§19 Principal Ideal Domains (PIDs)

§19.1 Polynomial Rings

Let $F$ be a field, and let $F[t]$ be the polynomial ring.

Theorem 19.1 If $I\subset F[t]$ is an ideal, then there exists $p(t)\in F[t]$ such that

I=(p(t)).

That is, every ideal is generated by a single element.

Proof: If

I=(0),

then we are done.

Thus we may assume that $I$ contains an element of degree $\geqslant0$ .

Let $p(t)$ be an element of minimal degree in $I$ :

p(t) = a_0+a_1t+\cdots+a_dt^d, \quad d>-\infty.

Since

p(t)\in I,

p(t)\subset I.

Now let

f(t)\in I.

Consider the division algorithm:

where

f(t)=b_nt^n+\cdots+b_0

and

Q_{n-1} = b_{n-1}-(a_d)^{-1}a_{n-1}b_n.

Then

f(t) = p(t)q(t)+r(t),

where

\deg(r(t)) < \deg(p(t)).

But $p(t)$ already has minimal degree among the elements of $I$ , so

r(t)=0.

~\tag*{$\square$}

Definition 19.1 If

f(t)\in F[t],

we say that $f(t)$ is irreducible or prime if

f(t)=a(t)b(t)

implies that either $a(t)$ or $b(t)$ is a constant polynomial.

That is, there is no polynomial of degree $d$ with

0<d<\deg f

that divides $f$ .

Theorem 19.2 Every $f(t)$ can be factored as a product of irreducible polynomials.

Proof: We prove this by induction.

For degree $0$ , i.e. constant polynomials, $f(t)$ is either $0$ or a unit, so the factorization is trivial.

Assume that every polynomial of degree less than $n$ can be factored into irreducible polynomials.

Now let $f(t)$ have degree $n$ .

If $f(t)$ is irreducible, we are done.
Otherwise,

f(t)=g(t)h(t),

where

\deg(g),\deg(h)<n.

By the induction hypothesis, both $g(t)$ and $h(t)$ can be factored into irreducible polynomials.

Therefore every $f(t)$ can be factored as a product of irreducible polynomials over the field $F$ .

~\tag*{$\square$}

§19.2 The Similarity Between the Ring of Integers $\mathbb{Z}$ and the Polynomial Ring $F[t]$

Exercise Let $R$ be a commutative ring, and let

x_1,\ldots,x_n

be a finite collection of elements.

Define the ideal generated by $x_1,\ldots,x_n$ and prove that it is an ideal.

Solution: Since there are $n$ elements of $R$ , they uniquely define a module homomorphism

R^{\oplus n}\to R.

We define the ideal generated by $x_1,\ldots,x_n$ to be the image of this homomorphism.

The image is a submodule of $R$ , and by definition a submodule of $R$ is an ideal.

Definition 19.2 We use

(x_1,\ldots,x_n)\subset R

to denote the ideal generated by the elements $x_1,\ldots,x_n$ .

Explicitly, it is the set of all elements of $R$ that can be written as

a_1x_1+\ldots+a_nx_n,

where

a_i\in R.

Let $F$ be a field.

Our goal is to show that $\mathbb{Z}$ and $F[t]$ are very similar rings.

At first sight, this may seem surprising, but we will explain what this means.

There is an important analogy between:

(1) the size of an integer, or the $\log$ of its size, and

(2) the degree of a polynomial.

For example, for any two integers $x,y\in\mathbb{Z}$ ,

\log(|xy|) = \log|x| + \log|y|.

On the other hand, for any two polynomials in $F[t]$ ,

\deg(fg) = \deg f+\deg g.

The size of integers allows us to use induction when proving statements about all integers.

Although we used $\log$ above, $\log$ preserves the ordering of numbers, so the multiplicative property above is still useful in inductive proofs.

Similarly, polynomial degree allows us to prove statements about all polynomials by induction.

§19.3 Review of Some Definitions

Definition 19.3 Let $R$ be a commutative ring.

A zero divisor is an element $x\in R$ such that

xy=0

for some

y\neq0.

Here are some simple examples.

(1) If $R$ contains more than one element, then $0$ is always a zero divisor because

0\cdot y=0

for every $y\in R$ .

We need $R$ to contain more than one element so that we can choose a nonzero $y$ .

(2) Let

R=\mathbb{Z}/n\mathbb{Z},

where $n$ is not prime.

Then we can choose two integers $x$ and $y$ such that

xy=n,

with neither $x$ nor $y$ equal to $\pm1$ .

Thus in $R$ , $\overline{x}$ and $\overline{y}$ are both zero divisors because

\overline{x}\neq0, \qquad \overline{y}\neq0,

but

\overline{xy} = \overline{n} = 0.

Definition 19.4 A commutative ring $R$ is called a principal ideal domain, or PID, if:

(1) for every ideal

I\subset R,

there exists some $x\in R$ such that

I=(x);

(2) the only zero divisor in $R$ is $0$ .

Remark The word “domain” means that there are no nonzero zero divisors.

You will sometimes hear the term integral domain, meaning a commutative ring with no nonzero zero divisors.

The “principal ideal” part means that every ideal is principal, i.e. generated by one element.

Example 19.2 The two most important examples of principal ideal domains are:

(1)

R=\mathbb{Z}.

We know that every subgroup of $\mathbb{Z}$ has the form

n\mathbb{Z},

where $n$ is an integer.

Moreover,

n\mathbb{Z} = (n),

because by definition every element of $n\mathbb{Z}$ is a multiple of $n$ by some integer $a$ .

Since every ideal is, in particular, a subgroup of $R$ , we conclude that every ideal in $\mathbb{Z}$ is principal.

(2)

R=F[t],

where $F$ is a field.

By the theorem, $F[t]$ is a PID.

Theorem 19.3 Let $F$ be a field. Then every ideal

I\subset F[t]

is generated by a single element.

Proof: Since $F$ is a field, the polynomial ring $F[t]$ has a division algorithm:

for any polynomials $f$ and $g$ with $g\neq0$ , there exist unique polynomials $q$ and $r$ such that

f=qg+r,

where either

r=0

\deg(r)<\deg(g).

Now let $I$ be an ideal of $F[t]$ .

I=\{0\},

then $I$ is generated by $0$ .

Suppose

I\neq\{0\}.

Consider the set of degrees of nonzero polynomials in $I$ :

D = \{\deg(f)\mid f\in I,\ f\neq0\}.

Since $D$ is a nonempty subset of $\mathbb{N}_0$ , it has a least element $d$ .

Choose

g\in I

with

\deg(g)=d.

We claim

I=(g).

First, since

g\in I,

(g)\subseteq I.

Conversely, for any

f\in I,

the division algorithm gives

f=qg+r,

where

\deg(r)<\deg(g)

r=0.

Since

f,qg\in I,

because $I$ is an ideal,

r=f-qg\in I.

If $r\neq0$ , then

\deg(r)<\deg(g)=d,

contradicting the minimality of $d$ .

Therefore,

r=0,

f=qg\in(g).

Hence

I=(g),

which proves that every ideal of $F[t]$ is principal.

~\tag*{$\square$}

§19.4 The Euclidean Algorithm

An important reason why $\mathbb{Z}$ and $F[t]$ are very similar rings is that both have a division-with-remainder algorithm, namely the Euclidean algorithm.

Recall the following proposition, which we have known since elementary school.

Theorem 19.4 (Division with Remainder for Integers)

Let $x$ be an integer and let $n$ be any integer.

Then there exist integers $q$ and $r$ such that

x=nq+r,

where

0\leqslant r<n.

Remark We used this proposition extensively when proving that the only subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ .

The analogous statement for polynomials replaces the size of an integer by the degree of a polynomial.

Theorem 19.5 Let $F$ be a field, and let

g\in F[t]

be a polynomial.

For every polynomial

f\in F[t],

there exist polynomials

q,r\in F[t]

such that

g=fq+r,

where

0\leqslant\deg r<\deg f.

Remark This means that we can always divide one polynomial $g$ by another polynomial $f$ and look at the remainder.

Proof: If

\deg g<\deg f,

then we are done by simply taking

g=f0+g.

That is, we cannot divide a polynomial of lower degree by one of higher degree in a nontrivial way, so the only division is the trivial one and the remainder is $g$ itself.

Thus we need only prove the case

\deg g\geqslant\deg f.

We proceed by induction on the degree of $g$ .

Fix the polynomial $f$ .

We already know that the statement holds for every $g$ satisfying

\deg g<\deg f.

This is the base case.

Assume that the statement holds for all $g$ with

\deg g\leqslant e-1.

We need to prove it for

\deg g=e.

Let

f = a_dt^d+\ldots+a_1t+a_0, \qquad a_d\neq0,

and

g = b_et^e+\ldots+b_1t+b_0, \qquad b_e\neq0,

so that $f$ and $g$ have degrees $d$ and $e$ , respectively.

Since $a_d$ and $b_e$ are nonzero elements of the field $F$ , there exists a unique number $q_{e-d}$ such that

q_{e-d}a_d = b_e.

Consider the polynomial

q_{e-d}f = b_et^d + q_{e-d}a_{d-1}t^{d-1} + \ldots + q_{e-d}a_1t + q_{e-d}a_0.

Multiplying this polynomial by $t^{e-d}$ gives

q_{e-d}t^{e-d}f = b_et^e + q_{e-d}a_{d-1}t^{e-1} + \ldots + q_{e-d}a_1t^{e-d+1} + q_{e-d}a_0t^{e-d}.

Notice that this polynomial has the same degree as $g$ , and $g$ has the same leading coefficient $b_e$ .

Therefore, we can subtract it from $g$ to obtain a polynomial of smaller degree:

g^{\prime} := g-q_{e-d}t^{e-d}f.

By the induction hypothesis on degree, there exist polynomials $Q$ and $r$ such that

g^{\prime} = Qf+r,

where

\deg r<\deg f.

Thus we can write

\begin{aligned} g &= \left(q_{e-d}t^{d-e}\right)f + g- \left(q_{e-d}t^{d-e}\right)f\\ &= \left(q_{e-d}t^{d-e}\right)f + g'\\ &= \left(q_{e-d}t^{d-e}\right)f + Qf+r\\ &= \left(q_{e-d}t^{d-e}+Q\right)f+r. \end{aligned}

Set

q = q_{e-d}t^{t-e}+Q.

Then

g=qf+r,

where

\deg r<\deg f.

This completes the proof.

~\tag*{$\square$}

§19.5 Prime Elements and Factorization in Principal Ideal Domains

Definition 19.5 An element $x\in R$ is called a unit if there exists $y\in R$ such that

xy=yx=1_R.

Example 19.3 In $\mathbb{Z}$ , the units are

\pm1.

Similarly, if $F$ is a field, then the units of $F$ are precisely the nonzero elements of $F$ .

Proposition 19.6 Let

R=F[t].

Then the units of $R$ are precisely the nonzero constant polynomials.

Proof: If

fg=1,

then

\deg f+\deg g = \deg1 = 0.

Therefore, both $f$ and $g$ must have degree $0$ , so both must be constant polynomials.

But the constant polynomials form the subring

F\subset F[t].

Thus two constant polynomials multiply to $1$ if and only if both are nonzero.

~\tag*{$\square$}

We now want to generalize the notion of prime numbers in $\mathbb{Z}$ to arbitrary rings.

Definition 19.6 In a ring $R$ , an element $x$ is called prime or irreducible if

(1) $x$ is not a unit;

(2) the only elements dividing $x$ are units or unit multiples of $x$ .

That is, if

x=ab,

for $a,b\in R$ , then either $a$ or $b$ must be a unit.

Example 19.4 Here are some examples of prime elements in rings.

(1) Let

R=\mathbb{Z}.

If $x$ is a prime number or the negative of a prime number, then the only numbers dividing $x$ are

\pm1

and

\pm x.

Thus the prime elements of $\mathbb{Z}$ are precisely the primes and their negatives.

Notice that zero is not a unit.

(2) Let

R=F[t].

The only units in $F[t]$ are nonzero constant polynomials.

Therefore, $f$ is prime or irreducible if and only if every polynomial dividing $f$ either has the same degree as $f$ or is a constant polynomial.

(3) For example, if

\deg f=1,

then $f$ is irreducible.

Indeed, if

gh=f,

then

\deg g+\deg h = \deg f = 1.

Thus either $\deg g=0$ or $\deg h=0$ .

Hence every linear polynomial is irreducible.

Theorem 19.7 (Unique Factorization in Principal Ideal Domains)

Let $R$ be a principal ideal domain.

For every nonzero element

x\in R,

there exists a finite collection of distinct prime elements

p_1,\ldots,p_k\in R

such that

x = p_1^{n_1} p_2^{n_2} \ldots p_k^{n_k}, \qquad n_i\geqslant1,

and for $i\neq j$ , $p_i$ and $p_j$ are not unit multiples of one another.

The exponents $n_i$ are unique, and the elements $p_i$ are unique up to order and multiplication by units.

Proof: Let

x\in R, \qquad x\neq0.

If $x$ is prime, then we are done: take

p_1=x.

Otherwise,

x=a_1b_1,

where $a_1$ and $b_1$ are nonunits in $R$ .

If both $a_1$ and $b_1$ are prime, then we are done.

Suppose $a_1$ is not prime.

Then

a_1=a_2b_2,

where $a_2$ and $b_2$ are nonunits.

What does this mean?

a_1\in(a_2),

(a_1)\subset(a_2).

Notice that this inclusion is strict:

(a_1)\neq(a_2).

Why?

Otherwise, we would have

\left(a_2\right) \subset \left(a_1\right) \Longrightarrow a_1 = ca_1b_2 = a_1cb_2 \Longrightarrow (1-cb_2)a_1 = 0 \Longrightarrow (1-cb_2) = 0.

Thus $b_2$ is a unit.

Notice that in the final implication, we used the fact that $R$ is a domain.

If $a_2$ is also not prime, then again we can write

a_2=a_3b_3

and obtain a strict inclusion

(a_2)\subset(a_3).

Continuing in this way, every time we write

a_i=a_{i+1}b_{i+1},

we obtain a chain of inclusions

\cdots \subset (a_i) \subset (a_{i+1}) \subset \cdots.

But as we saw earlier, at some stage $(a_i)$ must equal $(a_{i+1})$ , contradicting strict inclusion, because then

(a_{n+1}) \subset (a) \subset (a_n) \Longrightarrow (a_{n+1}) = (a_n).

Thus some $a_n$ must be prime after finitely many steps.

We have proved:

every nonzero element $x$ can be written as

x=p_1y_1 \tag{$\star$}

where $p_1$ is prime.

But we may have no control over $y_1$ .

We still need to prove that $x$ can be written as a finite product of prime elements.

If $y_1$ is not irreducible, write

y_1=p_2y_2,

where $p_2$ is prime, using $(\star)$ above.

If $y_2$ is not irreducible, continue.

This produces a chain of strict inclusions

(x) \subset (y_1) \subset (y_2) \subset \cdots.

If $y_n$ were nonprime at every stage, we would obtain a contradiction, because a PID has no such infinite ascending chain of ideals, as proved earlier.

Therefore, let

p_{n+1}=y_n.

Then

\begin{aligned} x &= p_1y_1\\ &= p_1p_2y_2\\ &= \cdots\\ &= p_1p_2\cdots p_ny_n\\ &= p_1p_2\cdots p_np_{n+1}. \end{aligned}

Thus every element $x$ can be written as a product of prime elements.

~\tag*{$\square$}

The key fact used in the proof is the following.

Proposition 19.8 Fix a principal ideal domain $R$ .

Suppose there is a sequence of ideals

I_1\subset I_2\subset\cdots.

Then there exists a finite integer $n$ such that

I_n = I_{n+1} = \cdots.

Proof: Let

I = \bigcup_jI_j.

Since $R$ is a principal ideal domain, there exists a single element $a$ such that

I=(a).

Since

a\in I,

by definition $a$ belongs to some finite stage $I_n$ .

Thus

(a) \subset I_n \subset (a).

Hence

I_n=(a).

For every $j$ , if

I_n \subset I_{n+j} \subset (a) = I_n,

then

I_n = I_{n+j}.

~\tag*{$\square$}

Remark A commutative ring $R$ satisfying the ascending chain condition above is called a Nötherian ring, in honor of the mathematician Emmy Nöther.

If you take courses related to algebraic geometry, you will encounter many more Nötherian rings.

Example 19.5 Here are some applications.

(1) Let

R=\mathbb{Z}.

We know that the prime elements of $\mathbb{Z}$ are the prime numbers and their negatives.

Therefore, the factorization theorem says that every integer

x\in\mathbb{Z}

can be written as a product of powers of primes:

x = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}.

If every $p_i$ is chosen to be a positive prime, this is usually called the prime factorization of $x$ .

However, in the context of the theorem, notice that we may replace $p_1$ and $p_2$ by $-p_1$ and $-p_2$ and still express $x$ as a product of powers of primes.

In this sense, the $p_i$ are unique only up to multiplication by units.

Of course, for integers we may choose the ordering so that

p_i<p_{i+1},

giving a preferred order, but this does not make sense in a general PID.

(2) If

R=F[t],

then the theorem says that every polynomial can be written as a product of irreducible polynomials $p_i$ :

f = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}.

(3) For example, if

F=\mathbb{C},

then every polynomial can be written as a product of linear polynomials:

f = (t-\alpha_1)^{n_1} \cdots (t-\alpha_k)^{n_k}.

Notice that over other fields, we may not be able to factor $f$ into linear polynomials.

Exercise Here are some exercises.

(1) Let $F$ be a field and let

g\in F[t].

Prove that

g(x)=0

if and only if

t-x

divides the polynomial $g(t)$ .

Hint: use the division algorithm.

(2) Fix a commutative ring $R$ , and fix

a,b\in R.

Prove that

(a)=(b)

if and only if

a=ub,

where $u$ is a unit.

(3) Let $R$ be a commutative ring. Prove that a unit cannot be a zero divisor.

What is the converse?

(4) Prove that every field is a principal ideal domain.

Solution:

(1) If

\deg g=0,

the statement is immediate because

g(x)=a_0=0

if and only if

g=0,

and

(t-x)\cdot0=0.

Thus $t-x$ divides $g$ .

Now suppose

\deg g\geqslant1.

Using the division algorithm, we may write

g = (t-x)q+r.

Then

g(x) = (x-x)q(x)+r(x) = 0q(r)+r(x) = r(x).

Thus

r(x)=0.

But

\deg r<\deg(t-x),

so $r(x)$ must be a polynomial of degree $0$ having $x$ as a root.

This means

r=0

as a polynomial, and therefore

g = (t-x)q.

(2) Since

a=ub,

we have

a\in(b).

Thus

(a)\subset(b).

Indeed, if

y=ra,

then

y=rub=(ru)b,

so every multiple of $a$ is also a multiple of $b$ .

Similarly,

u^{-1}a=b,

b\in(a),

and therefore

(b)\subset(a).

(3) Suppose $x$ is a unit.

Then there exists

y\in R

such that

xy=1.

For any

a\in R,

axy = a\cdot1 = a.

On the other hand, if

ax=0,

then

axy = 0y = 0.

Therefore,

a=0.

Hence $x$ cannot be a zero divisor.

(4) A commutative ring is a field if and only if its only ideals are

\{0\}

and

R

itself.

Clearly,

\{0\}

is principal because

\{0\}=(0).

Similarly,

R=(1)

for every ring.

Thus we only need to prove that there are no zero divisors other than $0$ .

Every nonzero element of a field has an inverse, so there are no nonzero zero divisors.

§19.6 Modules over Principal Ideal Domains

The following theorem shows that every finitely generated module over a principal ideal domain has a simple form.

If every ring had such a simple theory of modules, the algebraic world would be a very beautiful place.

Theorem 19.9 (Classification of Finitely Generated Modules over a Principal Ideal Domain)

Let $R$ be a principal ideal domain, and let $M$ be a finitely generated $R$ -module.

Then there exist finitely many prime elements

p_1,\ldots,p_k\in R

where $p_i$ may equal $p_j$ , and integers

n_0,n_1,\ldots,n_k

such that

M \cong R^{n_0} \oplus R/(p_1^{n_1}) \oplus R/(p_2^{n_2}) \oplus \cdots \oplus R/(p_k^{n_k}),

and this decomposition is unique up to reordering and replacing the $p_i$ by unit multiples.

Remark What does uniqueness mean explicitly?

Suppose we have another decomposition

M \cong R^{m_0} \oplus R/(q_1^{m_1}) \oplus \cdots \oplus R/(q_j^{m_j}),

where each $q_i$ is also prime.

Then:

(1)

m_0=n_0;

(2)

j=k;

and

(3) there is a reordering of the indices such that

n_i=m_i,

and $p_i$ and $q_i$ are unit multiples of one another.

It is particularly important to note that $p_i$ may equal $p_j$ even when $i\neq j$ .

In other words, modules are different from numbers.

Their decomposition is not simply unique prime factorization in which

p\cdot\cdots\cdot p

can always be combined into

p^k.

Repeated prime factors are significant.

Example 19.6 ( $R=F$ a field)

If $F$ is a field, what are its prime elements?

There are no prime elements because prime elements are nonzero nonunits.

Therefore, every finitely generated module over $F$ has the form

M \cong F^{n_0}.

This means that every finitely generated $F$ -module has a finite basis.

The integer $n_0$ is precisely the dimension of the vector space.

Example 19.7 ( $R=\mathbb{Z}$ )

What are the primes in $\mathbb{Z}$ ?

They are the numbers of the form

\pm p,

where $p$ is a prime number.

Notice that

(p)=(-p).

Therefore, the theorem above says that every finitely generated $\mathbb{Z}$ -module, i.e. every finitely generated Abelian group, has the form

M \cong \mathbb{Z}^{n_0} \oplus \mathbb{Z}/p_1^{n_1}\mathbb{Z} \oplus \ldots \oplus \mathbb{Z}/p_k^{n_k}\mathbb{Z}.

Uniqueness means, for example, that

\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \quad \left( p_1=p_2=2, ~n_0=0, ~n_1=n_2=1 \right)

and

\mathbb{Z}/4\mathbb{Z} \quad \left( p_1=2, ~n_0=0, ~n_1=2 \right)

are not isomorphic as $\mathbb{Z}$ -modules, i.e. they are not isomorphic Abelian groups.

We already knew this.

For example, $\mathbb{Z}/4\mathbb{Z}$ is cyclic, while the former group is not.

Notice that the former group is also an example in which

p_i=p_j

for $i\neq j$ .

Example 19.8 Still take

R=\mathbb{Z}.

We can now classify all Abelian groups of order $8$ :

Example 19.9 Another example:

Let

M = \mathbb{Z}/6\mathbb{Z}.

It is not in the standard form appearing in the theorem.

In fact,

M \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}.

Exercise Classify all Abelian groups of order

7\times7\times11\times11 = 5929.

Solution: We need to find all possible collections $p_i,n_i$ satisfying

|M|=5929

with

\begin{aligned} 5929 &= \left| \mathbb{Z}/(p_1^{n_1}) \oplus \cdots \oplus \mathbb{Z}/(p_k^{n_k}) \right|\\ &= p_1^{n_1}\cdots p_k^{n_k}. \end{aligned}

Notice:

\mathbb{Z}/p^2\mathbb{Z}

is not isomorphic to

\mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}.

Exercise Which of these groups are isomorphic to

\mathbb{Z}/5929\mathbb{Z}?

Solution: We have already proved that if

\gcd(m,n)=1,

then

\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/(mn)\mathbb{Z}.

Therefore,

\mathbb{Z}/49\mathbb{Z} \oplus \mathbb{Z}/121\mathbb{Z} \cong \mathbb{Z}/5929\mathbb{Z}.

Let $F$ be a field, and let $V$ be an $F$ -vector space, i.e. an $F$ -module.

Any $F$ -linear map

A:V\to V

defines an $F[t]$ -module structure on $V$ .

f = a_dt^d+\cdots+a_1t+a_0,

then define

fv := a_dA^d(v) + \cdots + a_1A(v) + a_0v,

where

A^i = \underbrace{ A\circ\cdots\circ A }_{i~\text{times}}.

Thus, suppose $V$ is a finite-dimensional vector space over $F$ .

Choose an $F$ -linear map

A:V\to V

making $V$ into an $F[t]$ -module.

Proposition 19.10 $V$ is a finitely generated $F[t]$ -module.

Proof: Let

v_1,\ldots,v_n

be a finite basis.

Then

V = \{ b_1v_1+\cdots+b_nv_n \mid b_1,\ldots,b_n\in F \}.

In particular, if

f_i=b_i

are constant polynomials, then

V = \{ f_1v_1+\cdots+f_nv_n \}.

Therefore,

F[t]^{\oplus n} \to V

is surjective.

~\tag*{$\square$}

Corollary 19.11 $V$ is isomorphic, as an $F[t]$ -module, to

F[t]/(p_1^{n_1}) \oplus \cdots \oplus F[t]/(p_k^{n_k}) \oplus F[t]^{n_0},

where

p_i\in F[t]

are irreducible,

n_i\geqslant1,

and

n_0\geqslant0.

Remark In this decomposition,

n_0=0.

Why?

Because $V$ is finite-dimensional as an $F$ -vector space, while $F[t]$ is not finite-dimensional.

Therefore, $V$ cannot contain a subspace isomorphic to $F[t]$ .

In general, determining irreducible polynomials can be difficult.

For example, deciding whether

x^3+2x^2+x+1

is irreducible over

\mathbb{Z}/p\mathbb{Z}

often requires checking the possibilities individually.

§19.7 When the Principal Ideal Domain Is a Polynomial Ring

One of the principal ideal domains mentioned earlier is

R=F[t].

So what are the prime elements of $F[t]$ ?

This is generally a complicated question.

A first necessary condition for $f$ to be prime is that it have no roots in $F$ .

Otherwise, as we saw earlier, $f$ is divisible by a linear polynomial, which is not a unit in $F[t]$ .

However, for certain special fields, the irreducible elements of $F[t]$ are easier to identify.

Definition 19.7 A field $F$ is called algebraically closed if every polynomial

f\in F[t]

has a root.

An obvious example is

F=\mathbb{C}.

There is an important theorem.

Theorem 19.12 Every field $F$ can be embedded into an algebraically closed field.

Remark Notice that not every field $F$ admits an injective ring homomorphism into $\mathbb{C}$ .

For example, let

F=\mathbb{Z}/2\mathbb{Z}.

Its multiplicative identity $\overline{1}$ satisfies

\overline{1}+\overline{1} = 0.

Any ring homomorphism

\phi: \mathbb{Z}/2\mathbb{Z} \to \mathbb{C}

must satisfy

\phi(\overline{1}) + \phi(\overline{1}) = \phi(0),

which is impossible in $\mathbb{C}$ , because a ring homomorphism must also satisfy

\phi(\overline{1}) = 1_{\mathbb{C}}.

In other words, there must exist some field different from $\mathbb{C}$ that contains roots of every polynomial and admits an injection from $\mathbb{Z}/2\mathbb{Z}$ .

Sounds mysterious, doesn't it?

Proposition 19.13 If $F$ is algebraically closed, then the only irreducible elements of $F[t]$ are nonzero linear polynomials.

Proof: We already know that every nonzero linear polynomial is irreducible, since whenever

f=ab,

either $a$ or $b$ must have degree $0$ .

Thus every factorization of a linear polynomial involves a unit.

On the other hand, if

\deg f\geqslant2,

then by the definition of algebraic closure, $f$ has a root.

Therefore, we may write

f = (t-x)q,

where

\deg q = \deg f-1.

Both $t-x$ and $q$ are nonunits because both have positive degree.

Therefore, no polynomial of degree greater than $1$ can be prime.

~\tag*{$\square$}

Corollary 19.14 If $F$ is algebraically closed, then every finitely generated module over $F$ is isomorphic to

F[t]^{n_0} \oplus F[t]/(t-\lambda_1)^{n_1} \oplus \cdots \oplus F[t]/(t-\lambda_k)^{n_k},

where

\lambda_i\in F

and

n_1,\ldots,n_k\geqslant1.

Why is this useful?

A very good example is an $F[t]$ -module, i.e. an $F$ -vector space equipped with a linear map

A:V\to V.

This helps us classify linear maps $A$ .

Corollary 19.15 If $F$ is algebraically closed, $V$ is a finite-dimensional $F$ -vector space, and

A:V\to V

is $F$ -linear, then

V \cong F[t]/(t-\alpha_1)^{n_1} \oplus \cdots \oplus F[t]/(t-\alpha_k)^{n_k},

where

\alpha_i\in F.

Remark If

f=a_1t-a_0,

then

a_1^{-1}f = t-a_1^{-1}a_0

assuming

a_1\neq0.

Therefore,

(f) = (a_1^{-1}f) = (t-a_1^{-1}a_0).

That is, we may always assume

a_1=1.

Let us look at some examples.

We want to study

F[t]/(t-\alpha)^n

both as an $F$ -module and as an $F[t]$ -module.

Notice that the $F[t]$ -module structure on

F[t]/(p^n)

is defined by

\begin{aligned} F[t]\times F[t]/(p^n) &\to F[t]/(p^n),\\ (f,\overline{g}) &\mapsto \overline{fg}. \end{aligned}

Proposition 19.16 If

\deg p=d,

then

F[t]/(p^n) \cong F^{n\cdot d}

as $F$ -vector spaces.

Proof: Every

f\in F[t]

can be written as

f = p^nq+r,

where

\deg r < \deg p^n = nd.

Since $r$ and $q$ are unique for fixed $p^n$ and $f$ , the function

\overline{f} \mapsto r, \quad F[t]/(p^n) \to \{ \text{polynomials of degree at most }nd-1 \} \cong F^{nd}

gives a bijection.

~\tag*{$\square$}

Example 19.11

V = F[t]/(t),

where

\alpha=0, \qquad n=1.

What is the corresponding $F[t]$ -action?

(1)

F[t]/(t) \cong \{\text{constant polynomials}\} \cong F.

(2)

t\cdot\overline{a_0} = \overline{ta_0} = \overline{0},

because

a_0t\in(t).

Thus the action of multiplication by $t$ is the map

\begin{aligned} A:F&\to F\\ a_0&\mapsto0. \end{aligned}

Example 19.11

V = F[t]/(t-\alpha).

The action of multiplication by $t$ corresponds to

\begin{aligned} A:&V\to V\\ &\begin{aligned} \overline{a_0} \mapsto \overline{ta_0} &= \overline{(t-\alpha)} \overline{a_0} + \alpha\overline{a_0}\\ &= \alpha\overline{a_0}. \end{aligned} \end{aligned}

That is,

\begin{aligned} A:F&\to F\\ a_0&\mapsto\alpha a_0. \end{aligned}

Example 19.12 Let

V = F[t]/(t-\alpha)^n.

Then $V$ has a basis

\overline{1}, \overline{t-\alpha}, \overline{(t-\alpha)^2}, \ldots, \overline{(t-\alpha)^{n-1}}.

Moreover,

\begin{aligned} tv_i = t\overline{(t-\alpha)}^i &= (t-\alpha) \overline{(t-\alpha)}^i + \alpha \overline{(t-\alpha)}^i\\ &= \overline{(t-\alpha)}^{i+1} + \alpha \overline{(t-\alpha)}^i\\ &= v_{i+1} + \alpha v_i. \end{aligned}

Therefore,

A = \begin{pmatrix} \alpha & 1 & 0 & \cdots & 0 \\ 0 & \alpha & 1 & \cdots & 0 \\ 0 & 0 & \alpha & \cdots & 0 \\ 0 & \cdots & 0 & \ddots & 1 \\ 0 & \cdots & \cdots & 0 & \alpha \end{pmatrix},

where the main diagonal consists of $\alpha$ , and the entries immediately above the diagonal are all $1$ .

Contents

§19 Principal Ideal Domains (PIDs)

§19.1 Polynomial Rings

§19.2 The Similarity Between the Ring of Integers Z\mathbb{Z}Z and the Polynomial Ring F[t]F[t]F[t]

§19.3 Review of Some Definitions

§19.4 The Euclidean Algorithm

§19.5 Prime Elements and Factorization in Principal Ideal Domains

§19.6 Modules over Principal Ideal Domains

§19.7 When the Principal Ideal Domain Is a Polynomial Ring

§19.2 The Similarity Between the Ring of Integers $\mathbb{Z}$ and the Polynomial Ring $F[t]$