---
title: "Modules over a Principal Ideal Domain"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 3
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain
---

# Chapter 3 — Modules over a Principal Ideal Domain

Linear algebra over a *ring* instead of a field: this small change of hypothesis produces one of algebra’s great unification theorems. A [module](#def-b3-modules-module) over $\Z$ is an abelian group; a [module](#def-b3-modules-module) over $K[X]$ is a vector space equipped with an endomorphism. The structure theorem for [finitely generated modules](#def-b3-modules-free) over a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) therefore classifies, in one stroke, all [finitely generated](#def-b3-modules-free) abelian groups *and* all endomorphisms up to similarity — Jordan’s reduction, which Year 2 obtained by delicate inductions, falls out as a corollary, together with its subtler sibling, the *rational* canonical form, valid over every field. The computational engine is the [Smith normal form](#thm-b3-modules-smith), an arithmetic of matrices worthy of Euclid.

Throughout, $A$ is a commutative ring, soon a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd); “[module](#def-b3-modules-module)” means $A$-module.

## 3.1 Modules, free modules

**Definition 3.1.**

An *$A$-module* is an abelian group $(M, +)$ with a scalar multiplication $A \times M \to M$ satisfying the vector-space axioms: $a(x + y) = ax + ay$, $(a + b)x = ax + bx$, $(ab)x = a(bx)$, $1x = x$. Submodules, quotients $M/N$, morphisms ($A$-linear maps), direct sums $\bigoplus_i M_i$, and the isomorphism theorems are defined and proved word for word as for vector spaces and abelian groups; in particular $M/\ker f \cong
\operatorname{im} f$ for a morphism $f$.

**Example 3.2.**

The three motivating cases.

1. $A = K$ a field: [modules](#def-b3-modules-module) are vector spaces.
2. $A = \Z$ : [modules](#def-b3-modules-module) are exactly abelian groups ( $nx$ is forced to be $x + \dots + x$ ), submodules are subgroups.
3. $A = K[X]$ : a [module](#def-b3-modules-module) is a $K$ -vector space $V$ together with the $K$ -linear map $u\colon x \mapsto X\cdot x$ — conversely, every pair $(V, u)$ with $u \in \mathcal L(V)$ becomes a $K[X]$ -module by $P \cdot x = P(u)(x)$ . The submodules are precisely the $u$ -stable subspaces.

An [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) of $A$ is exactly a submodule of $A$; a [quotient ring](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) $A/I$ is an $A$-module. Unlike vector spaces, [modules](#def-b3-modules-module) can have *[torsion](#def-b3-modules-torsion)*: in $\Z/6\Z$, the element $\bar 3 \ne 0$ is killed by $2 \neq 0$.

**Definition 3.3.**

$M$ is *finitely generated* if $M = Ax_1 + \dots + Ax_n$ for some $x_i$. $M$ is *free* of rank $n$ if $M \cong A^n$, i.e. if it has a *basis* (a generating family that is $A$-linearly independent). Every finitely generated $M$ is a quotient of a free [module](#def-b3-modules-module): $(a_1, \dots, a_n) \mapsto \sum a_ix_i$ maps $A^n$ onto $M$.

**Proposition 3.4 (Invariance of rank).**

If $A \neq 0$ and $A^m \cong A^n$, then $m = n$.

**Proof.** Pick a [maximal ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-primemaximal) $\mathfrak m$ of $A$ ([Theorem 2.8](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-krull)) and set $k = A/\mathfrak m$, a field. An isomorphism $f \colon A^m
\to A^n$ maps $\mathfrak m A^m$ into $\mathfrak m A^n$ (linearity), hence induces an isomorphism of quotients

$$
A^m/\mathfrak m A^m \;\cong\; A^n/\mathfrak m A^n,
\qquad\text{i.e.}\qquad k^m \cong k^n
$$

as $k$-vector spaces (the quotient $A^m/\mathfrak m A^m$ is killed by $\mathfrak m$, so the $A$-action factors through $k$; the images of the standard basis form a $k$-basis). Dimension theory over the field $k$ gives $m = n$. ∎

**Theorem 3.5 (Submodules of free modules).**

Let $A$ be a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) and $M \subseteq A^n$ a submodule. Then $M$ is [free](#def-b3-modules-free) of rank $\leq n$.

**Proof.** Induction on $n$. For $n = 1$: $M$ is an [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal), so $M = (0)$ ([free](#def-b3-modules-free) of rank $0$) or $M = dA \cong A$ ($x \mapsto dx$ is injective: domain). For $n > 1$: let $\pi \colon A^n \to A$ be the last coordinate. Then $\pi(M)$ is an [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal), $(0)$ or $dA$. If $(0)$: $M \subseteq A^{n-1} \times \{0\}$ and induction applies. Otherwise choose $x_0 \in M$ with $\pi(x_0) = d$. Every $x \in M$ writes uniquely

$$
x = \underbrace{\frac{\pi(x)}{d}}_{\in A}\, x_0 +
\Bigl(x - \tfrac{\pi(x)}d x_0\Bigr),
\qquad x - \tfrac{\pi(x)}d x_0 \in M \cap \ker\pi
$$

($\pi(x) \in dA$, so the coefficient is in $A$). Thus $M = Ax_0
\oplus (M \cap \ker \pi)$: the sum is direct since $\pi(ax_0) =
ad = 0$ forces $a = 0$. By induction $M \cap \ker\pi \subseteq
\ker \pi \cong A^{n-1}$ is [free](#def-b3-modules-free) of rank $\leq n - 1$; adjoining $x_0$ (independent from $\ker\pi$ as just seen) gives a basis of $M$ of cardinality $\leq n$. ∎

**Remark 3.6.**

Consequently, over a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) every [finitely generated module](#def-b3-modules-free) $M$ has a *finite presentation*: a surjection $\varphi\colon A^n \to M$ has [free](#def-b3-modules-free) kernel with basis $c_1, \dots, c_k$ ($k \leq n$), and $M
\cong A^n / \operatorname{im}(C)$ where $C \in M_{n,k}(A)$ is the matrix whose columns are the $c_j$. Understanding $M$ means understanding a *matrix over $A$* up to change of bases in source and target — the subject of the next section.

## 3.2 Smith normal form

**Definition 3.7.**

Two matrices $B, C \in M_{n,k}(A)$ are *equivalent* if $C =
QBP$ with $Q \in GL_n(A)$, $P \in GL_k(A)$ (invertible *over $A$*: determinant in $A^\times$). Equivalent presentation matrices define isomorphic [modules](#def-b3-modules-module) $A^n/
\operatorname{im}$ (change bases in $A^n$ and $A^k$).

**Theorem 3.8 (Smith normal form).**

Let $A$ be a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) and $B \in M_{n,k}(A)$. Then $B$ is equivalent to a diagonal matrix

$$
\operatorname{diag}(d_1, d_2, \dots, d_r, 0, \dots, 0),
\qquad d_1 \mid d_2 \mid \cdots \mid d_r \neq 0 ,
$$

and the $d_i$ are unique up to [associates](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility): $d_1 \cdots d_i$ is a gcd of the $i \times i$ minors of $B$ (in particular that gcd is an invariant of equivalence). The $d_i$ are the *invariant factors* of $B$.

**Proof.** *Existence.* If $B = 0$, done. Otherwise, consider the set of [ideals](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) $(b)$ generated by entries of matrices equivalent to $B$; since $A$ is [Noetherian](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-noetherian), choose a matrix $B'$ equivalent to $B$ and an entry $d$ of $B'$ with $(d)$ *maximal* in this set. Move $d$ to position $(1,1)$ by row and column swaps.

*Claim: $d$ divides every entry of $B'$.* First, column 1: if $b_{i1}$ is not a multiple of $d$, let $e = \gcd(d, b_{i1}) = ud +
vb_{i1}$ (Bézout), so $(e) \supsetneq (d)$. The $2 \times 2$ matrix trick: acting on rows $1$ and $i$ by

$$
\begin{pmatrix} u & v \\ -\dfrac{b_{i1}}e & \dfrac de
\end{pmatrix}
\in GL_2(A)
\qquad
\Bigl(\det = \tfrac{ud + v b_{i1}}e = 1\Bigr)
$$

produces an equivalent matrix with entry $e$ in position $(1,1)$: this contradicts maximality of $(d)$. So $d$ divides column $1$, and symmetrically row $1$. Subtracting multiples of row 1 and column 1 clears them: $B'$ is equivalent to $\begin{pmatrix} d &
0\\ 0 & B''\end{pmatrix}$. Next, $d$ divides every entry $b$ of $B''$: add the row of $b$ to row 1 (an elementary operation; the new first row contains $d$ and $b$-entries), and repeat the column-clearing argument: a non-multiple would again improve $(d)$. Now induct on the size: $B''$, all of whose entries are divisible by $d$, has a Smith form $\operatorname{diag}(d_2,
\dots)$ whose entries remain divisible by $d$ (every entry of any $QB''P$ is an $A$-combination of entries of $B''$); set $d_1 =
d$.

*Uniqueness.* Let $D_i(B)$ denote a gcd of all $i \times i$ minors. Row and column operations, and more generally multiplication by *any* matrix, cannot shrink the gcd: the $i \times i$ minors of $QB$ are $A$-combinations of those of $B$ (Cauchy–Binet expansion; or directly: each row of $QB$ is a combination of rows of $B$, and minors are multilinear in rows). So $D_i(QBP)$ and $D_i(B)$ divide each other: $D_i$ is an equivalence invariant. On the diagonal form, the nonzero $i
\times i$ minors are the products of $i$ of the $d_j$’s, and divisibility $d_1 \mid \dots \mid d_r$ makes $d_1 \cdots d_i$ the gcd. Hence $d_1 \cdots d_i = D_i(B)$ up to units, and $d_i =
D_i/D_{i-1}$ is determined. ∎

**Method 3.9.**

Over a [Euclidean domain](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) ($\Z$, $K[X]$), Smith reduction is an algorithm — no maximality argument needed: bring the entry of smallest Euclidean size to position $(1,1)$; if it fails to divide some entry of its row or column, a Euclidean division leaves a strictly smaller remainder there — swap it in and restart (termination: sizes decrease); when it divides its whole row and column, clear them; if it fails to divide an inner entry, add that row to row $1$ and restart; recurse on the inner block. In practice on integer matrices: compute $D_1 = \gcd$ of entries, $D_2$, … via minors for small sizes, or run the algorithm.

**Example 3.10 (A Smith reduction, in full).**

Reduce $M = \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9
\end{pmatrix}$ over $\Z$. The corner $1$ divides everything: clear its row and column ($L_2 \leftarrow L_2 - 4L_1$, $L_3
\leftarrow L_3 - 7L_1$, then $C_2 \leftarrow C_2 - 2C_1$, $C_3
\leftarrow C_3 - 3C_1$):

$$
M \sim \begin{pmatrix} 1 & 0 & 0\\ 0 & -3 & -6\\ 0 & -6 & -12
\end{pmatrix} .
$$

In the inner block, the corner $-3$ divides all entries: $L_3
\leftarrow L_3 - 2L_2$ and $C_3 \leftarrow C_3 - 2C_2$ clear it to $\operatorname{diag}(-3, 0)$. Adjusting signs (multiply a row by $-1$, a legal operation):

$$
M \sim \operatorname{diag}(1, 3, 0),
\qquad
\Z^3/M\Z^3 \cong \Z/3\Z \times \Z .
$$

Cross-check by determinantal divisors: $D_1 =
\gcd(\text{entries}) = 1$; every $2\times2$ minor of $M$ is a multiple of $3$ (e.g. $\det\bigl(\begin{smallmatrix}1 & 2\\ 4
& 5\end{smallmatrix}\bigr) = -3$) and one equals $-3$: $D_2 =
3$; $D_3 = \det M = 0$. Hence $d_1 = 1$, $d_2 = 3$, $d_3 = 0$: same answer. Two lessons: a zero invariant factor records the rank drop (the cokernel picks up a [free](#def-b3-modules-free) $\Z$ summand), and the divisibility chain $1 \mid 3 \mid 0$ is the Smith certificate — a diagonal reduction that violates the chain (say $\operatorname{diag}(2, 3)$, which the careless can produce from $\bigl(\begin{smallmatrix}2 & 0\\ 0 &
3\end{smallmatrix}\bigr)$ by stopping too early: correct Smith form $\operatorname{diag}(1, 6)$, as $D_1 = 1$ here!) is not finished.

![The sublattice L = ℤ(2,0) + ℤ(1,3) of ℤ2 (red points). The Smith normal form of ( smallmatrix 2 & 1\\ 0 & 3 smallmatrix ) is diag(1, 6): in the adapted basis f_1 = (1,3), f_2 = (0,1) of ℤ2, one has L = ℤ f_1 ℤ\,6f_2, so ℤ2/L ℤ/6ℤ — the index equals | | = 6, the area of the shaded fundamental domain.](https://one-course.com/images/onecourse/chapters/math-5/b3-modules/fig-3901c998d682.svg)

*The sublattice $L = \Z(2,0) + \Z(1,3)$ of $\Z^2$ (red points). The [Smith normal form](#thm-b3-modules-smith) of $\bigl(\begin{smallmatrix} 2 &
1\\ 0 & 3\end{smallmatrix}\bigr)$ is $\operatorname{diag}(1, 6)$: in the adapted basis $f_1 = (1,3)$, $f_2 = (0,1)$ of $\Z^2$, one has $L = \Z f_1 \oplus \Z\,6f_2$, so $\Z^2/L \cong \Z/6\Z$ — the index equals $\abs{\det} = 6$, the area of the shaded fundamental domain.*

## 3.3 The structure theorem

**Definition 3.11.**

Let $A$ be a domain and $M$ an $A$-module. The *torsion* submodule is

$$
T(M) = \{x \in M : ax = 0 \text{ for some } a \neq 0\}
$$

(a submodule: if $ax = by = 0$ then $ab(x + y) = 0$, $ab \ne 0$). $M$ is *[torsion-free](#def-b3-modules-free)* if $T(M) = 0$, a *torsion [module](#def-b3-modules-module)* if $T(M) = M$.

**Theorem 3.12 (Structure of finitely generated modules over a PID).**

Let $A$ be a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) and $M$ a [finitely generated](#def-b3-modules-free) $A$-module. There exist a unique $r \in \N$ and nonzero nonunits $d_1 \mid d_2 \mid
\cdots \mid d_s$, unique up to [associates](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility), with

$$
M \;\cong\; A^r \,\oplus\, A/(d_1) \oplus \cdots \oplus A/(d_s).
$$

Moreover $T(M) \cong A/(d_1)\oplus\dots\oplus A/(d_s)$ and $M/T(M)
\cong A^r$: a [finitely generated](#def-b3-modules-free) *[torsion-free](#def-b3-modules-torsion)* [module](#def-b3-modules-module) over a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) is [free](#def-b3-modules-free).

**Proof.** *Existence.* Present $M \cong A^n/\operatorname{im}(C)$ ([Remark 3.6](#rem-b3-modules-presentation)) and put $C$ in Smith form: after the two base changes, $M \cong A^n / (d_1A \times \dots
\times d_rA \times 0 \times \dots \times 0) \cong A/(d_1) \oplus
\dots \oplus A/(d_r) \oplus A^{\,n - r}$. Discard the factors where $d_i$ is a unit ($A/(d_i) = 0$); the divisibility chain survives.

*The [torsion](#def-b3-modules-torsion) identification.* In the decomposition, $A^r$ is [torsion-free](#def-b3-modules-torsion) (a domain has no zero divisors) and each $A/(d_i)$ is [torsion](#def-b3-modules-torsion) (killed by $d_i \ne 0$); a direct sum splits [torsion](#def-b3-modules-torsion) accordingly: $T(M) = \bigoplus_i A/(d_i)$ and $M/T(M) \cong A^r$.

*Uniqueness of $r$*: $M/T(M) \cong A^r$ depends only on $M$, and [Proposition 3.4](#prop-b3-modules-rank) pins $r$ down.

*Uniqueness of the $d_i$*: it suffices to treat the [torsion module](#def-b3-modules-torsion) $T = T(M)$. Decompose each $d_i$ into primes and split by the Chinese remainder theorem ([Theorem 2.9](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-crt); distinct primes generate [comaximal ideals](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-crt)):

$$
A/(d) \cong \bigoplus_{p \mid d} A/\bigl(p^{v_p(d)}\bigr):
\qquad
T \cong \bigoplus_{p} \bigoplus_{j} A/\bigl(p^{k_{p,j}}\bigr),
$$

the *elementary divisors* $p^{k_{p,j}}$. Conversely the $d_i$ are reconstructed from the multiset of [elementary divisors](#thm-b3-modules-structure) ($d_s$ = product of the highest power of each prime, etc.), so it suffices to prove the multiset $\{k_{p,j}\}_j$ is determined by $T$, for each prime $p$. Fix $p$; for $j \geq 1$ consider the $A/(p)$-vector spaces $p^{j-1}T/p^jT$. On a cyclic factor $A/(p^k)$:

$$
p^{j-1}\bigl(A/(p^k)\bigr)\big/p^{j}\bigl(A/(p^k)\bigr)
\cong
\begin{cases}
A/(p) & \text{if } j \leq k,\\
0 & \text{if } j > k,
\end{cases}
$$

and on a factor $A/(q^k)$, $q \neq p$: multiplication by $p$ is bijective there ($p$ invertible mod $q^k$: Bézout), so the quotient is $0$. Direct sums pass through: $\dim_{A/(p)}
p^{j-1}T/p^jT = \#\{i : k_{p,i} \geq j\}$. These intrinsic dimensions determine the multiset of exponents. ∎

**Corollary 3.13 (Finitely generated abelian groups).**

Every [finitely generated](#def-b3-modules-free) abelian group is $\Z^r \times
\Z/d_1\Z\times\dots\times\Z/d_s\Z$ with $d_1 \mid \dots \mid
d_s$, uniquely. Every *finite* abelian group is a product of cyclic groups of prime-power order, unique as a multiset.

**Example 3.14.**

The abelian groups of order $p^n$ correspond to *partitions* of $n$: for $p^4$: $\Z/p^4$, $\Z/p^3\times\Z/p$, $(\Z/p^2)^2$, $\Z/p^2 \times (\Z/p)^2$, $(\Z/p)^4$ — five groups, as $4$ has five partitions. Mixed orders multiply the counts prime by prime (CRT): there are $5 \times 2$ abelian groups of order $2^4 \cdot 3^2 = 144$.

## 3.4 Application: canonical forms of endomorphisms

Let $K$ be a field, $V$ a $K$-vector space of finite dimension $n$, and $u \in \mathcal L(V)$; make $V$ a $K[X]$-module via $P
\cdot x = P(u)(x)$ ([Example 3.2](#ex-b3-modules-examples)). This [module](#def-b3-modules-module) is [finitely generated](#def-b3-modules-free) (a $K$-basis generates) and [torsion](#def-b3-modules-torsion): for each $x$, the $n+1$ vectors $x, u(x), \dots, u^n(x)$ are $K$-dependent, providing a nonzero annihilating polynomial.

**Definition 3.15.**

For $P = X^m + a_{m-1}X^{m-1} + \dots + a_0$ monic, the *companion matrix* is

$$
C_P = \begin{pmatrix}
0 & & & -a_0\\
1 & \ddots & & -a_1\\
& \ddots & 0 & \vdots\\
& & 1 & -a_{m-1}
\end{pmatrix} :
$$

the matrix of “multiplication by $X$” on $K[X]/(P)$ in the basis $1, \bar X, \dots, \bar X^{m-1}$.

**Theorem 3.16 (Frobenius: rational canonical form).**

There is a unique sequence of monic nonconstant polynomials $P_1
\mid P_2 \mid \dots \mid P_s$ (the *similarity invariants* of $u$) such that, as $K[X]$-modules,

$$
V \cong K[X]/(P_1) \oplus \cdots \oplus K[X]/(P_s):
$$

in a suitable basis, $u$ has block-diagonal matrix $\operatorname{diag}(C_{P_1}, \dots, C_{P_s})$. Moreover:

1. $P_s = \mu_u$ (minimal polynomial) and $P_1\cdots P_s =  \chi_u$ (characteristic polynomial); in particular $\mu_u  \mid \chi_u$ *(Cayley–Hamilton re-proved)* and $\chi_u \mid \mu_u^{\,s}$ , so $\chi_u$ and $\mu_u$ have the same [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factors.
2. Two endomorphisms (or square matrices) are similar iff they have the same [similarity invariants](#thm-b3-modules-frobenius) .

**Proof.** The structure theorem ([Theorem 3.12](#thm-b3-modules-structure)) applied to the [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) $K[X]$: the [torsion module](#def-b3-modules-torsion) $V$ decomposes with [invariant factors](#thm-b3-modules-smith) $P_i$, normalized monic (units of $K[X]$ are $K^\times$); no [free](#def-b3-modules-free) part occurs ($V$ is [torsion](#def-b3-modules-torsion)). On each cyclic factor $K[X]/(P_i)$, multiplication by $X$ has matrix $C_{P_i}$ in the basis of powers of $\bar X$: concatenating bases gives the block form.

(1) The annihilator of $V = \bigoplus K[X]/(P_i)$ is $(P_1)\cap
\dots\cap(P_s) = (P_s)$ (divisibility chain: $P_s$ is a common multiple, and the class of $1$ in the last factor is killed exactly by $(P_s)$): $\mu_u = P_s$. For $\chi_u$: on a cyclic factor, $\chi_{C_P} = P$, by induction on $m = \deg P$. Expanding $\det(XI_m - C_P)$ along the first row (whose entries are $X$, then zeros, then $a_0$ in the last column):

$$
\det(XI_m - C_P) = X\,\det\bigl(XI_{m-1} - C_{\tilde P}\bigr)
+ (-1)^{1+m}\,a_0\,\det L ,
$$

where $\tilde P = X^{m-1} + a_{m-1}X^{m-2} + \dots + a_1$ (same shape, one size down) and $L$ is triangular with diagonal $(-1, \dots, -1)$, so $\det L = (-1)^{m-1}$. By induction the first term is $X\tilde P$, and the second is $a_0$: the total is $X\tilde P + a_0 = P$ (base case $m=1$: $\det(X + a_0) = P$). Determinants multiply over blocks: $\chi_u = \prod P_i$. Cayley–Hamilton: $\chi_u \in (\mu_u)$ since $P_s \mid$ each… conversely each $P_i \mid P_s$, so $\chi_u = \prod P_i$ divides $P_s^{\,s} = \mu_u^s$; and $\mu_u =
P_s$ divides $\chi_u$ as one of its factors.

(2) Similar endomorphisms are conjugate [module](#def-b3-modules-module) structures, hence have equal invariants (uniqueness in [Theorem 3.12](#thm-b3-modules-structure)); conversely equal invariants give isomorphic $K[X]$-modules, and a [module](#def-b3-modules-module) isomorphism is exactly a linear bijection intertwining the two endomorphisms: a similarity. ∎

**Corollary 3.17 (Similarity is insensitive to field extension).**

Let $K \subseteq L$ be fields and $M, N \in M_n(K)$. If $M$ and $N$ are similar over $L$, they are similar over $K$.

**Proof.** The [similarity invariants](#thm-b3-modules-frobenius) of $M$ are computed by Smith’s minor formula ([Theorem 3.8](#thm-b3-modules-smith)) applied to the presentation matrix $XI_n - M$ over $K[X]$ — indeed the $K[X]$-module $V_M = K^n$ has presentation $XI_n - M$: the map $K[X]^n \to V_M$, $(Q_i) \mapsto \sum Q_i(M)e_i$, is onto with kernel generated by the columns of $XI_n - M$ (a direct verification: modulo those columns, every element of $K[X]^n$ reduces to a constant vector, and constant vectors map bijectively; the weekend problem spells this out). Gcds of polynomials do not change under field extension: if $d$ is the monic gcd in $K[X]$ of a family $(f_j)$, Bézout gives $d = \sum u_jf_j$ with $u_j \in K[X]$, so every common divisor of the $f_j$ *in $L[X]$* divides $d$; as $d$ is itself a common divisor, it is the gcd in $L[X]$ too. Hence the [invariant factors](#thm-b3-modules-smith) of $XI_n - M$, quotients of successive minor gcds, are the same over $K$ and over $L$: $M, N$ have the same [similarity invariants](#thm-b3-modules-frobenius) over $L$ iff over $K$; conclude by [Theorem 3.16](#thm-b3-modules-frobenius)(2). ∎

**Theorem 3.18 (Jordan form, re-derived).**

Suppose $\chi_u$ splits over $K$ (e.g. $K = \C$). Applying to $V$ the elementary-divisor decomposition (proof of [Theorem 3.12](#thm-b3-modules-structure)) instead of [invariant factors](#thm-b3-modules-smith):

$$
V \cong \bigoplus_{\lambda, j} K[X]\big/\bigl((X -
\lambda)^{k_{\lambda,j}}\bigr),
$$

and in the basis $\bigl(\overline{(X-\lambda)^{k-1}}, \dots,
\overline{(X - \lambda)}, \bar 1\bigr)$ of each factor, $u$ acts as the Jordan block $J_k(\lambda)$: every endomorphism with split characteristic polynomial has a Jordan basis, and the multiset of blocks $(\lambda, k)$ is unique.

**Proof.** The [elementary divisors](#thm-b3-modules-structure) of the [torsion module](#def-b3-modules-torsion) $V$ are the $(X - \lambda)^k$ with $X - \lambda$ ranging over the [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factors of $\mu_u$ (which splits, since $\chi_u$ does and both have the same [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) factors, [Theorem 3.16](#thm-b3-modules-frobenius)). In $W = K[X]/((X-\lambda)^k)$, put $f_j = \overline{(X - \lambda)^{k-j}}$ for $j = 1, \dots, k$: then $(X - \lambda)f_j = f_{j-1}$ (with $f_0 = 0$), i.e. $u(f_j)
= \lambda f_j + f_{j-1}$: the matrix of $u$ on $(f_1, \dots,
f_k)$ is exactly $J_k(\lambda)$ (ones above the diagonal). Uniqueness of the multiset of [elementary divisors](#thm-b3-modules-structure) is [Theorem 3.12](#thm-b3-modules-structure). ∎

**Remark 3.19.**

The hierarchy of canonical forms is now transparent: the *rational* form exists over every field and detects similarity absolutely ([Corollary 3.17](#cor-b3-modules-descent)); the *Jordan* form is its refinement when $\chi_u$ splits. Year 2’s dimension-counting proofs of Jordan’s theorem are subsumed: all the combinatorics was the arithmetic of the [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) $K[X]$.

## 3.5 Exercises

**Exercise 3.1 ★.**

(a) Show that $\Q$ is not [finitely generated](#def-b3-modules-free) as a $\Z$-module. (b) Show that $\Q$ is [torsion-free](#def-b3-modules-torsion) but not [free](#def-b3-modules-free). (c) Why does neither statement contradict [Theorem 3.12](#thm-b3-modules-structure)?

**Solution of Exercise 3.1.**

(a) If $\Q = \Z q_1 + \dots + \Z q_k$, let $d$ be a common denominator of the $q_i$: every combination lies in $\frac1d\Z$, but $\frac1{2d} \notin \frac1d\Z$. Contradiction.

(b) [Torsion-free](#def-b3-modules-torsion): $nq = 0$ with $n \neq 0$ forces $q = 0$ in $\Q$. Not [free](#def-b3-modules-free): any two nonzero rationals $\frac ab, \frac cd$ satisfy the nontrivial relation $(bc)\frac ab - (ad)\frac cd =
0$, so a basis has at most one element; $\Q \cong \Z$ would make $\Q = \Z q$ cyclic, but $\frac q2 \notin \Z q$. (And $\Q \neq
0$.)

(c) [Theorem 3.12](#thm-b3-modules-structure) assumes *finite generation*, which (a) denies: no contradiction — rather, $\Q$ shows the hypothesis is necessary in the statement “[torsion-free](#def-b3-modules-torsion) $\Rightarrow$ [free](#def-b3-modules-free)”.

**Exercise 3.2 ★.**

List the abelian groups of order $360$ up to isomorphism, in both elementary-divisor and invariant-factor forms. How many abelian groups of order $p^5$ are there?

**Solution of Exercise 3.2.**

$360 = 2^3\cdot3^2\cdot5$. Partitions: of $3$: $(3), (2,1),
(1,1,1)$; of $2$: $(2), (1,1)$; of $1$: $(1)$. Hence $3 \times 2
\times 1 = 6$ groups. [Elementary divisors](#thm-b3-modules-structure) $\to$ [invariant factors](#thm-b3-modules-smith):

| $\Z/8 \times \Z/9 \times \Z/5$ | $\Z/360$ |
| --- | --- |
| $\Z/8 \times \Z/3 \times \Z/3 \times \Z/5$ | $\Z/3 \times \Z/120$ |
| $\Z/4 \times \Z/2 \times \Z/9 \times \Z/5$ | $\Z/2 \times \Z/180$ |
| $\Z/4 \times \Z/2 \times \Z/3 \times \Z/3 \times \Z/5$ | $\Z/6 \times \Z/60$ |
| $(\Z/2)^3 \times \Z/9 \times \Z/5$ | $\Z/2 \times \Z/2 \times \Z/90$ |
| $(\Z/2)^3 \times \Z/3 \times \Z/3 \times \Z/5$ | $\Z/2 \times \Z/6 \times \Z/30$ |

(To pass to [invariant factors](#thm-b3-modules-smith): the largest $d_s$ collects the highest prime power of each prime, and so on down.) Of order $p^5$: as many as partitions of $5$, namely $7$.

**Exercise 3.3 ★.**

Compute the [Smith normal form](#thm-b3-modules-smith) over $\Z$ of

$$
B = \begin{pmatrix} 2 & 4\\ 6 & 8 \end{pmatrix},
\qquad
C = \begin{pmatrix} 2 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 12
\end{pmatrix},
$$

and identify the abelian groups $\Z^2/B\Z^2$ and $\Z^3/C\Z^3$.

**Solution of Exercise 3.3.**

$B$: $D_1 = \gcd(2,4,6,8) = 2$; $D_2 = \abs{\det B} = \abs{16 -
24} = 8$. [Invariant factors](#thm-b3-modules-smith) $d_1 = 2$, $d_2 = 8/2 = 4$: Smith form $\operatorname{diag}(2, 4)$, and $\Z^2/B\Z^2 \cong \Z/2\Z
\times \Z/4\Z$.

$C$: diagonal but *not* Smith ($2 \nmid 3$). $D_1 =
\gcd(2,3,12) = 1$; $D_2 = \gcd(2\cdot3,\, 2\cdot12,\, 3\cdot12) =
\gcd(6, 24, 36) = 6$; $D_3 = 72$. So $d = (1, 6, 12)$ and $\Z^3/C\Z^3 \cong \Z/6\Z\times\Z/12\Z$ — consistently with the CRT: $\Z/2\times\Z/3\times\Z/12 \cong \Z/6\times\Z/12$.

**Exercise 3.4 ★★.**

Let $L \subseteq \Z^n$ be a subgroup of rank $n$ with basis the columns of $B \in M_n(\Z)$, $\det B \neq 0$. Show that $\Z^n/L$ is finite of cardinality $\abs{\det B}$, and that $\Z^n/L
\cong \prod_i \Z/d_i\Z$ for the [invariant factors](#thm-b3-modules-smith) $d_i$ of $B$. Illustrate with $L = \Z(2,0) + \Z(1,3)$.

**Solution of Exercise 3.4.**

Write $B = Q\,\operatorname{diag}(d_1, \dots, d_n)\,P$ with $Q, P
\in GL_n(\Z)$ ([Theorem 3.8](#thm-b3-modules-smith); no zero $d_i$ since $\det B \neq 0$). Then $\Z^n/B\Z^n \cong \Z^n/
\operatorname{diag}(d)\Z^n = \prod_i \Z/d_i\Z$ (the composed isomorphism $x \mapsto Q^{-1}x$ of $\Z^n$ maps $B\Z^n$ onto $\operatorname{diag}(d)P\Z^n = \operatorname{diag}(d)\Z^n$). Its cardinality is $\prod \abs{d_i} = \abs{\det
\operatorname{diag}(d)} = \abs{\det B}$, as $\det Q, \det P = \pm
1$. For $L = \Z(2,0) + \Z(1,3)$: $B = \bigl(\begin{smallmatrix}2
& 1\\ 0 & 3\end{smallmatrix}\bigr)$, $D_1 = 1$, $D_2 = 6$: $\Z^2/L
\cong \Z/6\Z$, of cardinality $\abs{\det B} = 6$.

**Exercise 3.5 ★★.**

Let $A$ be a domain. (a) Verify that $T(M)$ is a submodule and that $M/T(M)$ is [torsion-free](#def-b3-modules-torsion). (b) Show that the [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) $(X, Y)$ of $K[X,Y]$, as a $K[X,Y]$-module, is [torsion-free](#def-b3-modules-torsion) but not [free](#def-b3-modules-free): the structure theorem genuinely needs the [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) hypothesis.

**Solution of Exercise 3.5.**

(a) Submodule: done in [Definition 3.11](#def-b3-modules-torsion). If $a(x +
T(M)) = 0$ in $M/T(M)$ with $a \neq 0$, then $ax \in T(M)$: $bax
= 0$ for some $b \neq 0$, and $ba \neq 0$ (domain), so $x \in
T(M)$: the class is zero. $M/T(M)$ is [torsion-free](#def-b3-modules-torsion).

(b) $(X, Y) \subseteq K[X,Y]$ is [torsion-free](#def-b3-modules-torsion) (a submodule of the domain $K[X,Y]$ acting on itself). Suppose it were [free](#def-b3-modules-free); any two elements $P, Q$ satisfy $Q\cdot P - P \cdot Q = 0$, a nontrivial relation when $P \ne Q$ are nonzero, so a basis has one element: $(X, Y) = (P)$ principal — contradicting [Exercise 2.6](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#exo-b3-rings-6)(a). [Torsion-free](#def-b3-modules-torsion) and [finitely generated](#def-b3-modules-free) ($X, Y$ generate), yet not [free](#def-b3-modules-free): over the [non-PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd) $K[X,Y]$, the structure theorem fails.

**Exercise 3.6 ★★.**

(a) Show that $2\Z$ has no direct complement in the $\Z$-module $\Z$: submodules of [free modules](#def-b3-modules-free) are [free](#def-b3-modules-free) ([Theorem 3.5](#thm-b3-modules-submodule)), but direct summands they need not be. (b) Show that if $M \subseteq A^n$ ($A$ a [PID](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-pidufd)) satisfies: $A^n/M$ is [torsion-free](#def-b3-modules-torsion), then $M$ *is* a direct summand.

**Solution of Exercise 3.6.**

(a) If $\Z = 2\Z \oplus C$, the projection $\Z \to \Z/2\Z$ restricts to an isomorphism $C \cong \Z/2\Z$: $C$ would be a subgroup of $\Z$ whose nonzero element $x$ satisfies $2x \in C
\cap 2\Z = 0$. But $\Z$ is [torsion-free](#def-b3-modules-torsion): $C = 0$, forcing $\Z =
2\Z$ — false.

(b) $A^n/M$ is [finitely generated](#def-b3-modules-free) and [torsion-free](#def-b3-modules-torsion), hence [free](#def-b3-modules-free) ([Theorem 3.12](#thm-b3-modules-structure)): $A^n/M \cong A^r$ with basis $f_1, \dots, f_r$. Choose preimages $y_i \in A^n$ of the $f_i$ and set $F = Ay_1 + \dots + Ay_r$. Every $x \in A^n$ has $\pi(x)
= \sum a_if_i$, so $x - \sum a_iy_i \in M$: $A^n = M + F$. If $\sum a_iy_i \in M$, applying $\pi$ gives $\sum a_if_i = 0$, hence all $a_i = 0$ (basis): $M \cap F = 0$. So $A^n = M \oplus
F$.

**Exercise 3.7 ★★.**

Solve in $\Z^2$ the system

$$
\begin{cases}
2x + 4y \equiv b_1 \pmod{20},\\
6x + 8y \equiv b_2 \pmod{20},
\end{cases}
$$

for which pairs $(b_1, b_2)$ solutions exist, using the Smith form of [Exercise 3.3](#exo-b3-modules-3) (invertible changes of variables on both sides).

**Solution of Exercise 3.7.**

The reduction of [Exercise 3.3](#exo-b3-modules-3) was effective: with

$$
L = \begin{pmatrix} 1 & 0\\ -3 & 1\end{pmatrix},
\qquad
R = \begin{pmatrix} 1 & 2\\ 0 & -1 \end{pmatrix},
\qquad
LBR = \begin{pmatrix} 2 & 0\\ 0 & 4\end{pmatrix}
$$

(row operation $R_2 \leftarrow R_2 - 3R_1$, column operations $C_2 \leftarrow C_2 - 2C_1$ then $C_2 \leftarrow -C_2$). Setting $y = R^{-1}x$ (a bijection of $(\Z/20\Z)^2$, $R$ being invertible over $\Z$), the system $Bx \equiv b \pmod{20}$ is equivalent to

$$
2y_1 \equiv b_1, \qquad 4y_2 \equiv -3b_1 + b_2 \pmod{20}.
$$

The congruence $ky \equiv c \pmod{20}$ is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) iff $\gcd(k, 20) \mid c$: solutions exist iff $2 \mid b_1$ and $4
\mid b_2 - 3b_1$, i.e. *$b_1$ even and $b_2 \equiv 3b_1
\pmod 4$*. When [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) there are $2 \times 4 = 8$ solutions modulo $20$.

**Exercise 3.8 ★★.**

(a) Determine all [similarity invariants](#thm-b3-modules-frobenius) and possible [Jordan forms](#thm-b3-modules-jordan) of a nilpotent $4 \times 4$ matrix, sorted by the partition of $4$ they realize. (b) Exhibit two $4\times4$ complex matrices with the same characteristic *and* minimal polynomials that are not similar, and prove that for $n \leq 3$ this cannot happen.

**Solution of Exercise 3.8.**

(a) A nilpotent $u$ has $\mu_u = X^k$; the [elementary divisors](#thm-b3-modules-structure) are $X^{k_1} \geq \dots$, one Jordan block $J_{k_i}(0)$ per part of a partition of $4$:

| partition | [Jordan form](#thm-b3-modules-jordan) | [invariant factors](#thm-b3-modules-smith) |
| --- | --- | --- |
| $(4)$ | $J_4$ | $X^4$ |
| $(3,1)$ | $J_3 \oplus J_1$ | $X,\ X^3$ |
| $(2,2)$ | $J_2 \oplus J_2$ | $X^2,\ X^2$ |
| $(2,1,1)$ | $J_2 \oplus J_1 \oplus J_1$ | $X,\ X,\ X^2$ |
| $(1,1,1,1)$ | $0$ | $X, X, X, X$ |

(b) Take $u = J_2\oplus J_2$ and $v = J_2 \oplus J_1 \oplus J_1$: both have $\chi = X^4$, $\mu = X^2$, but different [invariant factors](#thm-b3-modules-smith) — not similar ([Theorem 3.16](#thm-b3-modules-frobenius)); one can also compare ranks: $\operatorname{rk} u = 2 \neq 1 =
\operatorname{rk} v$. For $n \leq 3$: $\chi$ and $\mu$ determine, for each eigenvalue $\lambda$ (over a splitting field), the total size $m_\lambda \leq 3$ of the $\lambda$-blocks and the largest block $r_\lambda$; a partition of $m \leq 3$ is determined by its largest part ($m = 3, r = 2$ forces $(2,1)$, etc.). So the [elementary divisors](#thm-b3-modules-structure) coincide, and [Corollary 3.17](#cor-b3-modules-descent) descends the similarity to the base field.

**Exercise 3.9 ★★★.**

Let $u \in \mathcal L(V)$, $\dim V = n$. Show that the following are equivalent: (i) $V$ is a cyclic $K[X]$-module (there is $x$ with $V = K[u]x$, a *cyclic vector*); (ii) $\mu_u = \chi_u$; (iii) $s = 1$ in [Theorem 3.16](#thm-b3-modules-frobenius). Deduce that a [companion matrix](#def-b3-modules-companion) has a cyclic vector, and determine when a diagonal matrix has one.

**Solution of Exercise 3.9.**

(i)$\Rightarrow$(ii): if $V = K[u]x$, then $V \cong
K[X]/\operatorname{Ann}(x)$, and $\operatorname{Ann}(x) = (\mu_u)$ (a polynomial kills $x$ iff it kills all of $V = K[u]x$, since $P(u)Q(u)x = Q(u)P(u)x$). So $n = \dim V = \deg \mu_u$; as $\mu_u \mid \chi_u$ and $\deg\chi_u = n$, monicity gives $\mu_u =
\chi_u$.

(ii)$\Rightarrow$(iii): $\deg\chi_u = \sum_i \deg P_i$ and $\mu_u = P_s$ ([Theorem 3.16](#thm-b3-modules-frobenius)); equality of degrees forces $s = 1$.

(iii)$\Rightarrow$(i): $V \cong K[X]/(P_1)$ is cyclic, generated by the preimage of $\bar 1$.

A [companion matrix](#def-b3-modules-companion) is the case $V = K[X]/(P)$ itself: $x = \bar
1$, i.e. $e_1$, is cyclic. For a diagonal matrix $\operatorname{diag}(\lambda_1, \dots, \lambda_n)$: $\chi = \prod
(X - \lambda_i)$, $\mu = \prod_{\lambda \text{ distinct}} (X -
\lambda)$; they agree iff the $\lambda_i$ are pairwise distinct: a diagonal matrix has a cyclic vector iff its diagonal entries are pairwise distinct (then $x = (1, \dots, 1)$ works: Vandermonde).

**Exercise 3.10 ★★★.**

For $M \in M_n(\Z)$ viewed as an endomorphism of $\Z^n$, prove the index formula: if $\det M \ne 0$, then $[\Z^n : M\Z^n] = \abs{\det M}$, and deduce that $M \in GL_n(\Z)$ iff $\det M = \pm 1$. Application: the group $\Z^2$ has exactly $\sigma_1(m) = \sum_{d \mid m} d$ subgroups of index $m$. *(Count matrices in Hermite form $\bigl(\begin{smallmatrix}
a & b\\ 0 & d\end{smallmatrix}\bigr)$, $ad = m$, $0 \leq b <
d$.)*

**Solution of Exercise 3.10.**

Smith: $M = Q\operatorname{diag}(d_1,\dots,d_n)P$; [Exercise 3.4](#exo-b3-modules-4) gives $[\Z^n : M\Z^n] = \prod\abs{d_i} =
\abs{\det M}$. If $\det M = \pm 1$: the adjugate formula $M^{-1} = (\det M)^{-1}\,{}^{t}\!\operatorname{com}(M)$ has integer entries, so $M \in GL_n(\Z)$; conversely $MM^{-1} = I$ gives $\det M \cdot \det M^{-1} = 1$ in $\Z$, so $\det M = \pm1$.

Subgroups of index $m$ in $\Z^2$: such a subgroup $L$ has rank $2$ (finite index) and a unique basis in *Hermite normal form* $\bigl(\begin{smallmatrix} a & b\\ 0 &
d\end{smallmatrix}\bigr)$: $d$ is characterized by $L \cap (\{0\}
\times \Z) = \{0\} \times d\Z$, $a$ by $\pi_1(L) = a\Z$ (first coordinates), and $b$ is then unique modulo $d$; normalize $a, d
> 0$ and $0 \leq b < d$. The index is $ad = m$. Counting: for each divisor $d \mid m$ ($a = m/d$), there are $d$ choices of $b$: total $\sum_{d \mid m} d = \sigma_1(m)$.

**Exercise 3.11 ★★.**

(Equations $x^k = e$ in abelian groups) Let $G$ be a finite abelian group with [invariant factors](#thm-b3-modules-smith) $d_1 \mid d_2 \mid \dots
\mid d_s$. (a) Show that for every $k \geq 1$,

$$
\#\{x \in G : x^k = e\} \;=\; \prod_{i=1}^{s}\gcd(k, d_i) .
$$

(b) Deduce: a finite abelian group is cyclic if and only if for every $k$, the equation $x^k = e$ has at most $k$ solutions. (c) Recover the cyclicity of finite subgroups of $K^\times$ ($K$ a field, [Chapter 4](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#ch-b3-galois)): why does the polynomial $X^k - 1$ guarantee the criterion of (b)?

**Solution of Exercise 3.11.**

(a) By the structure theorem, $G \cong \prod_i\Z/d_i\Z$, and $x^k = e$ decouples coordinatewise. In $\Z/d\Z$: $kx \equiv 0
\pmod d$ has exactly $\gcd(k, d)$ solutions ($x$ must be a multiple of $d/\gcd(k,d)$, and there are $\gcd(k, d)$ of those). Multiply over the factors.

(b) If $G = \Z/d_s\Z$ is cyclic ($s = 1$), the count is $\gcd(k, d_s) \leq k$. If $s \geq 2$: take $k = d_1$; the count is $\prod_i\gcd(d_1, d_i) = d_1^{\,s} > d_1$ (each $\gcd$ equals $d_1$ by the divisibility chain): the equation $x^{d_1} = e$ has more than $d_1$ solutions.

(c) In a field, $X^k - 1$ has at most $k$ roots ([Chapter 2](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#ch-b3-rings): a nonzero polynomial of degree $k$ over a domain), so every finite subgroup $G \leq K^\times$ satisfies the criterion of (b): $G$ is cyclic — the one-line structural proof of the cyclicity of $\mathbb F_q^\times$, complementing the counting proof of [Chapter 4](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#ch-b3-galois).

**Exercise 3.12 ★★★.**

(Elementary matrices generate) (a) Show that $M \in M_n(\Z)$ is invertible in $M_n(\Z)$ iff $\det M = \pm1$. (b) Show that $SL_2(\Z)$ is generated by the two elementary matrices $E = \bigl(\begin{smallmatrix}1 & 1\\ 0 &
1\end{smallmatrix}\bigr)$ and $F = \bigl(\begin{smallmatrix}1 & 0\\ 1 &
1\end{smallmatrix}\bigr)$. *(Run the Euclidean algorithm on the first column of $M
\in SL_2(\Z)$ by left multiplications by powers of $E, F$, reaching $\pm\bigl(\begin{smallmatrix}1 & *\\ 0 &
1\end{smallmatrix}\bigr)$; finish by hand — note $-I = (EF^{-1}E)^2$.)* (c) Explain the connection with Smith reduction: over $\Z$, row and column operations of determinant $1$ suffice to diagonalize, up to signs.

**Solution of Exercise 3.12.**

(a) If $MN = I$ with integer $N$: $\det M\det N = 1$ with both integers, so $\det M = \pm1$. Conversely if $\det M =
\pm1$, the cofactor formula $M^{-1} =
\frac1{\det M}\,{}^t\!\operatorname{com}(M)$ has integer entries.

(b) Left multiplication by $E^{-k}$ subtracts $k$ times row $2$ from row $1$; by $F^{-k}$, $k$ times row $1$ from row $2$. Given $M = \bigl(\begin{smallmatrix}a & *\\ c &
*\end{smallmatrix}\bigr) \in SL_2(\Z)$, the first column $(a, c)$ is a unimodular vector ($\gcd(a, c) = 1$: it divides $\det M = 1$). Run Euclid on $(a, c)$ by these row operations: after finitely many steps the column becomes $(\pm1, 0)$. The matrix is now $\pm
\bigl(\begin{smallmatrix}1 & b\\ 0 & 1\end{smallmatrix}
\bigr) = \pm E^{b}$ (the determinant stayed $1$). It remains to write $-I$ in the generators: $(EF^{-1}E)^2 =
\bigl(\begin{smallmatrix}0 & 1\\ -1 &
0\end{smallmatrix}\bigr)^2 = -I$ (check the square of the rotation matrix). Unwinding, $M$ is a word in $E^{\pm1},
F^{\pm1}$.

(c) The Smith algorithm ([Method 3.9](#met-b3-modules-smith)) uses exactly such row and column operations (plus swaps and sign changes, themselves products of elementary operations up to determinant sign): over $\Z$, every matrix is $U\,D\,V$ with $U, V$ products of elementary matrices and $D$ the Smith form — (b) is the $2\times2$, determinant-$1$ instance of the general fact that $E$-type matrices generate $SL_n(\Z)$.

## 3.6 Problem: the commutant and the double commutant

**Problem 3.1.**

Weekend problem — rational form, commutant, bicommutant

Let $K$ be a field, $V$ a $K$-vector space of dimension $n \geq
1$, and $u \in \mathcal L(V)$. We study the *commutant*

$$
\mathcal C(u) = \{v \in \mathcal L(V) : uv = vu\},
$$

a subalgebra of $\mathcal L(V)$ containing $K[u] = \{P(u) : P \in
K[X]\}$, and we prove Frobenius’ dimension formula and the double commutant theorem: $\mathcal C(\mathcal C(u)) = K[u]$. Throughout, $V$ is the $K[X]$-module defined by $u$, with [invariant factors](#thm-b3-modules-smith) $P_1 \mid \cdots \mid P_s$ and cyclic decomposition $V =
\bigoplus_{i=1}^s V_i$, $V_i = K[u]\,x_i \cong K[X]/(P_i)$, $n_i
= \deg P_i$ ([Theorem 3.16](#thm-b3-modules-frobenius)).

**Part I — The presentation matrix $XI - M$, and warm-ups.**

1. Let $M \in M_n(K)$ and let $\varphi \colon K[X]^n \to V_M  = K^n$ send $(Q_1, \dots, Q_n)$ to $\sum_i Q_i(M)e_i$ . Show that $\varphi$ is a surjective morphism of $K[X]$ -modules and that every column of $XI_n - M$ lies in $\ker\varphi$ .
2. Show that, modulo the columns of $XI_n - M$ , every element of $K[X]^n$ is congruent to a constant vector *(reduce degrees using $Xe_i \equiv \sum_j m_{ji}e_j$)* , and deduce $\ker\varphi = (XI_n - M)\,K[X]^n$ : the [module](#def-b3-modules-module) $V_M$ has presentation matrix $XI_n - M$ . Recover [Corollary 3.17](#cor-b3-modules-descent) ’s starting point: the [similarity invariants](#thm-b3-modules-frobenius) of $M$ are the nonunit [invariant factors](#thm-b3-modules-smith) of $XI_n - M$ .
3. Compute the [similarity invariants](#thm-b3-modules-frobenius) of: a scalar matrix $\lambda I_n$ ; a diagonal matrix with distinct diagonal entries; the $n \times n$ Jordan block $J_n(0)$ ; $\operatorname{diag}(J_2(0), J_1(0))$ for $n = 3$ .
4. Show that $\dim K[u] = \deg \mu_u = n_s$ .

**Part II — Morphisms between cyclic [modules](#def-b3-modules-module).**

5. Let $P, Q$ be monic nonconstant. Show that a $K[X]$ -morphism $f \colon K[X]/(P) \to K[X]/(Q)$ is determined by $f(\bar 1)$ , and that $\bar c \in K[X]/(Q)$ can serve as $f(\bar 1)$ iff $P\bar c = 0$ in $K[X]/(Q)$ .
6. Deduce $$\operatorname{Hom}_{K[X]}\bigl(K[X]/(P),\, K[X]/(Q)\bigr)  \;\cong\; K[X]\big/\bigl(\gcd(P, Q)\bigr),$$ of dimension $\deg \gcd(P, Q)$ over $K$. *(Show that the solutions $\bar c$ of $P\bar c = 0$ in $K[X]/(Q)$ form the cyclic submodule generated by $\overline{Q/\gcd(P,Q)}$.)*
7. Prove *Frobenius’ formula*: $$\dim_K \mathcal C(u) = \sum_{i,j=1}^{s} \deg\gcd(P_i,  P_j) = \sum_{i=1}^{s} (2s - 2i + 1)\, n_i .$$ *(A commuting $v$ is exactly a $K[X]$-endomorphism of $V$; decompose $\operatorname{End}(\bigoplus_i V_i)$ as matrices of morphisms $V_j \to V_i$ and use the divisibility chain.)*
8. Deduce $\dim \mathcal C(u) \geq n$ , with equality iff $u$ is cyclic ( $s = 1$ ), and compute $\dim\mathcal C(u)$ for $u = \lambda\,\mathrm{id}$ : both extremes of the formula.
9. Verify Frobenius’ formula directly for $\operatorname{diag}(J_2(0), J_1(0))$ by computing the commutant explicitly as $3\times3$ matrices.

**Part III — The double commutant theorem.** Let $w \in \mathcal C(\mathcal C(u))$; we prove $w \in K[u]$.

10. Show $K[u] \subseteq \mathcal C(\mathcal C(u))$ , and that every $w \in \mathcal C(\mathcal C(u))$ commutes with $u$ — so the inclusion to be proved, $\mathcal C(\mathcal C(u)) \subseteq K[u]$ , is a genuine sharpening of $w \in \mathcal C(u)$ .
11. Suppose first that $u$ is *cyclic* , $V = K[u]x$ . Show directly that $\mathcal C(u) = K[u]$ *(evaluate a commuting $v$ on $x$: $v(x) = P(u)x$ for some $P$, and compare $v$ with $P(u)$ on the basis $u^k x$)* , and conclude the theorem in this case.
12. Back to the general case. For each $i$ , let $\pi_i\colon  V \to V_i$ be the projection along the other summands. Show $\pi_i \in \mathcal C(u)$ , and deduce that $w$ preserves each $V_i$ and commutes with $u_i =  u\restriction_{V_i}$ ; conclude via question 11 applied to the cyclic $u_i$ : there are polynomials $Q_i$ with $w\restriction_{V_i} = Q_i(u)\restriction_{V_i}$ .
13. It remains to glue the $Q_i$ into one polynomial. For $i  \leq j$ (so $P_i \mid P_j$ ), show that $\eta_{ij} \colon  V_j \to V_i$ , $R(u)x_j \mapsto R(u)x_i$ , is a well-defined $K[X]$ -morphism *(what must be checked is that $R(u)x_j = 0$ implies $R(u)x_i = 0$)* , and that $\tilde\eta_{ij} = \eta_{ij}\circ\pi_j$ , extended by $0$ on the other summands, lies in $\mathcal C(u)$ .
14. Using $w\tilde\eta_{ij} = \tilde\eta_{ij}w$, show $Q_i  \equiv Q_j \pmod{P_i}$ for $i \leq j$. Deduce that $Q =  Q_s$ satisfies $Q \equiv Q_i \pmod {P_i}$ for all $i$, hence $w = Q(u)$ on every $V_i$, hence on $V$: $$\boxed{\ \mathcal C(\mathcal C(u)) = K[u].\ }$$
15. (Coda) Deduce from the theorem: if $v$ commutes with *every* matrix commuting with $u$ , and $u$ is cyclic, then $v$ is a polynomial in $u$ ; and give an example showing $\mathcal C(u) = K[u]$ fails for $u =  \mathrm{id}$ , $n \geq 2$ — where exactly does cyclicity enter?

**Part IV — Dividends of the [similarity invariants](#thm-b3-modules-frobenius).** The rational canonical form is a machine; here are five of its classical outputs.

16. (Transpose) Show that every $M \in M_n(K)$ is similar to its transpose ${}^tM$ . *(The operations that bring $XI - M$ to Smith form, transposed, bring $XI - {}^tM$ to the same Smith form: equal [similarity invariants](#thm-b3-modules-frobenius).)*
17. (Descent of similarity) Let $K \subseteq L$ be a field extension and $M, N \in M_n(K)$ . Show that if $M$ and $N$ are similar over $L$ , they are similar over $K$ . *(The Smith form of $XI - M$ computed in $K[X]$ is still a Smith form in $L[X]$ — why do the [invariant factors](#thm-b3-modules-smith) not change?)* Consequence worth memorizing: two real matrices conjugate in $GL_n(\C)$ are conjugate in $GL_n(\R)$ .
18. (Nilpotent classification) Let $u$ be nilpotent. Show that the number of blocks of size $\geq k$ in its decomposition into nilpotent Jordan blocks equals $\operatorname{rk}u^{k-1} - \operatorname{rk}u^k$ , and deduce: nilpotent classes of $M_n(K)$ , for any field $K$ , are in bijection with the partitions of $n$ . How many nilpotent classes in $M_5(K)$ ?
19. (A concrete pair) Determine the [similarity invariants](#thm-b3-modules-frobenius) of the derivation $D\colon P \mapsto P'$ acting on the space $K_{n-1}[X]$ of polynomials of degree $< n$ : (a) for $K =  \Q$ ; (b) for $K = \mathbb F_p$ with $p < n$ *(in characteristic $p$, $(X^p)' = 0$: compute $\ker D^k$ and use question 18)* .
20. ( [Conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of $GL_2(\mathbb F_q)$ ) Using [invariant factors](#thm-b3-modules-smith) , show that every class of $GL_2(\mathbb F_q)$ is of exactly one of four types: central $aI$ ; diagonalizable with two distinct eigenvalues $a \neq b$ in $\mathbb F_q^\times$ ; non-semisimple with minimal polynomial $(X - a)^2$ ; cyclic with [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) characteristic polynomial.
21. Count the classes of each type and conclude: $GL_2(\mathbb F_q)$ has exactly $q^2 - 1$ [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) . *(Count monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) quadratics over $\mathbb F_q$; unordered pairs $\{a, b\}$; remember invertibility constrains constant terms.)*
22. (Cyclic is generic) Show that $M \in M_2(\mathbb F_q)$ fails to be cyclic iff $M$ is scalar, and deduce that a uniformly random $2\times2$ matrix over $\mathbb F_q$ is cyclic with probability $1 - q^{-3}$ . State the analogous heuristic for $M_n$ and large $q$ (no proof required): non-cyclic matrices are rare — which is why [Problem 3.1](#pb-b3-modules-1) ’s Part III needed real work only past the generic case.

**Part V — Complements.**

23. ( [Center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of the commutant) Show that the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of the algebra $\mathcal C(u)$ is exactly $K[u]$ *(combine the two inclusions of Part III)* . Deduce that $\mathcal C(u)$ is commutative iff $u$ is cyclic — recovering the equality case of question 8 by a purely structural route.
24. (Which dimensions occur?) Deduce from Frobenius’ formula that $\dim\mathcal C(u) \equiv n \pmod 2$ for every $u$ . Then determine the exact set of values taken by $\dim \mathcal C(u)$ as $u$ ranges over $\mathcal L(V)$ with $\dim V = 4$ : show it is $\{4, 6,  8, 10, 16\}$ *(enumerate the degree sequences $n_1  \leq \dots \leq n_s$ summing to $4$ and realize each by a nilpotent)* . In particular $12$ and $14$ , though of the right parity, are not attained: the parity constraint is necessary but not sufficient.
25. ([Class equation](https://one-course.com/books/math/5/en/chapter/1-group-theory#cor-b3-groups-classeq) of $GL_2(\mathbb F_3)$) For $q = 3$, compute the size of each [conjugacy class](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of question 20 via orbit–stabilizer: the centralizer of a *cyclic* $M$ in $GL_2(\mathbb F_q)$ is the unit group of $K[M]$ (question 11). Identify $K[M]$ in the three noncentral types, list the three monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) quadratics over $\mathbb F_3$, and verify the [class equation](https://one-course.com/books/math/5/en/chapter/1-group-theory#cor-b3-groups-classeq) $$48 = \abs{GL_2(\mathbb F_3)}  = 2\cdot1 + 1\cdot12 + 2\cdot8 + 3\cdot6,$$ with $2 + 1 + 2 + 3 = 8 = q^2 - 1$ classes, as predicted by question 21.

**Solution of Problem 3.1.**

**1.** $\varphi$ is additive, and $K[X]$-linear: $\varphi(X
\cdot (Q_i)_i) = \sum_i (XQ_i)(M)e_i = M\sum_i Q_i(M)e_i = X
\cdot \varphi\bigl((Q_i)_i\bigr)$, the [module](#def-b3-modules-module) structure of $V_M$ being $X \cdot v = Mv$. It is surjective: constant vectors give all of $K^n$. Column $j$ of $XI - M$ is $Xe_j - \sum_i
m_{ij}e_i$, whose image is $Me_j - \sum_i m_{ij}e_i = 0$.

**2.** Modulo the columns, $Xe_j \equiv \sum_i m_{ij}e_i$: any vector of polynomials reduces, by induction on the top degree, to a *constant* vector $c \in K^n$. If the original vector is in $\ker\varphi$, then $\varphi(c) = c = 0$ (on constants, $\varphi$ is the identification $K^n = V_M$), so the vector lies in the column span: $\ker\varphi = (XI_n -
M)K[X]^n$. Hence $V_M \cong K[X]^n/(XI - M)K[X]^n$, and Smith over $K[X]$ (all [invariant factors](#thm-b3-modules-smith) nonzero, their product being $\det(XI - M) = \chi_M$) gives $V_M \cong \bigoplus_i
K[X]/(f_i)$: the nonconstant $f_i$ are the [similarity invariants](#thm-b3-modules-frobenius), computable as quotients of minor gcds ([Theorem 3.8](#thm-b3-modules-smith)).

**3.** $\lambda I_n$: $XI - \lambda I$ is already Smith: invariants $(X - \lambda, \dots, X - \lambda)$, $n$ of them. Distinct diagonal entries: $V \cong \bigoplus_i K[X]/(X -
\lambda_i)$ with pairwise [comaximal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#thm-b3-rings-crt) moduli, so CRT compresses to the single cyclic $K[X]/\bigl(\prod_i(X - \lambda_i)\bigr)$: one invariant, $\chi$. $J_n(0)$: $\mu = X^n = \chi$ forces a single invariant $X^n$. $\operatorname{diag}(J_2(0), J_1(0))$: [elementary divisors](#thm-b3-modules-structure) $X^2, X$: invariants $P_1 = X \mid P_2 =
X^2$.

**4.** $P \mapsto P(u)$ maps $K[X]$ onto $K[u]$, with kernel $(\mu_u)$ by definition of the minimal polynomial: $K[u]
\cong K[X]/(\mu_u)$, of dimension $\deg\mu_u = \deg P_s = n_s$.

**5.** $K[X]$-linearity forces $f(\bar Q) = f(Q\cdot\bar 1)
= Q\,f(\bar 1)$. The class $\bar 1$ satisfies $P \bar 1 = 0$, so $P f(\bar 1) = 0$ is necessary. Conversely if $P\bar c = 0$, then $f(\bar Q) = Q\bar c$ is well defined ($Q \equiv Q' \bmod P
\Rightarrow (Q - Q')\bar c \in$ multiples of $P\bar c = 0$) and $K[X]$-linear.

**6.** Let $g = \gcd(P, Q)$, $P = gP'$, $Q = gQ'$ with $\gcd(P', Q') = 1$. In $K[X]/(Q)$: $P\bar c = 0 \iff Q \mid Pc
\iff Q' \mid P'c \iff Q' \mid c$ (Euclid, $\gcd(P', Q') = 1$). So the admissible $\bar c$ form the submodule generated by $\bar{Q'}$, whose annihilator is $\{R : Q \mid RQ'\} = (g)$: that submodule is $\cong K[X]/(g)$. With question 5, $\operatorname{Hom}(K[X]/(P), K[X]/(Q)) \cong K[X]/(\gcd(P,Q))$, of dimension $\deg\gcd(P,Q)$.

**7.** $v$ commutes with $u$ iff $v$ commutes with every $P(u)$, iff $v$ is $K[X]$-linear: $\mathcal C(u) =
\operatorname{End}_{K[X]}(V)$. Writing morphisms of $V =
\bigoplus_j V_j$ as matrices $(f_{ij})$, $f_{ij} \in
\operatorname{Hom}(V_j, V_i)$ (compose with injections and projections), question 6 gives

$$
\dim \mathcal C(u) = \sum_{i,j} \deg\gcd(P_i, P_j)
= \sum_{i,j} n_{\min(i,j)}
= \sum_{k=1}^{s} \bigl(2(s - k) + 1\bigr)\,n_k ,
$$

using the divisibility chain ($\gcd(P_i, P_j) =
P_{\min(i,j)}$) and, for the last step, that $\min(i,j) = k$ happens for exactly $2(s - k) + 1$ pairs $(i, j)$.

**8.** Since $2(s-k)+1 \geq 1$, $\dim\mathcal C(u) \geq
\sum_k n_k = n$, with equality iff $s = 1$, i.e. iff $u$ is cyclic ([Exercise 3.9](#exo-b3-modules-9)). For $u = \lambda\,\mathrm{id}$: $s = n$, all $n_k = 1$: $\dim = \sum_{k=1}^n (2(n-k)+1) = n^2$ — correct, since $\mathcal C(\lambda\,\mathrm{id}) = \mathcal
L(V)$.

**9.** Formula: invariants $(X, X^2)$, so $s = 2$, $n_1 =
1$, $n_2 = 2$: $\dim = 3\cdot1 + 1\cdot2 = 5$. Directly: in the basis $(e_1, e_2, e_3)$ with $u e_2 = e_1$, $ue_1 = ue_3 = 0$, writing $Au = uA$ for $A = (a_{ij})$ yields the conditions $a_{21} = a_{23} = a_{31} = 0$ and $a_{11} = a_{22}$: five [free](#def-b3-modules-free) parameters $a_{11}{=}a_{22}, a_{12}, a_{13}, a_{32}, a_{33}$.

**10.** A polynomial $P(u)$ commutes with anything that commutes with $u$ (it is a sum of powers of $u$): $K[u]
\subseteq \mathcal C(\mathcal C(u))$. And $u \in \mathcal C(u)$, so any $w \in \mathcal C(\mathcal C(u))$ commutes with $u$.

**11.** Let $V = K[u]x$ and $v \in \mathcal C(u)$. Write $v(x) = P(u)x$ (cyclicity). For $y = Q(u)x$ arbitrary: $v(y) =
vQ(u)x = Q(u)v(x) = Q(u)P(u)x = P(u)y$. So $v = P(u)$: $\mathcal
C(u) = K[u]$. Then $\mathcal C(\mathcal C(u)) = \mathcal
C(K[u]) = \mathcal C(u) = K[u]$ (commuting with all of $K[u]$ is the same as commuting with $u$). The theorem holds in the cyclic case.

**12.** $\pi_i$ is $K[X]$-linear (the decomposition is a direct sum of submodules), so $\pi_i \in \mathcal C(u)$, and $w$ commutes with it: $w(V_i) = w\pi_i(V) = \pi_i w(V) \subseteq
V_i$. The restriction $w_i = w\restriction_{V_i}$ commutes with the cyclic $u_i = u\restriction_{V_i}$ (question 10), and $w_i
\in \mathcal C(u_i) = K[u_i]$ (question 11): $w_i = Q_i(u_i) =
Q_i(u)\restriction_{V_i}$ for some $Q_i \in K[X]$.

**13.** Well-definedness of $\eta_{ij}(R(u)x_j) = R(u)x_i$: if $R(u)x_j = 0$ then $P_j \mid R$, and $P_i \mid P_j$ gives $P_i \mid R$, so $R(u)x_i = 0$ ($\operatorname{Ann}(x_i) =
(P_i)$). $\eta_{ij}$ is then $K[X]$-linear by construction, and $\tilde\eta_{ij} = \eta_{ij}\pi_j$ is a composition of $K[X]$-morphisms $V \to V$: $\tilde\eta_{ij} \in \mathcal
C(u)$.

**14.** Evaluate $w\tilde\eta_{ij} = \tilde\eta_{ij}w$ at $x_j$: the left side is $w(x_i) = Q_i(u)x_i$; the right side is $\eta_{ij}\bigl(Q_j(u)x_j\bigr) = Q_j(u)x_i$. Hence $(Q_i -
Q_j)(u)\,x_i = 0$: $P_i \mid Q_i - Q_j$ for all $i \leq j$. In particular, with $Q = Q_s$: $Q \equiv Q_i \pmod{P_i}$, so $Q(u)$ and $Q_i(u)$ agree on $V_i$ (which $P_i(u)$ kills). Therefore $w
= Q(u)$ on each $V_i$, hence on $V$: $\mathcal C(\mathcal C(u))
\subseteq K[u]$, and with question 10, $\mathcal C(\mathcal
C(u)) = K[u]$.

**15.** The first assertion is questions 10–14 (or, for cyclic $u$, question 11 alone). For $u = \mathrm{id}$, $n \geq
2$: $\mathcal C(u) = \mathcal L(V)$ has dimension $n^2$, while $\dim K[u] = \deg\mu_u = 1$. So $\mathcal C(u) = K[u]$ fails badly; yet the bicommutant theorem holds ($\mathcal C(\mathcal L(V)) = K\,\mathrm{id} = K[u]$: the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of the matrix algebra is the scalars). Cyclicity is what makes the *single* commutant already polynomial; the *double* commutant is polynomial always.

**16.** If $P(XI - M)Q = S$ is a Smith reduction ($P, Q$ invertible over $K[X]$), transposing gives ${}^tQ\,(XI -
{}^tM)\,{}^tP = {}^tS = S$: same Smith form, so $XI - M$ and $XI - {}^tM$ have the same [invariant factors](#thm-b3-modules-smith), i.e. $M$ and ${}^tM$ have the same [similarity invariants](#thm-b3-modules-frobenius) ([Corollary 3.17](#cor-b3-modules-descent)): they are similar.

**17.** The [similarity invariants](#thm-b3-modules-frobenius) of $M$ over $L$ are the [invariant factors](#thm-b3-modules-smith) of $XI - M$ in $L[X]$. A Smith reduction of $XI - M$ over $K[X]$ — invertible $P, Q$ over $K[X]$, diagonal with the divisibility chain — is *also* a valid Smith reduction over $L[X]$ ($P, Q$ stay invertible: their determinants are nonzero constants), and monic [invariant factors](#thm-b3-modules-smith) are unique: the [invariant factors](#thm-b3-modules-smith) computed over $K$ and over $L$ coincide. So $M \sim_L N$ iff they have the same [invariant factors](#thm-b3-modules-smith) iff $M \sim_K N$. In particular $\C$-conjugate real matrices are $\R$-conjugate — a statement often proved analytically (specialize an invertible $P + \iu
Q$), here structurally.

**18.** Decompose $u = \bigoplus J_{m_t}(0)$ into nilpotent Jordan blocks. In one block of size $m$, $\operatorname{rk}J_m^k = \max(m - k, 0)$, so $\operatorname{rk}J_m^{k-1} - \operatorname{rk}J_m^k = 1$ if $m
\geq k$, $0$ otherwise. Summing over blocks: $\operatorname{rk}u^{k-1} - \operatorname{rk}u^k =
\#\{t : m_t \geq k\}$. The rank sequence therefore determines the multiset $(m_t)$ — a partition of $n$ — and conversely each partition is realized: nilpotent classes $\leftrightarrow$ partitions of $n$, over every field. $M_5$: $p(5) = 7$ classes ($5$; $4{+}1$; $3{+}2$; $3{+}1{+}1$; $2{+}2{+}1$; $2{+}1{+}1{+}1$; $1^5$).

**19.** (a) Over $\Q$ (or any characteristic-$0$ field), $D^n = 0$, $D^{n-1}(X^{n-1}) = (n-1)!\, \neq 0$: $D$ is nilpotent of index $n$ on an $n$-dimensional space, hence cyclic with single invariant $X^n$ ($x = X^{n-1}$ generates: its iterated derivatives span). (b) Over $\mathbb F_p$ with $p
< n$: $D^p = 0$, because the $p$-th derivative of every monomial $X^m$ carries the factor $m(m-1)\cdots(m-p+1)$, a product of $p$ consecutive integers, hence $\equiv 0 \bmod p$. Write $n = ap + r$, $0 \leq r < p$. Then $\ker D^k$ is spanned by the monomials $X^m$ with $D^kX^m = 0$; counting exponents $m < n$ by their residue mod $p$: $\dim\ker D^k = ak + \min(r,
k)$ for $0 \leq k \leq p$, so $\operatorname{rk}D^{k-1} -
\operatorname{rk}D^k = \dim\ker D^k - \dim\ker D^{k-1} = a +
\mathbf 1_{k \leq r}$. By question 18, the partition has $a$ blocks of size exactly $p$ and (if $r > 0$) one block of size $r$: [similarity invariants](#thm-b3-modules-frobenius) $X^r \mid X^p \mid \dots \mid X^p$. Characteristic changes the canonical form of the most familiar operator in mathematics.

**20.** $M \in GL_2$ has $s \in \{1, 2\}$ [invariant factors](#thm-b3-modules-smith). If $s = 2$: $P_1 = P_2 = X - a$ ($a \neq 0$: invertibility), i.e. $M = aI$, central. If $s = 1$: $M$ is cyclic with characteristic $=$ minimal polynomial $\chi$ of degree $2$, and the classes correspond to the possible $\chi$ with $\chi(0) \neq 0$: $\chi$ split with distinct roots $a \neq
b$ (companion $\sim$ diagonal); $\chi = (X - a)^2$ (companion, non-semisimple); $\chi$ [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility). Exactly one type each — the [invariant factors](#thm-b3-modules-smith) are a complete invariant.

**21.** Central: $q - 1$ choices of $a$. Distinct split eigenvalues: unordered pairs $\{a, b\} \subseteq
\mathbb F_q^\times$, $a \neq b$: $\binom{q-1}2$ classes. Minimal $(X-a)^2$: $q - 1$ classes. [Irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) quadratics with nonzero constant term: all [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) quadratics qualify (their roots are nonzero), and there are $\frac{q^2 - q}2$ monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) quadratics (the $q^2$ monic quadratics minus the $\binom q2 + q = \frac{q^2+q}2$ split ones). Total:

$$
(q - 1) + \frac{(q-1)(q-2)}2 + (q - 1) + \frac{q^2 - q}2
= q^2 - 1 .
$$

**22.** If $M$ is not cyclic, $s = 2$ and $M$ is scalar (question 20’s dichotomy holds in $M_2$, invertible or not: two [invariant factors](#thm-b3-modules-smith) of degree $1$ with $P_1 \mid P_2$ and $P_1 =
P_2$ forces $M = aI$). Scalars number $q$ among the $q^4$ matrices: cyclic probability $1 - q^{-3}$. In general the non-cyclic locus of $M_n$ is where the $(n-1)\times(n-1)$ minors of $XI - M$ share a factor — a proper algebraic condition — so its proportion is $O(1/q)$-small for large $q$: matrices with $\chi = \mu$ are the rule, and Part III’s gluing argument is the price paid for the exceptions.

**23.** An element of the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of $\mathcal C(u)$ lies in $\mathcal C(u)$ and commutes with every element of $\mathcal C(u)$, i.e. lies in $\mathcal C(\mathcal C(u)) =
K[u]$ (question 14). Conversely $K[u] \subseteq \mathcal C(u)$, and every $P(u)$ commutes with every $v \in \mathcal C(u)$ (such a $v$ commutes with $u$, hence with each power of $u$): $K[u]$ is central in $\mathcal C(u)$. Hence $Z(\mathcal C(u))
= K[u]$. Consequently $\mathcal C(u)$ is commutative iff $\mathcal C(u) = Z(\mathcal C(u)) = K[u]$; in that case $\dim\mathcal C(u) = \dim K[u] = n_s \leq n$, while question 8 gives $\dim\mathcal C(u) \geq n$: so $n_s = n$ and $s = 1$, i.e. $u$ is cyclic. Conversely, for $u$ cyclic question 11 gives $\mathcal C(u) = K[u]$, commutative. Structurally: a matrix algebra equal to its own [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) is exactly a polynomial algebra $K[u]$ of a cyclic $u$.

**24.** Each coefficient $2s - 2i + 1$ in Frobenius’ formula is odd, so

$$
\dim\mathcal C(u) = \sum_{i=1}^s(2s - 2i + 1)\,n_i
\equiv \sum_{i=1}^s n_i = n \pmod 2 .
$$

For $n = 4$, the possible degree sequences $n_1 \leq \dots
\leq n_s$ of the [invariant factors](#thm-b3-modules-smith), summing to $4$, are $(4)$, $(1, 3)$, $(2, 2)$, $(1, 1, 2)$, $(1, 1, 1, 1)$; every one is realized, e.g. by the nilpotent with $P_i = X^{n_i}$ (the divisibility chain holds automatically). The formula gives, respectively,

$$
1\cdot4 = 4, \quad 3 + 3 = 6, \quad 6 + 2 = 8, \quad
5 + 3 + 2 = 10, \quad 7 + 5 + 3 + 1 = 16 .
$$

So the value set is $\{4, 6, 8, 10, 16\}$: even numbers of the right parity, but $12$ and $14$ never occur — between the almost-cyclic sequences and the scalar’s $n^2$ there is a gap.

**25.** $\abs{GL_2(\mathbb F_3)} = (q^2 - 1)(q^2 - q) =
8 \cdot 6 = 48$. Central type: $I$ and $2I$, two classes of size $1$. In the three other types $M$ is cyclic (question 20), so its centralizer in $GL_2$ is the group of invertible elements of $\mathcal C(M) = K[M]$ (question 11), and class size $=48/\abs{K[M]^\times}$ by orbit–stabilizer. Distinct split eigenvalues: only the pair $\{1, 2\}$, one class; $K[M]
\cong \mathbb F_3 \times \mathbb F_3$ (CRT on $\chi = (X-1)
(X-2)$), units $2 \cdot 2 = 4$, size $48/4 = 12$. Minimal $(X - a)^2$, $a \in \{1, 2\}$: two classes; $K[M] \cong
\mathbb F_3[X]/((X-a)^2)$, units $q^2 - q = 6$ (constant term of the unit $\neq 0$ after centering), size $48/6 = 8$. [Irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) $\chi$: the monic [irreducible](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-divisibility) quadratics over $\mathbb F_3$ number $(9 - 3)/2 = 3$, namely

$$
X^2 + 1, \qquad X^2 + X + 2, \qquad X^2 + 2X + 2,
$$

(no roots in $\mathbb F_3$: check $0, 1, 2$); three classes, $K[M] \cong \mathbb F_9$, units $q^2 - 1 = 8$, size $48/8 =
6$. [Class equation](https://one-course.com/books/math/5/en/chapter/1-group-theory#cor-b3-groups-classeq): $2\cdot1 + 1\cdot12 + 2\cdot8 + 3\cdot6 =
2 + 12 + 16 + 18 = 48$; and $2 + 1 + 2 + 3 = 8 = q^2 - 1$ classes, matching question 21.