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2.2. EUCLIDEAN DOMAINS 53
2.2 Euclidean Domains
The sets of integers and of polynomials (for any field of coefficients) have:
(a) Addition that associates and commutes. (b) An additive identity element 0 and additive inverses of everything. (c) Multiplication that associates, commutes and distributes with addition. (d) A multiplicative identity element 1. (e) A cancellation rule: if a 6 = 0 and ab = ac, then b = c. (f) Division with remainders. Any set D with addition and multiplication rules that has all the properties (a)-(e) above is called an integral domain. A field is one kind of integral domain, and the integers and polynomials are another. Condition (f) will be part of the definition of a Euclidean domain.
Definition: An element a ∈ D of an integral domain is called a unit if it has a multiplicative inverse element, which we denote a−^1 or 1/a. There is always at least one unit in any integral domain, namely the multiplicative identity 1.
Note: Units are the things we call “not interesting” when we factor.
Examples: (a) In a field F , all the elements except 0 are units.
(b) In F [x], the constant polynomials are the units (Corollary 2.1.2). (c) 1 and −1 are the integer units.
Definition: A function:
deg : D − { 0 } → R+^ ∪ { 0 }
is called a degree function if it has the following properties:
(i) deg converts multiplication to addition:
deg(ab) = deg(a) + deg(b)
(ii) deg detects the units of the integral domain:
deg(a) = 0 if and only if a is a unit
Example: The degree of a polynomial in §2.1 is a degree function:
deg(a(x)) = the ordinary degree of a(x)
This is what Proposition 2.1.1 and Corollary 2.1.2 tell us. Notice that the range of this degree function is the set of whole numbers.
To define the degree of an integer, I need to remind you of the:
54 CHAPTER 2. POLYNOMIALS
Natural Logarithm: This is defined for all positive real numbers by:
ln(x) =
∫ (^) x
1
t
dt
from which it follows immediately that ln(1) = 0 and
ln(x) < ln(y) whenever x < y
(in other words, ln(x) is an increasing function of x).
If x and y are fixed positive real numbers, then: ∫ (^) xy
x
t
dt =
∫ (^) y
1
xs
(xds) =
∫ (^) y
1
s
ds = ln(y)
using the substitution t = xs (and dt = xds). But then:
ln(xy) =
∫ (^) xy
1
t
dt =
∫ (^) x
1
t
dt +
∫ (^) xy
x
t
dt = ln(x) + ln(y)
Proposition 2.2.1. The “natural log of the absolute value:”
deg(a) = ln(|a|)
is a degree function for the integers.
Proof: If a = 0, then deg(a) = ln(0) is undefined. Otherwise |a| ≥ 1, and then deg(a) = ln(|a|) ≥ 0, so deg has the required domain and range.
Next, −1 and 1 are the only integers with ln(|a|) = 0, and these are the integer units. This gives Property (ii). And finally,
ln(|ab|) = ln(|a||b|) = ln(|a|) + ln(|b|)
is what we require for Property (i). So ln(|a|) is a degree function.
Remark: The smallest range of this degree function is {0 = ln(1), ln(2), ln(3), ...} which is not the set of whole numbers, but like the set of natural numbers and the set of whole numbers, this set does satisfy the well-ordered axiom. This will be important for us later.
Definition: An integral domain D with degree function is called a Euclidean domain if it has division with remainders: For all a, b ∈ D − { 0 }, either:
(a) a = bq for some q, so b divides a (b is a factor of a), or else: (b) a = bq + r with deg(r) < deg(b), and r is the remainder.
Examples: (a) F [x] is a Euclidean domain, with the ordinary degree function.
(b) Z is a Euclidean domain with log(|a|) as its degree function.
56 CHAPTER 2. POLYNOMIALS
Next: rk− 1 = rkqk+1 + rk+ = (rk+1qk+2)qk+1 + rk+ = rk+1(qk+2qk+1 + 1)
shows that rk+1 divides rk− 1. As we work our way up and substitute, we see that rk+1 divides all the remainders, and it divides a and b as well, so that rk+ is a common divisor of a and b (and all other remainders, too!).
To see that rk+1 has greatest degree among all the common divisors, we work our way down Euclid’s algorithm. The first equation:
a = bq 1 + r 1
can be rewritten as r 1 = a + (−q 1 )b
showing that r 1 is a linear combination of a and b. Then:
r 2 = b + (−q 2 )r 1 = b + (−q 2 )(a + (−q 1 )b) = (−q 2 )a + (1 + q 1 q 2 )b
so r 2 is a linear combination of a and b, too, and as we work our way down, every remainder is a linear combination of a and b, down to:
rk+1 = ua + vb
for some pair of elements u, v ∈ D.
Now if d is any common divisor of a and b, then a = dq and b = dq′, and:
rk+1 = udq + vdq′^ = d(uq + vq′) so d divides rk+
But then deg(d) + deg(uq + vq′) = deg(rk+1)
so deg(d) ≤ deg(rk+1). Thus rk+1 has the possible greatest degree of any common divisor of a and b!
Definition: A common divisor of greatest degree will be called a gcd.
Two Examples: First, an integer example. Start with 750 and 144.
750 = 144(5) + 30 144 = 30(5) + (−6) 30 = (−6)(−5)
First we go up Euclid’s algorithm and substitute:
30 = (−6)(−5) 144 = 30(5) + (−6) = (−6)(−5)(5) + (−6) = (−6)(−24) 750 = 144(5) + 30 = (−6)(−24)(5) + (−6)(−5) = (−6)(−125)
to see that −6 is a common divisor of 144 and 750.
2.2. EUCLIDEAN DOMAINS 57
Then we go down Euclid’s algorithm: 30 = 750 + 144(−5) − 6 = 144 + 30(−5) = 144 + (750 + 144(−5))(−5) = 750(−5) + 144(26)
to see that −6 is a linear combination of 750 and 144.
Next, a polynomial example. Start with x^4 − 1 and x^3 + x in Q[x].
x^4 − 1 = (x^3 + x)(x) + (−x^2 − 1) x^3 + x = (−x^2 − 1)(−x)
First we go up Euclid’s algorithm and substitute:
x^3 + x = (−x^2 − 1)(−x) x^4 − 1 = (x^3 + x)(x) + (−x^2 − 1) = (−x^2 − 1)(−x)(x) + (−x^2 − 1) (−x^2 − 1)(−x^2 + 1)
to see that −x^2 − 1 is a common divisor. Then we go down:
(−x^2 − 1) = (x^4 − 1) + (x^3 + x)(−x)
to see that −x^2 − 1 is a linear combination of the polynomials.
Note: Unlike the natural numbers, gcd’s in Z and F [x] are not unique. In the first example, 6 would have been a perfectly good gcd, and in the second, x^2 +1, or even 12 x^2 + 12 would have been possible gcd’s.
Proposition 2.2.2. Every gcd of a and b is a linear combination of a and b.
Proof: Start with the linear combination from Euclid’s algorithm:
rk+1 = ua + vb
If d is any gcd, then d divides rk+1 (see the proof of Euclid’s algorithm above). So rk+1 = dq. But deg(rk+1) = deg(d) (because both of them are gcds). This tells us deg(q) = 0, so q is a unit. That means d = rk+1/q, and:
d = (u/q)a + (v/q)b
is a linear combination of a and b.
Example: We said 6 is a gcd of 750 and 144. We multiply:
−6 = 750(−5) + 144(26)
from Euclid’s algorithm by the unit −1 to get:
6 = 750(5) + 144(−26)
Definition: An element p of positive degree in a Euclidean domain is prime if its only factors of smaller degree are units.
Example: In F [x], the primes are, of course, the prime polynomials. The integer primes are p and −p, where p are the natural number primes.
2.2. EUCLIDEAN DOMAINS 59
Otherwise p 1 divides p′ 2 (p′ 3 · · · p′ m), and continuing in this fashion, eventually p 1 is associated to one of the p′’s. Similarly, every pi is associated to one of the p′ j ’s, and reversing the argument, every p′ j is associated to one of the pi’s.
Example: There are two possible prime factorizations of 15:
(3)(5) = 15 = (−3)(−5)
and 3 is associated to −3 and 5 is associated to −5.
There are many prime factorizations of x^2 − 1 in Q[x]. Examples:
(x − 1)(x + 1) = x^2 − 1 = (
x +
)(2x − 2)
and x − 1 is associated to 2x − 2 and x + 1 is associated to 12 x + 12.
Finishing up the proof of Proposition 1.2.5: We needed to show that there is only one fraction in lowest terms representing each rational number. That is, we need to know that if: a b
a′ b′
and both are in lowest terms, then a = a′^ and b = b′. If any of them is 1 or −1, the result is obvious. Otherwise we factorize them:
a = p 1 · · · pn, a′^ = p′ 1 · · · p′ m, b = q 1 · · · ql, b′^ = q′ 1 · · · q′ k
and then: ab′^ = p 1 · · · pn · q 1 ′ · · · q′ k = q 1 · · · ql · p′ 1 · · · p′ m = ba′
and we can assume that all the q’s and q′s are positive, since b and b′^ are positive. But remember that a and b have no common factors, so every q must be associated, in fact equal to one of the q′’s. And a′^ and b′^ have no common factors, so each q′^ is equal one of the q’s. But then b = b′^ and then a = a′ (cancellation law!) and we’re done.
The same argument gives another useful result:
Proposition 2.2.6. If a/b is in lowest terms, and
a b
a′ b′
then a divides a′^ and b divides b′.
Proof: Again we factorize. And again, we conclude as above that each q is equal to one of the q′’s. This is enough to let us conclude that b divides b′. Of course there may be more q′’s than q’s, so it may be that b 6 = b′. But anyway, let b′^ = bc. Then ba′^ = ab′^ = abc cancels to give a′^ = ac, so a divides a′^ as well (with the same quotient c).
60 CHAPTER 2. POLYNOMIALS
2.2.1 Euclidean Domain Exercises
6-1 Suppose D is an integral domain and ab = 0. Prove that a = 0 or b = 0.
6-2 Exactly one of the following is a degree function for the integers. Figure out which it is, and explain why the others don’t qualify.
(a) The “absolute value minus one” function:
deg(a) = |a| − 1
(b) The zero function: deg(a) = 0 (c) The “natural log of the square” function:
deg(a) = ln(a^2 )
6-3 For each of the following pairs of integers:
(i) Find a gcd. (ii) Express your gcd as a linear combination of the integers.
(a) 37 and 100 (b) − 77 and 91 (c) 777, 777 and 100, 100
6-4 For each of the following pairs of polynomials (in Q[x]):
(i) Find a gcd. (ii) Express your gcd as a linear combination of the polynomials.
(a) x^5 and x^3 + 1 (b) x^12 − 1 and x^8 − x^6 + x^2 − 1
6-5 Consider again the Gaussian integers Z[i] = {a + bi} from §1.4.
(a) Show that log(|a + bi|) = log(
a^2 + b^2 ) is a degree function. There is a long division for Gaussian integers! Given a + bi and c + di, with deg(c+di) < deg(a+bi), let p+qi be the closest Gaussian integer to the complex number: a + bi c + di
(a + bi)(c − di) c^2 + d^2
ac + bd c^2 + d^2
bc − ad c^2 + d^2
i
Then p + qi is the quotient Gaussian integer.
Next, define r + si by:
a + bi = (c + di)(p + qi) + (r + si)
This is the remainder, which does satisfy deg(r + si) < deg(c + di).
(b) Long divide the Gaussian integer 10 + 5i by 2 + 3i. (c) Find a gcd of 5 + 5i and 4 + 2i.
