AATA Irreducible Polynomials

Section 17.3 Irreducible Polynomials

A nonconstant polynomial

f (x) \in F [x]

is irreducible over a field

F

f (x)

cannot be expressed as a product of two polynomials

g (x)

and

h (x)

F [x],

where the degrees of

g (x)

and

h (x)

are both smaller than the degree of

f (x) .

Irreducible polynomials function as the “prime numbers” of polynomial rings.

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Example 17.11.

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The polynomial

x^{2} - 2 \in Q [x]

is irreducible since it cannot be factored any further over the rational numbers. Similarly,

x^{2} + 1

is irreducible over the real numbers.

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Example 17.12.

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The polynomial

p (x) = x^{3} + x^{2} + 2

is irreducible over

Z_{3} [x] .

Suppose that this polynomial was reducible over

Z_{3} [x] .

By the division algorithm there would have to be a factor of the form

x - a,

where

a

is some element in

Z_{3} [x] .

Hence, it would have to be true that

p (a) = 0 .

However,

\begin{aligned} p (0) & = 2 \\ p (1) & = 1 \\ p (2) & = 2 . \end{aligned}

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Therefore,

p (x)

has no zeros in

Z_{3}

and must be irreducible.

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Lemma 17.13.

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Let

p (x) \in Q [x] .

Then

p (x) = \frac{r}{s} (a_{0} + a_{1} x + \dots + a_{n} x^{n}),

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where

r, s, a_{0}, \dots, a_{n}

are integers, the

a_{i}

’s are relatively prime, and

r

and

s

are relatively prime.

Proof.

Suppose that

p (x) = \frac{b_{0}}{c_{0}} + \frac{b_{1}}{c_{1}} x + \dots + \frac{b_{n}}{c_{n}} x^{n},

where the

b_{i}

’s and the

c_{i}

’s are integers. We can rewrite

p (x)

p (x) = \frac{1}{c_{0} \dots c_{n}} (d_{0} + d_{1} x + \dots + d_{n} x^{n}),

where

d_{0}, \dots, d_{n}

are integers. Let

d

be the greatest common divisor of

d_{0}, \dots, d_{n} .

Then

p (x) = \frac{d}{c_{0} \dots c_{n}} (a_{0} + a_{1} x + \dots + a_{n} x^{n}),

where

d_{i} = d a_{i}

and the

a_{i}

’s are relatively prime. Reducing

d / (c_{0} \dots c_{n})

to its lowest terms, we can write

p (x) = \frac{r}{s} (a_{0} + a_{1} x + \dots + a_{n} x^{n}),

where

gcd (r, s) = 1 .

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Theorem 17.14. Gauss’s Lemma.

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Let

p (x) \in Z [x]

be a monic polynomial such that

p (x)

factors into a product of two polynomials

α (x)

and

β (x)

Q [x],

where the degrees of both

α (x)

and

β (x)

are less than the degree of

p (x) .

Then

p (x) = a (x) b (x),

where

a (x)

and

b (x)

are monic polynomials in

Z [x]

with

\deg α (x) = \deg a (x)

and

\deg β (x) = \deg b (x) .

Proof.

By Lemma 17.13, we can assume that

\begin{aligned} α (x) & = \frac{c_{1}}{d_{1}} (a_{0} + a_{1} x + \dots + a_{m} x^{m}) = \frac{c_{1}}{d_{1}} α_{1} (x) \\ β (x) & = \frac{c_{2}}{d_{2}} (b_{0} + b_{1} x + \dots + b_{n} x^{n}) = \frac{c_{2}}{d_{2}} β_{1} (x), \end{aligned}

where the

a_{i}

’s are relatively prime and the

b_{i}

’s are relatively prime. Consequently,

p (x) = α (x) β (x) = \frac{c_{1} c_{2}}{d_{1} d_{2}} α_{1} (x) β_{1} (x) = \frac{c}{d} α_{1} (x) β_{1} (x),

where

c / d

is the product of

c_{1} / d_{1}

and

c_{2} / d_{2}

expressed in lowest terms. Hence,

d p (x) = c α_{1} (x) β_{1} (x) .

d = 1,

then

c a_{m} b_{n} = 1

since

p (x)

is a monic polynomial. Hence, either

c = 1

c = - 1 .

c = 1,

then either

a_{m} = b_{n} = 1

a_{m} = b_{n} = - 1 .

In the first case

p (x) = α_{1} (x) β_{1} (x),

where

α_{1} (x)

and

β_{1} (x)

are monic polynomials with

\deg α (x) = \deg α_{1} (x)

and

\deg β (x) = \deg β_{1} (x) .

In the second case

a (x) = - α_{1} (x)

and

b (x) = - β_{1} (x)

are the correct monic polynomials since

p (x) = (- α_{1} (x)) (- β_{1} (x)) = a (x) b (x) .

The case in which

c = - 1

can be handled similarly.

Now suppose that

d \neq 1 .

Since

gcd (c, d) = 1,

there exists a prime

p

such that

p ∣ d

and

p ∤ c .

Also, since the coefficients of

α_{1} (x)

are relatively prime, there exists a coefficient

a_{i}

such that

p ∤ a_{i} .

Similarly, there exists a coefficient

b_{j}

β_{1} (x)

such that

p ∤ b_{j} .

Let

α_{1}^{'} (x)

and

β_{1}^{'} (x)

be the polynomials in

Z_{p} [x]

obtained by reducing the coefficients of

α_{1} (x)

and

β_{1} (x)

modulo

p .

Since

p ∣ d,

α_{1}^{'} (x) β_{1}^{'} (x) = 0

Z_{p} [x] .

However, this is impossible since neither

α_{1}^{'} (x)

nor

β_{1}^{'} (x)

is the zero polynomial and

Z_{p} [x]

is an integral domain. Therefore,

d = 1

and the theorem is proven.

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Corollary 17.15.

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Let

p (x) = x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0}

be a polynomial with coefficients in

Z

and

a_{0} \neq 0 .

p (x)

has a zero in

Q,

then

p (x)

also has a zero

α

Z .

Furthermore,

α

divides

a_{0} .

Proof.

Let

p (x)

have a zero

a \in Q .

Then

p (x)

must have a linear factor

x - a .

By Gauss’s Lemma,

p (x)

has a factorization with a linear factor in

Z [x] .

Hence, for some

α \in Z

p (x) = (x - α) (x^{n - 1} + \dots - a_{0} / α) .

Thus

a_{0} / α \in Z

and so

α ∣ a_{0} .

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Example 17.16.

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Let

p (x) = x^{4} - 2 x^{3} + x + 1 .

We shall show that

p (x)

is irreducible over

Q [x] .

Assume that

p (x)

is reducible. Then either

p (x)

has a linear factor, say

p (x) = (x - α) q (x),

where

q (x)

is a polynomial of degree three, or

p (x)

has two quadratic factors.

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p (x)

has a linear factor in

Q [x],

then it has a zero in

Z .

By Corollary 17.15, any zero must divide 1 and therefore must be

\pm 1;

however,

p (1) = 1

and

p (- 1) = 3 .

Consequently, we have eliminated the possibility that

p (x)

has any linear factors.

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Therefore, if

p (x)

is reducible it must factor into two quadratic polynomials, say

\begin{aligned} p (x) & = (x^{2} + a x + b) (x^{2} + c x + d) \\ = x^{4} + (a + c) x^{3} + (a c + b + d) x^{2} + (a d + b c) x + b d, \end{aligned}

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where each factor is in

Z [x]

by Gauss’s Lemma. Hence,

\begin{aligned} a + c & = - 2 \\ a c + b + d & = 0 \\ a d + b c & = 1 \\ b d & = 1 . \end{aligned}

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Since

b d = 1,

either

b = d = 1

b = d = - 1 .

In either case

b = d

and so

a d + b c = b (a + c) = 1 .

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Since

a + c = - 2,

we know that

- 2 b = 1 .

This is impossible since

b

is an integer. Therefore,

p (x)

must be irreducible over

Q .

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Theorem 17.17. Eisenstein’s Criterion.

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Let

p

be a prime and suppose that

f (x) = a_{n} x^{n} + \dots + a_{0} \in Z [x] .

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p ∣ a_{i}

for

i = 0, 1, \dots, n - 1,

but

p ∤ a_{n}

and

p^{2} ∤ a_{0},

then

f (x)

is irreducible over

Q .

Proof.

By Gauss’s Lemma (see Theorem 17.17 and Lemma 18.26), we need only show that

f (x)

does not factor into polynomials of lower degree in

Z [x] .

Let

f (x) = (b_{r} x^{r} + \dots + b_{0}) (c_{s} x^{s} + \dots + c_{0})

be a factorization in

Z [x],

with

b_{r}

and

c_{s}

not equal to zero and

r, s < n .

Since

p^{2}

does not divide

a_{0} = b_{0} c_{0},

either

b_{0}

c_{0}

is not divisible by

p .

Suppose that

p ∤ b_{0}

and

p ∣ c_{0} .

Since

p ∤ a_{n}

and

a_{n} = b_{r} c_{s},

neither

b_{r}

nor

c_{s}

is divisible by

p .

Let

m

be the smallest value of

k

such that

p ∤ c_{k} .

Then

a_{m} = b_{0} c_{m} + b_{1} c_{m - 1} + \dots + b_{m} c_{0}

is not divisible by

p,

since each term on the right-hand side of the equation is divisible by

p

except for

b_{0} c_{m} .

Therefore,

m = n

since

a_{i}

is divisible by

p

for

m < n .

Hence,

f (x)

cannot be factored into polynomials of lower degree and therefore must be irreducible.

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Example 17.18.

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The polynomial

f (x) = 16 x^{5} - 9 x^{4} + 3 x^{2} + 6 x - 21

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is easily seen to be irreducible over

Q

by Eisenstein’s Criterion if we let

p = 3 .

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Eisenstein’s Criterion is more useful in constructing irreducible polynomials of a certain degree over

Q

than in determining the irreducibility of an arbitrary polynomial in

Q [x] :

given an arbitrary polynomial, it is not very likely that we can apply Eisenstein’s Criterion. The real value of Theorem 17.17 is that we now have an easy method of generating irreducible polynomials of any degree.

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Subsection Ideals in $F [x]$

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Let

F

be a field. Recall that a principal ideal in

F [x]

is an ideal

⟨ p (x) ⟩

generated by some polynomial

p (x);

that is,

⟨ p (x) ⟩ = {p (x) q (x) : q (x) \in F [x]} .

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Example 17.19.

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The polynomial

x^{2}

F [x]

generates the ideal

⟨ x^{2} ⟩

consisting of all polynomials with no constant term or term of degree

1 .

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Theorem 17.20.

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F

is a field, then every ideal in

F [x]

is a principal ideal.

Proof.

Let

I

be an ideal of

F [x] .

I

is the zero ideal, the theorem is easily true. Suppose that

I

is a nontrivial ideal in

F [x],

and let

p (x) \in I

be a nonzero element of minimal degree. If

\deg p (x) = 0,

then

p (x)

is a nonzero constant and 1 must be in

I .

Since 1 generates all of

F [x],

⟨ 1 ⟩ = I = F [x]

and

I

is again a principal ideal.

Now assume that

\deg p (x) \geq 1

and let

f (x)

be any element in

I .

By the division algorithm there exist

q (x)

and

r (x)

F [x]

such that

f (x) = p (x) q (x) + r (x)

and

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

Since

f (x), p (x) \in I

and

I

is an ideal,

r (x) = f (x) - p (x) q (x)

is also in

I .

However, since we chose

p (x)

to be of minimal degree,

r (x)

must be the zero polynomial. Since we can write any element

f (x)

I

p (x) q (x)

for some

q (x) \in F [x],

it must be the case that

I = ⟨ p (x) ⟩ .

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Example 17.21.

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It is not the case that every ideal in the ring

F [x, y]

is a principal ideal. Consider the ideal of

F [x, y]

generated by the polynomials

x

and

y .

This is the ideal of

F [x, y]

consisting of all polynomials with no constant term. Since both

x

and

y

are in the ideal, no single polynomial can generate the entire ideal.

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Theorem 17.22.

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Let

F

be a field and suppose that

p (x) \in F [x] .

Then the ideal generated by

p (x)

is maximal if and only if

p (x)

is irreducible.

Proof.

Suppose that

p (x)

generates a maximal ideal of

F [x] .

Then

⟨ p (x) ⟩

is also a prime ideal of

F [x] .

Since a maximal ideal must be properly contained inside

F [x],

p (x)

cannot be a constant polynomial. Let us assume that

p (x)

factors into two polynomials of lesser degree, say

p (x) = f (x) g (x) .

Since

⟨ p (x) ⟩

is a prime ideal one of these factors, say

f (x),

is in

⟨ p (x) ⟩

and therefore be a multiple of

p (x) .

But this would imply that

⟨ p (x) ⟩ \subset ⟨ f (x) ⟩,

which is impossible since

⟨ p (x) ⟩

is maximal.

Conversely, suppose that

p (x)

is irreducible over

F [x] .

Let

I

be an ideal in

F [x]

containing

⟨ p (x) ⟩ .

By Theorem 17.20,

I

is a principal ideal; hence,

I = ⟨ f (x) ⟩

for some

f (x) \in F [x] .

Since

p (x) \in I,

it must be the case that

p (x) = f (x) g (x)

for some

g (x) \in F [x] .

However,

p (x)

is irreducible; hence, either

f (x)

g (x)

is a constant polynomial. If

f (x)

is constant, then

I = F [x]

and we are done. If

g (x)

is constant, then

f (x)

is a constant multiple of

I

and

I = ⟨ p (x) ⟩ .

Thus, there are no proper ideals of

F [x]

that properly contain

⟨ p (x) ⟩ .

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Subsection Historical Note

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Throughout history, the solution of polynomial equations has been a challenging problem. The Babylonians knew how to solve the equation

a x^{2} + b x + c = 0 .

Omar Khayyam (1048–1131) devised methods of solving cubic equations through the use of geometric constructions and conic sections. The algebraic solution of the general cubic equation

a x^{3} + b x^{2} + c x + d = 0

was not discovered until the sixteenth century. An Italian mathematician, Luca Pacioli (ca. 1445–1509), wrote in Summa de Arithmetica that the solution of the cubic was impossible. This was taken as a challenge by the rest of the mathematical community.

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Scipione del Ferro (1465–1526), of the University of Bologna, solved the “depressed cubic,”

a x^{3} + c x + d = 0 .

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He kept his solution an absolute secret. This may seem surprising today, when mathematicians are usually very eager to publish their results, but in the days of the Italian Renaissance secrecy was customary. Academic appointments were not easy to secure and depended on the ability to prevail in public contests. Such challenges could be issued at any time. Consequently, any major new discovery was a valuable weapon in such a contest. If an opponent presented a list of problems to be solved, del Ferro could in turn present a list of depressed cubics. He kept the secret of his discovery throughout his life, passing it on only on his deathbed to his student Antonio Fior (ca. 1506–?).

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Although Fior was not the equal of his teacher, he immediately issued a challenge to Niccolo Fontana (1499–1557). Fontana was known as Tartaglia (the Stammerer). As a youth he had suffered a blow from the sword of a French soldier during an attack on his village. He survived the savage wound, but his speech was permanently impaired. Tartaglia sent Fior a list of 30 various mathematical problems; Fior countered by sending Tartaglia a list of 30 depressed cubics. Tartaglia would either solve all 30 of the problems or absolutely fail. After much effort Tartaglia finally succeeded in solving the depressed cubic and defeated Fior, who faded into obscurity.

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At this point another mathematician, Gerolamo Cardano (1501–1576), entered the story. Cardano wrote to Tartaglia, begging him for the solution to the depressed cubic. Tartaglia refused several of his requests, then finally revealed the solution to Cardano after the latter swore an oath not to publish the secret or to pass it on to anyone else. Using the knowledge that he had obtained from Tartaglia, Cardano eventually solved the general cubic

a x^{3} + b x^{2} + c x + d = 0 .

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Cardano shared the secret with his student, Ludovico Ferrari (1522–1565), who solved the general quartic equation,

a x^{4} + b x^{3} + c x^{2} + d x + e = 0 .

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In 1543, Cardano and Ferrari examined del Ferro’s papers and discovered that he had also solved the depressed cubic. Cardano felt that this relieved him of his obligation to Tartaglia, so he proceeded to publish the solutions in Ars Magna (1545), in which he gave credit to del Ferro for solving the special case of the cubic. This resulted in a bitter dispute between Cardano and Tartaglia, who published the story of the oath a year later.

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Section 17.3 Irreducible Polynomials

Example 17.11.

Example 17.12.

Lemma 17.13.

Proof.

Theorem 17.14. Gauss’s Lemma.

Proof.

Corollary 17.15.

Proof.

Example 17.16.

Theorem 17.17. Eisenstein’s Criterion.

Proof.

Example 17.18.

Subsection Ideals in F[x]

Example 17.19.

Theorem 17.20.

Proof.

Example 17.21.

Theorem 17.22.

Proof.

Subsection Historical Note

Subsection Ideals in $F [x]$