LECTURE NOTES 9/19
BENJAMIN BAKKER
The goal of these notes is to generalize our proof of the Fundamental Theorem of Arithmetic to other
rings. Before we understand what that means, consider the following set:
Definition 1. Z[i] = {a + bi|a, b ∈ Z} ⊂ C is the Gaussian integers.
We can imagine the Gaussian integers as the points on a square lattice in the complex plane. Note that
Z[i] is closed under addition and multiplication in the sense that the sum or product of any two Gaussian
integers is a Gaussian integer:
(a + bi) + (c + di) = (a + c) + (b + d)i ∈ Z[i]
(a + bi)(c + di) = (ac − bd) + (ad + bc)i ∈ Z[i]
In fact, addition and multiplication in Z[i] behave much that same way as in Z:
Informal Definition. A set R with addition and multiplication satisfying the basic arithmetic laws is called
a ring.
Example 1. Z, Q, R, C, Z[t], Q[t] = {polynomials with rational coefficients} are all rings.
I don’t want to get into what a ring actually is, so for now we will work with subrings of C:
Definition 2. A subset R ⊂ C is a subring if:
(a) 0, 1, −1 ∈ R
(b) R is closed under addition and multiplication.
Thus, R will inherit all the usual arithmetic axioms for C; in particular, subrings of C are rings. Note
that for any subring R ⊂ C contains Z.
√
√
Example 2. Z ⊂ Q√⊂ R ⊂ C are all subrings, as is Z[i] ⊂ C. Z[ −5] = {a + b −5|a, b ∈ Z} ⊂ C is also a
subring; R = {a + b 3 2|a, b ∈ Z} ⊂ C is not.
We will work mostly with subrings of C, but occasionally we will want to consider some that aren’t (mainly
Q[t]). I will indicate when definitions or propositions hold for arbitrary rings or just subrings of C, but all
will apply to Q[t] (and any other polynomial rings we consider for now).
Now that we have generalized the integers Z to subrings R ⊂ C, we want to generalize the notion of
prime.
Definition 3. Let R be a ring. For a, b ∈ R, we say a|b (“a divides b”) if there is some c ∈ R with ac = b.
Note that in the integers we had a|b and b|a ⇒ a = ±b. The reason for this was multiplying by ±1 can’t
change any divisibility relationship because we can always multiply the reciprocal to undo it, i.e. a|b if and
only if ±a| ± b. In general, there may be many elements whose reciprocals are in the ring:
Definition 4. Let R be a ring. u ∈ R is a unit if u−1 ∈ R, i.e. if there is some y ∈ R such that xy = 1.
The set of units in R is R∗ , and we say x ∼ y (“x is associated to y”) if there is some unit u ∈ R such that
x = uy.
Note that x ∼ 1 if and only if x is a unit. We then have:
Proposition 1. For R ⊂ C a subring, x, y ∈ R. If x|y and y|x, then x ∼ y.
Proof. If x|y, there is z ∈ R such that xz = y. Similarly, there is z 0 ∈ R such that yz 0 = x. Thus xzz 0 = x.
If x = 0, then y = 0 since x|y, and thus x ∼ y. Otherwise, zz 0 = 1 and z is a unit, so x ∼ y.
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Example 3. Z∗ = {1, −1}; C∗ = C∗ \ 0.
Proposition 2.
(a) Z[i]∗ = {1, −1, i, −i}.
∗
(b) Q[t] = Q \ 0, i.e. the nonzero degree zero (constant) polynomials.
A particularly important role is played by the absolute value squared | · |2 function on Z[i]. Note that for
any Gaussian integer a + bi, |a + bi|2 = a2 + b2 ∈ Z≥0 .
(a) Certainly 1, −1, i, −i are units. Conversely, if x ∈ Z[i]∗ , for some y ∈ Z[i], xy = 1, so
|x| |y|2 = 1, so |x|2 = 1.
(b) Suppose f = a0 + · · · an tn ∈ Q[t] (an 6= 0) is a unit, i.e. there exists g = b0 + · · · bm tm ∈ Q[t]
(bm 6= 0) such that f g = 1. Clearly the coefficient of tm+n is an bm 6= 0 (in other words, deg(f g) =
deg(f ) + deg(g)), so this means n = m = 0 and f and g are degree zero. Since every nonzero degree
zero polynomial is invertible, we’re done.
Proof.
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Corollary 1.
(a) x ∈ Z[i] is a unit if and only if |x|2 = 1.
(b) For x, y ∈ Z[i], if x|y and |x|2 = |y|2 then x ∼ y.
WARNING: it is not true that if |x|2 = |y|2 , x ∼ y—we need either x|y or y|x to conclude this. For
example, |3 + 4i|2 = |5|2 , but certainly 3 + 4i 5
Now we can generalize the notion of prime:
Definition 5. Let R ⊂ C be a subring. x ∈ R is irreducible if
(a) x is neither zero nor a unit
(b) if y|x the either y ∼ x or y ∼ 1.
Any x ∈ R, is divisible by any associate of x and any associate of 1, so this says there are no nontrivial
factors.
Example 4. The irreducibles of Z are ±p for p a prime number. The irreducibles of Z are called prime
integers.
Proposition 3. If x ∈ Z[i] has |x|2 a prime number, then x is irreducible.
Proof. x is clearly not zero or a unit. If y|x, then |y|2 divides |x|2 , so if |x|2 is prime y is a trivial divisor, by
the above Corollary.
Example 5. 1 + i ∈ Z[i] is irreducible. 2 is not, since 2 = (1 + i)(1 − i), and |(1 + i)|2 = 2 whereas |2|2 = 4,
so (1 + i) is a nontrivial divisor. 3 is irreducible in Z[i] since a nontrivial factor x = a + bi|3 must have
|x|2 = a2 + b2 a nontrivial factor of |3|2 = 9, but a2 + b2 = 3 has no solutions.
On the other hand, we have
Proposition 4. Let p ∈ Z be a prime number. p considered as an element of Z[i] is irreducible if and only
if p = a2 + b2 has no integral solution (a solution with a, b ∈ Z).
Proof. (⇐) If x = a + bi is a nontrivial factor of p, then as above |x|2 = a2 + b2 is a nontrivial factor of p2
so there is a solution to a2 + b2 = p.
(⇒) If a2 + b2 = p is a solution, then a + bi is a nontrivial factor of p.
√
Similarly p ∈ Z a prime number is irreducible in Z[ −d] if and only if a2 + db2 = p has an integral
solution.
We would like for there to be a unique factorization for arbitrary subrings of C, but this is not true:
√
Example 6. In Z[ −5],
√
√
6 = 2 · 3 = (1 + −5) · (1 − −5)
2 and√3 are irreducible because 2 = a2 + 5b2 doesn’t have integral solutions, and neither does 3 = a2 + 5b2 .
(1 + −5) is irreducible since it’s norm squared is 6, so a nontrivial factor would have norm squared 2 or 3
which is impossible.
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The problem is that irreducibility is only the right generalization of prime to insure that factorizations
exist; to insure they are unique, recall we needed the theorem
Theorem 1. (Euclid) Let p ∈ N, p > 1. p is a prime number if and only if for all a, b ∈ Z, p|ab ⇒ p|a or
p|b.
This leads to a different generalization of the notion of prime, unfortunately still called prime:
Definition 6. R ⊂ C a subring. x ∈ R is prime if
(a) x is neither 0 nor a unit
(b) for all a, b ∈ R, if x|ab then x|a or x|b
At least one direction of Euclid’s theorem is true:
Lemma 1. Every prime in R is irreducible.
Proof. Suppose x is prime and y|x. Then x = yz, so x|yz, and x|y or x|z. In the former case, we already
know y|x, so x ∼ y; in the latter, z = wx, so z = wyz implies that 1 = wz (z cannot be zero since x isn’t),
in which case y ∼ 1.
The missing ingredient for Euclid’s theorem to be true—and thus for unique factorization to hold—is
the other direction, for which it is crucial to have the Euclidean property. We can formalize the Euclidean
property as follows:
Definition 7. A subring R ⊂ C is a Euclidean domain if there is a size function σ : R \ 0 → Z≥0 such that:
(1) For any a, b ∈ R neither zero, σ(ab) ≥ σ(a), and if σ(ab) = σ(a) then b ∈ R∗ .
(2) For any a, b ∈ R with b 6= 0, there exist q, r ∈ R with r either 0 or with σ(r) < σ(b) such that
a = bq + r
Note that we don’t require q, r to be unique like in the case of Z.
Example 7.
(a) For R = Z we can just take σ(n) = |n|.
(b) Not surprisingly, for R = Z[i] we can take σ(x) = |x|2 .
(c) For R = Q[t], we can take σ(f ) = deg f .
We will prove that examples (b) and (c) above are Euclidean domains in a bit.
Now the key insight is that everything we have done involving gcds in Z can be basically word for word
carried over to any Euclidean domain R. The only slight annoyance is that whereas for Z we could remove
the ambiguity caused by units by requiring positivity, in general there may be more units than {1, −1}.
Definition 8. Let R be a Euclidean domain with size function σ, and let a, b ∈ R not both zero. A gcd(a, b)
is an element which divides both a and b, and such that
σ(gcd(a, b)) = max{n ∈ Z≥0 |n = σ(x) for some x ∈ R with x|a and x|b}
i.e. it has maximum size of all common divisors.
A gcd(a, b) certainly exists since the above set is nonempty (σ(1) is in it), and therefore has a maximum
element, and therefore some common divisor realizes the maximum size. Most properties of gcds follow from:
Lemma 2. Let R be a Euclidean domain with size function σ, and choose any a, b ∈ R not both zero. Then
(a) For a, b ∈ R, if a|b then σ(a) ≤ σ(b). If equality holds, then a ∼ b.
(b) gcd(a, b) is defined up to units, i.e. for any two d, d0 ∈ R common divisors of maximum size, d ∼ d0 .
(c) For any c ∈ R, ax + by = c has a solution with x, y ∈ R if and only if gcd(a, b)|c.
Proof.
(a) Obvious from part (1) of the definition of a Euclidean domain.
(b) We claim that a nonzero element d ∈ R of the form ax + by (where x, y ∈ R) with smallest size is a
gcd(a, b). Certainly any gcd(a, b) divides ax + by for any x, y ∈ R and therefore divides d. Dividing
a by d,
a = qd + r
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for r = 0 or σ(r) < σ(d). In the latter case
r = a − qd = a − q(ax + by) = a(1 − qx) + b(−y)
has smaller size than d, contradicting the choice of d, so r = 0 and d|a. Similarly, d|b, so d is a
common divisor of a, b, and therefore σ(gcd(a, b)) ≥ σ(d) for any gcd(a, b). By part (a), d is an
associate of every gcd(a, b).
(c) (⇒) As in (b), since gcd(a, b)|a and gcd(a, b)|b, then gcd(a, b)|ax + by.
(⇐) By the proof of (b), a solution to gcd(a, b) = ax + by always exists, so if gcd(a, b)|c then by
multiplying we obtain a solution to ax + by = c.
All of the properties of gcd(a, b) that we knew from before are still true, with the same proofs, though we
have reordered them a bit.
Proposition 5. Let R be a Euclidean domain with size function σ, and take any a, b ∈ R not both zero.
(a) For any q ∈ R, gcd(a, b) ∼ gcd(a, b + qa).
(b) For any q ∈ R, gcd(qa, qb) ∼ q · gcd(a, b).
(c) For any q ∈ R, q| gcd(a, b) if and only if q|a and q|b.
Proof.
(a) Certainly any common divisor of a, b is a common divisor of a, b + qa; similarly, any common
divisor of a, b + qa is a common divisor of a, (b + qa) − qa so gcd(a, b) ∼ gcd(a, b + qa).
(b) ax+by = gcd(a, b) has a solution, so (qa)x+(qb)y = q gcd(a, b) does, and by (a) gcd(qa, qb)|q gcd(a, b).
On the other hand, q gcd(a, b) is certainly a common divisor of qa, qb and thus σ(q gcd(a, b)) ≤
σ(gcd(qa, qb)). By the above lemma, q gcd(a, b) ∼ gcd(qa, qb)
(c) (⇐) Certainly since gcd(a, b)|a and gcd(a, b)|b, then if q| gcd(a, b) it divides a and b.
(⇒) If q is a common divisor of a and b, suppose a = qx and b = qy. Then q gcd(x, y) ∼ gcd(a, b),
and thus q| gcd(a, b).
We now have Euclid’s theorem, which will give us the fundamental theorem of arithmetic:
Theorem 2. (Euclid) Let R be a Euclidean domain. Every irreducible element is prime.
Proof. Let x ∈ R be irreducible. Suppose x|ab. Then gcd(x, a)|x, and by the irreducibility of x either
gcd(x, a) ∼ x or gcd(x, a) ∼ 1. In the former case, x|a; in the latter case, we then have gcd(xb, ab) = b, but
since x is a common factor of xb, ab, x|b.
Corollary 2. Let R be a Euclidean domain, and x ∈ R an irreducible. If x|a1 · · · ak , then x|ai for some i.
Just as in class we now have:
Theorem 3. (Fundamental Theorem of Arithmetic) Let R be a Euclidean domain. Every x ∈ R can be
written as a product of irreducibles; given two such factorizations x = p1 · · · pk = q1 · · · q` with the pi s and
qi s irreducible, after reordering the qi s we have pi ∼ qi for all i.
Example 8. So for example in Z[i],
6 = 3 · (1 + i) · (1 − i) = (−1 + i) · (3i) · (−1 + i)
but 3 ∼ 3i and (1 + i) ∼ (−1 + i) ∼ (1 − i)
Proof. First let’s prove existence. First note that if n ∈ Z≥0 is the minimum size of the elements of R, then
any element x ∈ R with σ(x) = n is irreducible, for if y|x we have σ(y) ≤ σ(x), and since σ(y) can’t be less
than σ(x), y ∼ x. Given x ∈ R, if x is not irreducible, it has a nontrivial divisor y; note that σ(y) < σ(x), or
else y ∼ x and y wouldn’t be a nontrivial divisor. Continuing by induction, eventually we hit the minimum
size, and thus x is a product of irreducibles.
Now for uniqueness; again we’ll prove it by induction on size like in class for Z. Of course the factorization
of 1 is unique up to units. Suppose
x = p1 · · · pk = q1 · · · q`
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are two factorizations of x into irreducibles. By the Corollary to Euclid’s theorem, p1 |qi for some i; after
reordering we may assume p1 = q1 . Thus
x
= p2 · · · pk = q2 · · · q`
p1
are two irreducible factorizations of px1 , and since σ( px1 ) < σ(x) by part (1) of the definition (if there was
equality p1 would have to be a unit, a contradiction) by induction we may assume the statement of the
theorem is true, and thus after reordering the qi s again, qi ∼ pi for all i (and k = `).
Property (a) in the Proposition immediately gives us an (extended) Euclidean algorithm for any Euclidean
domain. Here are some worked out examples:
Example 9.
(a) R = Z[i]. Suppose we want to compute gcd(6 + 5i, 2 + 3i).
gcd(6 + 5i, 2 + 3i)
6 + 5i = 2(2 + 3i) + (2 − i) 13 = |2 + 3i|2 > |2 − i|2 = 5
= gcd(2 + 3i, 2 − i) 2 + 3i = (i)(2 − i) + (1 + i) 5 = |2 − i|2 > |1 + i|2 = 2
= gcd(2 − i, 2i)
2 − i = (−i)(1 + i) + 1
2 = |1 + i|2 > |1|2 = 1
= gcd(1 + i, 1)
1 + i = (1 + i)(1) + 0
=1
Running this backward to compute a solution to (6 + 5i)x + (2 + 3i)y = 1,
2 − i = (−i)(1 + i) + 1
+i 2 + 3i = (i)(2 − i) + (1 + i)
+2 6 + 5i = 2(2 + 3i) + (2 − i)
gives
(6 + 5i)(2) + (−4 + i)(2 + 3i) = 1
(b) R = Q[t]. Suppose we want to compute gcd(t5 + t4 + 1, t3 + t2 + 1).
gcd(t5 + t4 + 1, t3 + t2 + 1)
= gcd(t3 + t2 + 1, −t2 + 1)
= gcd(−t2 + 1, t + 2)
= gcd(t + 2, −3)
= −3 ∼ 1
t5 + t4 + 1 = t2 (t3 + t2 + 1) − t2 + 1
t3 + t2 + 1 = (−t − 1)(−t2 + 1) + t + 2
−t2 + 1 = (−t + 2)(t + 2) − 3
t + 2 = ( t+2
−3 )(−3) + 0
3 = deg b > deg r = 2
2>1
1>0
and running it backward to get a solution of (t5 + t4 + 1)f + (t3 + t2 + 1)g = −3,
−t2 + 1 = (−t + 2)(t + 2) − 3
3
+(t − 2) t + t2 + 1 = (−t − 1)(−t2 + 1) + t + 2
+(t2 − t − 1) t5 + t4 + 1 = t2 (t3 + t2 + 1) − t2 + 1
gives us
(t5 + t4 + 1)(t2 − t − 1) + (t3 + t2 + 1)((t − 2) − t2 (t2 − t − 1)) = −3
or simplifying
(t5 + t4 + 1)(t2 − t − 1) + (t3 + t2 + 1)(−t4 + t3 + t2 + t − 2) = −3
After putting it off for some time, let’s finally show
Proposition 6.
(a) R = Z[i] with size function | · |2 is a Euclidean domain.
(b) R = Q[t] with size function deg is a Euclidean domain.
Proof.
(a) We did this in class; the argument is geometric so it would be difficult to reproduce here.
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(b) Given f, g ∈ R neither zero, clearly deg(f g) = deg(f ) + deg(g) ≥ deg(g). If there is equality then
deg(f ) = 0, and since we know f 6= 0, then f is a unit. This proves part (1) of the definition.
Suppose given f, g ∈ R with g 6= 0. We will show by induction on deg(f ) (keeping g fixed) that
there must exist q, r ∈ R such that
f = qg + r
with either r = 0 or deg(r) > deg(g). First, the base case is if deg(f ) < deg(g); take q = 0, r = f . If
deg(f ) ≥ deg(g), let m = deg(f ) and n = deg(g), and choose a ∈ Q \ 0 so that atm−n g has the same
leading coefficient as f (i.e. the coefficients of tm are the same). Thus, deg(f − atm−n g) < deg(f ),
so by strong induction we may assume there exist q, r ∈ R with either r = 0 or deg(r) < deg(g) such
that
(f − atm−n g) = qg + r
Now just take q 0 = atm−n + q, so
f = q0 g + r
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