism

\in\End(M)

is the zero map. In other words,

\Ann_{F[t]}(M)

is nontrivial.

Since $t$ acts as $T\in\End(M)$ , we can construct a ring homomorphism $\varphi:F[t]\to\End(M)$ that maps linear combinations of $t$ ( $\sum_i a_it^i$ ) to their corresponding endomorphism ( $\sum_i a_iT^i$ ).
The kernel of $\varphi$ represents all polynomials $\in F[t]$ that map to the zero map in $\End(M)$ . Thus $\Ann_{F[t]}(M)=\ker\varphi$ , and it is enough to show that $\varphi$ has a non-trivial kernel.
Given that $\End(M)$ is finitely generated, and that polynomial rings $F[t]$ are always infinitely generated (by $\{1,t,t^2,\ldots\}$ , $\varphi$ 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, $\varphi$ has a non-trivial kernel.

Corollary: Every finitely generated

F[t]

-module

M

(where

t

acts as an endomorphism

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

Theorem: Torsion

R

-modules

M

have no nontrivial free submodules.

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

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

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

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

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

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

In this section, we find a basis in which to express $T$ .

Given a finite-dimensional $F$ -vector space $V$ , consider the endomorphism $T:V\to V$ . As before, define a $F[t]$ -module $M$ as $V$ where the action of $t$ is to apply $T$ . By studying this module, we can study properties of $T$ itself.

Since $M$ is a torsion module over a PID, apply the structure theorem to break it into a direct sum of torsion cyclic modules $\bigoplus_i F[t]/(f_i)$ where $f_1\mid f_2\mid\ldots\mid f_n$ . Note that each $f_i$ generates the annihilator for their corresponding cyclic module $F[t]/(f_i)$ .

Of interest is finding a polynomial for $T$ , say $t^3+t+1$ , for which the corresponding $T^3+T+I$ is the zero map. That is, $t^3+t+1$ annihilates $M$ . But so does $2t^3+2t+2$ , and $(t^3+t+1)(t+1)$ . Since $F[t]$ is a PID, the easiest way to simplify this situation is to say we’re looking for the generator of the ideal in $F[t]$ that annihilates $M$ , which divides all such polynomials. The fact that it divides all annihilating polynomials helps us identify this generator. Since the coefficients are in a field $F$ , this generating polynomial must be monic, and it must be unique because any two monic polynomials that divide each other are equal . Call this generator of $\Ann_{F[t]}(M)$ the minimal polynomial of $T$ , denoted $m_T$ .

Theorem: The minimal polynomial of any

T:V\to V

is nontrivial.

As before, define $M$ as the $F[t]$ -module $V$ where the action of $t$ is to apply $T$ . We know that the annihilator of $M$ is the nontrivial kernel of the homomorphism $\varphi:F[t]\to\End(M)$ . Since the kernel is nontrivial, its generator (the minimal polynomial of $T$ ) is nontrivial. (Even if $T$ is the zero map, its minimal polynomial $m_T$ would be $t$ .)

Theorem: The minimal polynomial of any

T:V\to V

is exactly the largest invariant factor of the corresponding

F[t]

-module

M

where

t

acts as

T

Recall that the structure theorem of modules over a PID states that $M$ can be decomposed into $M\iso F[t]/(f_1)\oplus\ldots\oplus F[t]/(f_n)$ where the invariant factors $f_i$ are elements of $F[t]$ that divide each other: $f_1\mid\ldots\mid f_n$ . By definition, each $f_i$ annihilates its component $F[t]/(f_i)$ , and because all of these annihilators divide $f_n$ , $f_n$ must annihilate all of the components, and therefore annihilates $M$ itself. This makes $(f_n)$ the annihilator of $M$ , so its generator $f_n$ is the minimal polynomial of $T$ .

Why are we concerned with finding some $t^3+t+1$ for which the corresponding $T^3+T+I$ is zero? The reason is because if we can find such a polynomial, then we can construct a basis for each component of $M$ and use it to express $T$ as a matrix.

Let $f_i$ be the invariant factors of $M$ . You can observe that, like the minimal polynomial, each $f_i$ is a monic polynomial $\in F[t]$ and each $f_i$ annihilates the $F[t]/(f_i)$ component of $M$ . Let’s study how $T$ acts on a particular component $F[t]/(f_i)$ using the basis $\{1,t,t^2,\ldots,t_{k-1}\}$ (where $k$ is the finite dimension of $F[t]/(f_i)$ , using the fact that $M$ is finite dimensional). Then $T$ acts with respect to this basis by $\left[\begin{matrix}1\\0\\0\\\vdots\\0\end{matrix}\right] \mapsto \left[\begin{matrix}0\\1\\0\\\vdots\\0\end{matrix}\right] \mapsto \left[\begin{matrix}0\\0\\1\\\vdots\\0\end{matrix}\right] \mapsto \quad\cdots\quad \mapsto \left[\begin{matrix}0\\0\\0\\\vdots\\1\end{matrix}\right] \mapsto \left[\begin{matrix}-a_0\\-a_1\\-a_2\\\vdots\\-a_{k-1}\end{matrix}\right]$ where each $a_j$ is the $j$ th coefficient of $f_i=a_0+a_1t+a_2t^2+\ldots+a_{k-1}t^{k-1}+t^k$ . Note that the last map is because in $F[t]/(f_i)$ we mod by $f_i=0$ , so $0=a_0+a_1t+a_2t^2+\ldots+a_{k-1}t^{k-1}+t^k$ becomes $t_k=-a_0-a_1t-a_2t^2-\ldots-a_{k-1}t^{k-1}$

Using the above, we obtain a matrix for $T$ under this basis:

$T=TI=T\left[\begin{matrix} 1&0&0&\ldots&0&0\\ 0&1&0&\ldots&0&0\\ 0&0&1&\ldots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\ldots&1&0\\ 0&0&0&\ldots&0&1 \end{matrix}\right] =\left[\begin{matrix} 0&0&0&\ldots&0&-b_0\\ 1&0&0&\ldots&0&-b_1\\ 0&1&0&\ldots&0&-b_2\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\ldots&0&-b_{k-2}\\ 0&0&0&\ldots&1&-b_{k-1} \end{matrix}\right]$

Call this the companion matrix $C_{f_i}$ for the polynomial $f_i$ . By taking the companion matrix for each of the invariant factors $f_i$ of $M$ we obtain a matrix for $T$ under some basis, in block-diagonal form: $\left[\begin{matrix} C_{f_1}\\ &C_{f_2}\\ &&C_{f_3}\\ &&&\ddots\\ &&&&C_{f_r} \end{matrix}\right]$ assuming there’s $r$ invariant factors. This block-diagonal representation of $T$ built using companion matrices is known as the rational canonical form of $T$ . Since we can always find the basis that puts $T$ into this form, every $T$ has a rational canonical form, determined completely by the invariant factors of its corresponding $F[t]$ -module $M$ .

Two endomorphisms $S,T:V\to V$ are similar if there is an isomorphism $U$ such that $S=UTU^{-1}$ . Since isomorphisms represent a change in basis, another way to state this is that two endomorphisms are similar if they are the same under a change-in-basis.

Theorem: Two endomorphisms

S,T:V\to V

are similar iff they have the same rational canonical form.

If $S,T$ are similar by the isomorphism $U$ , then let $M_S,M_T$ be their corresponding $F[t]$ -modules. We just need to show that $M_S\cong M_T$ . To see this, note that $U(t_{M_T})=UT(1)=SU(1)=t_{M_S}U(1)=t_{M_S}$ , so $U$ maps $t_{M_T}$ to $t_{M_S}$ (and vice versa), which extends $U$ to a $F[t]$ -module isomorphism between $M_S$ and $M_T$ . This means $M_S\cong M_T$ . In particular, their invariant factors are the same, implying the same rational canonical forms.

The reverse proof is simple. If their rational canonical forms are the same then the two endomorphisms are equal under some basis, thus they are similar.

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

In $R$ -module $M$ , an eigenvalue $\lambda$ of an $R$ -module endomorphism $T:M\to M$ is one where there exists a nonzero element $m\in M$ such that $Tm=\lambda m$ .

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

Lemma: An

R

-module endomorphism

T:M\to M

has an eigenvalue

\lambda

T-\lambda I

is nilpotent.

If $T-\lambda I$ is nilpotent, then $(T-\lambda I)^k=0$ for some $k\ge 1$ .
If $k=1$ , then we have $T-\lambda I=0$ implying $T=\lambda I$ , which directly shows that $\lambda$ is an eigenvalue.
Otherwise if $k\ge 2$ , being nilpotent means $(T-\lambda I)^{k-1}\ne 0$ . So there is some element $m\in M$ such that $n=(T-\lambda I)^{k-1}m\ne 0$ . But then we have $(T-\lambda I)n=(T-\lambda I)\left[(T-\lambda I)^{k-1}m\right]=(T-\lambda I)^km=0$ Since there exists an element $n\in M$ where $(T-\lambda I)n=0$ , this proves $Tn=\lambda n$ thus $\lambda$ is an eigenvalue of $T$ .

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

Lemma: The eigenvalues of an endomorphism of a direct sum

T:M\oplus N\to M\oplus N

are exactly the eigenvalues of

T_M

and

T_N

combined (where

T_M

denotes

T

restricted to

M

It is enough to show that the eigenvalues of $T_M$ are eigenvalues of $T$ , since the $N$ case has an identical proof. If $T_M$ has an eigenvalue $\lambda$ , then $T_Mm=\lambda m$ for some $m\in M$ . Thus in $M\oplus N$ , we have $T(m,0)=(T_Mm,T_N0)=(\lambda m,0)=\lambda(m,0)$ proving that the eigenvalue $\lambda$ of $M$ is an eigenvalue of $M\oplus N$ .

Theorem: For a torsion component

F[t]/(f)

of a

F[t]

-module

M

, the roots

\lambda

of its minimal polynomial

f

correspond to eigenvalues of

T

(the action of

t

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

Theorem: The minimal polynomial for a

F[t]

-module

M

is exactly the largest invariant factor

f_n

M

If the decomposition is $M\iso\bigoplus_iF[t]/(f_i)$ , then since the roots of each $f_i$ are eigenvalues of $M$ , the minimal polynomial must contain a factor $(t-\lambda)$ for each eigenvalue $\lambda$ for each component. This means the minimal polynomial is the LCM of all $f_i$ . Since the decomposition has $f_1\mid f_2\mid\ldots\mid f_n$ , we can observe that $f_n$ is a multiple of every $f_i$ , making $f_n$ the minimal polynomial of $M$ .

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

Theorem: The characteristic polynomial for a

F[t]

-module

M

is exactly the product of all invariant factors

f_i

M

This follows directly from knowing that the roots of each $f_i$ are eigenvalues of $M$ . Then the product of each $f_i$ captures all eigenvalues of $M$ .

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

There is another way to compute this characteristic polynomial:

Theorem: Over a finitely generated

F[t]

-module

M

, the characteristic polynomial of an endomorphism

T:M\to M

is exactly

\det(T-tI)

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

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

Corollary: All

F

-matrices

M

have at least one eigenvalue iff

F

is an algebraically closed field.

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

Theorem: Given a

F[t]

-module

M

where

t

acts as an endomorphism

M\to M

M

is cyclic iff the minimal and characteristic polynomials coincide.

Let $M\iso\bigoplus_iF[t]/(f_i)$ by the structure theorem. Since the minimal polynomial is exactly $f_n$ and the characteristic polynomial is exactly $\prod_i f_i$ , they can only coincide when there is only one invariant factor. But that indicates that the decomposition includes only one cyclic submodule $F[t]/(f_i)$ , meaning the original module $M$ must be cyclic as well.
The proof in the other direction is trivial: if $M$ is cyclic $M\iso F[t]/(f)$ , then both $f_n$ and $\prod_i f_i$ are equal to $f$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

Express the $R$ -module as a direct sum of cyclic modules $R/(f_i)$ where $f_i$ is the minimal polynomial of that cyclic component.
Split each $R/(f_i)$ into a direct sum of powers of linear factors (like $R/(t-\alpha)^2\oplus R/(t-\beta)^3$ ) via factoring the $f_i$ , 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 $F[t]$ -matrix $T$ that defines scalar multiplication by $t$ in a $F[t]$ -module.

< Back to category Exploration 2: Module actions (permalink)
Exploration 1: Modules and vector spaces Exploration 3: Representation theory

Exploration 2: Module actions

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

In this section, we find a basis in which to express TTT.

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

In this section, we display the module action of polynomial rings.

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

In this section, we find a basis in which to express $T$ .