Irreducibility.
Autumn 2010 Problem 5.
Let Z[i] be the ring of Gaussian integers. Then 3 is prime in Z[i] but
5 is not. Moreover, if a prime p is not prime in Z[i], then either p = 2 or
p ≡ 1 mod 4.
Proof. Let N : Z[i] → Z be the map defined by N(a + bi) = a2 + b2 . It’s easy to
verify that N is a multiplicative homomorphism.
Let (a + bi)(c + di) = 3. Then N(a + bi) = a2 + b2 is either 1, 3 or 9. If
N(a + bi) = 1, then either a = ±1 and b = 0 or b = ±1 and a = 0. In either case,
a + bi is a unit. If N(a + bi) = 3, then a2 + b2 = 3, but this is impossible. If
N(a + bi) = 9 = a2 + b2 , then either a = ±3 and b = 0 or b = ±3 and a = 0. In
either case, it forces c + di to be a unit. Hence 3 is prime.
Note that 5 = (1 + 2i)(1 − 2i) and N(1 + 2i) = N(1 − 2i) = 5, that is, neither of
1 ± 2i are units (if a + bi is a unit, then N(a + bi) = ±1). Hence 5 is not prime.
Let p be an odd rational prime that is not prime in Z[i]. Let a + bi and c + di
be non-units such that p = (a + bi)(c + di). Since p2 = (a2 + b2 )(c2 + d2 ), it
must be that a2 + b2 = p and c2 + d2 = p. Then a + bi must be prime. Indeed,
if a + bi = (α + βi)(γ + δi), then p = (α2 + β2 )(γ2 + δ2 ). So either α2 + β2 = 1
implying that it’s a unit, or the same can be said of the other factor.
So the ideal (a + bi) is a prime ideal. Consider the following chain of ideals
(p) ⊂ (a + bi) ⊂ Z[i]. Note that [Z[i] : (p)] = p2 , so [Z[i] : (a + bi)] = p. So
the quotient ring Z[i]/(a + bi) is a field isomorphic to Z/pZ. Notice that this
field contains a root of x2 + 1, namely i + (a + bi). So it contains an element of
multiplicative order 4. The order of the multiplicative subgroup of Z[i]/(a + bi) is
p − 1. Hence 4 | p − 1, that is, p ≡ 1 mod 4.
For fun, we show that the second statement is, in fact, an “if and only if”. Note
2 = (−i)(1 + i)2 . Now let p be an odd rational prime that is prime in Z[i]. As
before, we have
[Z[i] : (p)] = p2
i.e., Z[i]/(p) is a degree 2 field extension over Fp . Obviously, we have
Z[i]/(p) ' Fp (i)
where i is a root of x2 + 1. Since Fp (i)/Fp is non-trivial, it follows that x2 + 1 is
irreducible in Fp [x]. Hence, F×
p contains no elements of order 4, so 4 - p − 1. It
must be that p ≡ 3 mod 4.
Autumn 2009 Problem 4.
√
Let R = Z[ −m] where m is a square-free odd integer with m > 3. Then
1
2
• The units of R are ±1.
√
• 2 and 1 + −m are irreducible in R.
• R is not a unique factorization domain.
√
√
Proof. It’s easy to show that N : Z[ −m] → Z defined by a + b −m 7→ a2 + b2 m
√
is a multiplicative homomorphism. So if = a + b −m ∈ R is a unit, then
N() = ±1. So a2 + b2 m = 1, hence, a = ±1 and b = 0.
√
√
Now, suppose 2 = (a + b −m)(c + d −m) factors into two non-units. Note
that
4 = (a2 + b2 m)(c2 + d2 m),
so a2 + b2 m = 2 and c2 + d2 m = 2 since otherwise a2 + b2 m = 1 implies that
√
a+b −m is a unit. Now, if b 6= 0, then 2 = a2 +b2 m > m > 2 which is impossible.
So b = 0. But then a2 = 2, a contradiction. Hence 2 must be irreducible.
√
√
√
Suppose 1 + −m = (a + b −m)(c + d −m) factors into two non-units. Note
√
that both b and d cannot be zero since otherwise 1 + −m ∈ Z, a contradiction.
Without loss of generality, assume b 6= 0. Taking norms we have 1 + m = (a2 +
b2 m)(c2 + d2 m) and, as before, a2 + b2 m < 1 + m since otherwise c2 + d2 m = 1
making it a unit. Suppose a 6= 0. Then 0 6 a2 − 1 < m − b2 m, so
0 6 (a + 1)(a − 1) < m(1 − b)(1 + b) 6 0
which is impossible. So a = 0 implying that b = ±1. So
√
√
√
√
1 + −m = (± −m)(c + d −m) = ±(c −m − dm).
√
But then ±dm = 1 which is impossible since m is odd. So 1 + −m is irreducible.
√
√
Now, let σ : R → R be defined by a + b −m 7→ a − b −m. It’s easy to verify
√
that σ is a ring automorphism. Hence 1 − −m is irreducible, as well. Then
1 + m = (1 −
√
√
m+1
−m)(1 + −m) = 2 ·
2
yields two factorizations of 1 + m since m is odd. But the first factorization is a
factorization into irreducibles neither of which is the irreducible 2. Hence R is not
a unique factorization domain.
Spring 2000 Problem 5.
If f(x) ∈ Z[x] is reducible in Q[x], then it is reducible in Z[x]. On the
other hand, the converse is false.
Proof. Note that Z[x] is a UFD and p is an irreducible element. So (p) is a prime
ideal. Since g(x)h(x) ∈ (p) it follows that either g(x) ∈ (p) or h(x) ∈ (p).
Alternatively, let g(x) = an xn + · · · + a1 x + a0 and h(x) = bn xn + · · · + b1 x + b0
(we allow 0-valued coefficients). Let i be the first (in ascension) index for which p
3
does not divide both ai and bi . Without loss of generality, suppose p - bi and that
i < n (otherwise there’s nothing to show). Then
g(x)h(x)
≡
(an xn + · · · + ai xi )(bn xn + · · · + bi xi )
≡
x2i (an xn−1 + · · · ai )(bn xn + · · · + bi )
≡
0
mod p
mod p
mod p.
Now, multiplication by x2i has no effect whatsoever on the coefficients of any polynomial. It follows that we might as well have assumed that i = 0 from the beginning.
Let ci be the coefficients of g(x)h(x). Then
c0
= a0 b0
c1
= a0 b1 + a1 b0
c2
= a0 b2 + a1 b1 + a2 b0
..
.
cn
= a0 bn + a1 bn−1 + · · · + an−1 b1 + an b0 ,
and so on. Now, since we’re assuming p - b0 and p | c0 , it follows that p | a0 .
Suppose we’ve already shown that p | a0 , a1 , . . . , ak for some positive integer k < n.
Since p | ck+1 = a0 bk+1 + · · · ak b1 + ak+1 b0 , it follows that p | ak+1 b0 , so p | ak+1 .
Hence, by induction, p | ak for all k 6 n, so p | g(x).
Let f(x) = G(x)H(x) where G, H ∈ Q[x]. Let d1 , d2 ∈ Z such that d1 G(x) =
g(x) ∈ Z[x] and d2 H(x) = h(x) ∈ Z[x] (e.g. take di to be the least common multiple
of all the denominators in the coefficients of H and G). Then df(x) = g(x)h(x).
Let p be a prime dividing d. Then g(x)h(x) ≡ 0 mod p, so, by part (a), either
g ∈ (p) or h ∈ (p). Either way, we may cancel out a factor of p from both sides
of df(x) = g(x)h(x). We may repeat this process until all prime factors of d are
eliminated thus arriving at uf(x) = g̃(x)h̃(x) where u = ±1 and g̃, h̃ ∈ Z[x]. So f
factors in Z[x].
Now, on the other hand, note that 2x is reducible in Z[x]. But 2 is a unit in Q,
hence 2x is irreducible in Q[x]. So the converse is false.
Autumn 2001 Problem 2.
Let I be an ideal of Q[x]. Then Q[x]/I ' Q if and only if I = (x − a)Q[x]
for some a ∈ Q.
Proof. Fix a ∈ Q. Define ϕ : Q[x] → Q by f(x) 7→ f(a). Note that ker ϕ = {f(x) ∈
Q[x] : a is a root of f}. Since Q[x] is a Euclidean domain, for f ∈ ker ϕ, write
f(x) = (x − a)q(x) + r where r ∈ Q. Since f(a) = 0 + r = 0, it follows that r = 0.
So (x − a) | f. So ker ϕ ⊆ (x − a)Q[x]. The other inclusion is trivial, moreover, the
4
surjectivity of ϕ is easy to verify. Hence, by the isomorphism theorems for rings,
Q[x]/(x − a)Q[x] ' Q.
Now, conversely, let ϕ : Q[x]/I → Q be an isomorphism. Consider the chain of
maps
π
ϕ
Q[x] −
→ Q[x]/I −
→ Q,
where π is the natural projection. Let φ = ϕ ◦ π. Then it’s clear that ker φ = I and
φ is a surjective homomorphism. Since φ is a homomorphism it follows that φ(1)
is either 0 or 1 (since φ(1)φ(1) = φ(1)). If φ(1) = 0. Then φ(f(x)) = φ(1 · f(x)) =
φ(1) · φ(f(x)) = 0, so φ isn’t surjective. So it must be the case that φ(1) = 1.
So φ fixes Q. Let φ(x) = a for some a ∈ Q. Then, since φ is a homomorphism,
φ : f(x) 7→ f(a) for any given f ∈ Q[x]. So ker φ = (x − a)Q[x] = I.
Spring 2006 Problem3.
In the ring Q[x, y] there exists a finite set of generators for the ideal
I := {f(x, y) ∈ Q[x, y] : f(i, i) = 0}.
Proof. Note that (x − y), (x2 + 1) ∈ I. Consider the following diagram
π1
π2
Q[x, y]
Q[x, y]/(x − y) ' Q[x]
Q[x]/(x2 + 1) ' Q(i)
I
π1 (I)
π2 (π1 (I))
(x − y, x2 + 1)
(x2 + 1)
0
(x − y)
0
0
Since Q(i) is a field, it has no proper ideals. It follows that π2 (π1 (I)) = 0, so
π1 (I) = (x2 + 1), so I = (x − y, x2 + 1).