Download Valuations and discrete valuation rings, PID`s

Discrete valuation rings
Suppose F is a field. A discrete valuation on
F is a function v : F ∗ − {0} → Z such that:
1. v is surjective.
2. v(ab) = v(a) + v(b).
3. v(a + b) ≥ min(v(a), v(b)) if a + b 6= 0.
Proposition: The set
R = {0} ∪ {r ∈ R : v(r) ≥ 0}
is a ring, which we will call the valuation ring
of v.
Def: An integral domain will be called a discrete valuation ring if it is the valuation ring
of a discrete valuation of its quotient field.
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Proof of the Proposition: We have to show
that R is closed under subtraction and multiplication, and that 1 ∈ R.
Note first that
v(1) = v(1 · 1) = v(1) + v(1)
forces v(1) = 0, so 1 ∈ R. Then
v(−1) + v(−1) = v(−1 · −1) = v(1) = 0
so v(−1) = 0 and −1 ∈ R.
If a, b ∈ R, then v(a) ≥ 0 and v(b) ≥ 0. We get
v(ab) = v(a) + v(b) ≥ 0
so ab ∈ R. Also,
v(a−b) ≥ v(a)+v(−b) = v(a)+v(−1)+v(b) ≥ 0
so a − b ∈ R and we’re done.
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Theorem: Discrete valuation rings (D.V.R.’s)
are Euclidean.
Proof: We define N : R → Z≥0 by N (0) = 0
and N (r) = v(r) if 0 6= r ∈ R.
To show the Euclidean property, suppose a, b ∈
R and 0 6= b. We have to find q, r ∈ R with
a = qb + r
and r = 0 or N (r) < N (b).
If v(a) ≥ v(b) then v(a/b) = v(a) − v(b) ≥ 0 so
q = a/b ∈ R and we can let r = 0.
Suppose v(a) < v(b). This case is easy: let
q = 0 and r = a.
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Example: Suppose p is a rational prime. There
is a discrete valuation
vp : Q∗ → Z
defined by letting
a
vp( ) = ordp(a) − ordp(b)
b
for all non-zero integers a, b, where ordp(a) is
the exponent of the highest power of p which
divides a. The valuation ring of vp is just the
localization
a
Z(p) = { ∈ Q : a, b ∈ Z, p 6 |b}
b
It will be a homework problem on the next
homework assignment to check that every discrete valuation of Q has the form vp for some
prime p.
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Example: Suppose F = C(t) is the rational
function field in one variable. If α ∈ C, we can
define a discrete valuation
vα : C(t)∗ → Z
by
vα(f (x)) = n
if
s(x)
n
f (x) = (x − α) ·
t(x)
with
s(α) 6= 0 6= t(α).
The discrete valuation ring of vα is the localization
h(x)
C[x](x−α) = {
: h(x), g(x) ∈ C[x], g(α) 6= 0}
g(x)
and all the discrete valuations of C(t) have the
form vα for some α ∈ C.
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Example: If p is a prime, the p-adic integers
Zp form the valuation ring of the discrete valuation
vp : Q∗p → Z
defined in the following way.
Let vp(r) = n if rp−n is a unit of Zp.
This is equivalent to saying that r can be represented in the form
r = pn
∞
X
aipi
i=0
with ai ∈ {0, . . . , p − 1} for all i and a0 6= 0.
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Greatest common divisors
Def: Suppose R is a commutative ring and
that a, b ∈ R. An element d ∈ R is called a
G.C.D. (greatest common divisor) for a and b
if:
1. d|a in the sense that a = dd0 for some d0 ∈
R, and d|b.
2. If d0 is any other element of R such that
d0|a and d0|b then d0|d.
Write gcd(a, b) for the set of all G.C.D.’s for a
and b.
Example: The elements d of R = Z which are
G.C.D.’s for a = 2 and b = 3 are ±1. This
illustrates the fact if d ∈ gcd(a, b) then du ∈
gcd(a, b) for all units u ∈ R∗.
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Theorem: Suppose that a, b ∈ R and that the
ideal Ra + Rb equals Rd for some d ∈ R. Then
d ∈ gcd(a, b).
Proof: If d0|a and d0|b then Ra + Rd ⊂ Rd0.
So Rd = Ra + Rb ⊂ Rd0 and d = f d0 for some
f ∈ R. This means d0|d in R.
Warning: The converse is not true. For example, suppose R = Z[x], a = 2 and b = x. We
claim that d = 1 is in gcd(a, b), but Ra + Rb 6=
Rd.
It’s clear that 1|a and 1|b. So to show 1 ∈
gcd(a, b) it will suffice to show that if d0|a and
d0|b in R, then d0|1, which is the same as saying
d0 ∈ R∗.
If d0|2, then d0 must be a constant polynomial,
and d ∈ {±1, ±2}. But d0|x implies d0 6∈ {±2}.
So d0 = ±1 is a unit in R, and 1 ∈ gcd(2, x).
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The ideal Ra + Rb = R · 2 + R · x is not equal to
R · 1 = R, though, since the constant term of
every element of R · 2 + R · x is an even integer.
Corollary: If R is a P.I.D, then every pair of
elements a, b of R has a G.C.D. If one of a or
b is nonzero, then all the elements of gcd(a, b)
differ from each other by multiplication by a
unit.
Proof: Given a, b ∈ R we know Ra + Rb = Rd
for some d ∈ R since R is a P.I.D, and d ∈
gcd(a, b) by the previous Theorem. If of a or b
is not zero, then d must be non-zero as well.
If d0 is some other element of gcd(a, b), then
d0|d and d|d0. So d0 = ud and d = vd0 for some
u, v ∈ R. Then d = uvd so d(1 − uv) = 0. Since
d 6= 0 and R is an integral domain, this forces
uv = 1, so u, v ∈ R∗.
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The Euclidean algorithm
Theorem: Suppose R is a Euclidean domain
with Euclidean norm N : R → Z≥0. Let a and
b be elements of R, and suppose 0 6= b. Construct a sequence of pairs (qi, ri) of elements
of R using the Euclidean property of N :
a = qb + r1
with
r1 = 0 or N (r1) < N (b)
b = q1 r1 + r2
with
r2 = 0 or N (r2) < N (r1)
r1 = q2r2 +r3
with
r3 = 0 or N (r3) < N (r2)
···
rk = qk+1rk+1 + 0
The last non-zero remainder rk+1 in this sequence is in gcd(a, b).
Note: The sequence terminates because the
norms N (ri) form a decreasing sequence of
non-negative integers.
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Proof: We know that R is a P.I.D., so gcd(a, b)
is non-empty. An element c of R divides both
a and b if and only if it divides both b and
r1 = a − qb. So by induction,
gcd(a, b) = gcd(b, r1) = gcl(r1, r2) = · · · gcd(rk , rk+1)
However, by assumption, rk+1 divides rk , so
we find
rk+1 ∈ gcd(rk , rk+1) = gcd(a, b)
Note: We can write the equations in the Euclidean algorithm in reverse order and use them
to find c, d ∈ R such that ca + db = rk+1. For
example,
rk−1 − qk rk
rk−1 − qk (rk−2 − qk−1rk−1)
−qk rk−2 + (1 + qk qk−1)rk−1
−qk rk−2 + (1 + qk qk−1)(rk−3 − qk−2rk−2)
···
= ca + db
rk+1 =
=
=
=
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Principal ideal domains
We’ve seen that Euclidean domains are P.I.D.’s.
At the end of section 8.1 of Dummit and Foote’s
book, there is a proof that the ring
!
√
1 + −19
R=Z+Z
2
is a P.I.D. which is not Euclidean with respect
to any norm.
One way to determine that a ring R is not a
P.I.D. is to use the following result:
Theorem: Suppose R is a P.I.D.. Then every
non-zero prime ideal of R is maximal.
Example: The ring C[x, y] = A is not a P.I.D..
This is because the ideal P = Ax is prime,
because A/P is isomorphic to the polynomial
ring C[y], which is an integral domain. But P
is not maximal, since it’s property contained in
the maximal ideal Ax + Ay.
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Proof of the Theorem: Let P be a non-zero
prime ideal. Then P = Ra for some 0 6= a ∈ R
since R is a P.I.D..
Suppose P is not a maximal ideal. Then P is
properly contained in some maximal ideal M .
Write M = Rm for some m, again using that
R is a P.I.D.. Then P = Ra ⊂ M = Rm implies
a = m · m0 for some m0 ∈ R. If we can show
m, m0 6∈ P , this will contradict the fact that P
is prime and we’ll be done.
We have m 6∈ P since P = Ra 6= Rm = M .
If m0 ∈ P then m0 = za for some z ∈ R, and
a = m · m0 = mza. But then a · (1 − mz) = 0 so
a 6= 0 implies 1 − mz = 0 since R is an integral
domain. But then m is a unit, so M = Rm =
R, which is impossible since M is a maximal
ideal.
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Def: Supppose R is a commutative ring. A
function N : R → Z≥0 is a Dedekind Hasse
norm on R if for all a, b ∈ R, either a ∈ Rb
or there is an element r ∈ Ra + Rb such that
N (r) < N (b).
Comment: Any Euclidean norm is a Dedekind
Hasse norm.
Theorem: R is a principal ideal domain if
and only if it is an integral domain and has
a Dedekind Hasse norm.
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Proof: In one direction, suppose that R has a
Dedekind Hasse norm N . Let A be any nonzero ideal of R. Let b be an element of A
of minimal norm. We claim A = Rb. If not,
there is an a ∈ A not in Rb. The property of
D-H norms then says there is an element r ∈
Ra + Rb ⊂ A with N (r) < N (b), contradicting
the definition of b.
We’ll postpone proving the other part of the
Theorem until after we’ve discussed unique factorization domains.
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Unique Factorization Domains
Def: Suppose R is an integral domain.
1. An element r ∈ R is irreducible if r 6= 0, r is
not a unit, and whenever r = ab for some
a, b ∈ R, one of a or b is a unit.
2. An element r ∈ R is called a prime element
if Rr is a prime ideal.
3. If a, b ∈ R − {0} and a = bu for some unit
u ∈ R∗, say that a and b are associate. This
defines an equivalence relation on R.
Note: A non-zero prime element r is irreducible. This is because r = ab and Rr prime
implies a ∈ Rr or b ∈ Rr. If a = ur for some
u ∈ R, then a = ur = uab, so a(1 − ub) = 0.
But a 6= 0 since r 6= 0, so 1 − ub = 0 and b is a
unit.
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Def: An integral domain R is a U.F.D. (unique
factorization domain) if the following is true for
each non-zero r ∈ R:
1. There are (possibly not distinct) irreducible
elements p1, · · · , pn of R and a unit u ∈ R∗
such that
r=u
n
Y
pi
i=1
2. The factorization in (1) is unique up to
rearranging the pi, replacing them by associate elements of R and replacing u by a
different unit u0.
Example: Z is a U.F.D.. Each irreducible is a
prime element; these have the form ±p.
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Theorem: If R is a P.I.D. then it is a U.F.D..
Proof: We first have to show that each nonzero r ∈ R has a factorization into a finite product of irreducible elements.
If r can’t be written as a finite product of irreducibles, then there has to be a factorization
r = st in which neither s or t are units. Furthermore one of s or t can’t be written as a
finite product of irreducibles. The ideal Rr is
properly contained in each of Rs and Rt. For
example, if Rr = R(st) = Rs, we would get
s = stu for some u, and s(1 − tu) = 0 would
imply t is a unit, a contradiction.
Let r1 be an element of {s, t} which is not
a finite product of irreducibles. Then Rr is
properly contained in Rr1. On replacing Rr by
Rr1 and continuing this argument, we produce
an infinite chain of ideals
Rr ⊂ Rr1 ⊂ Rr2 ⊂ · · ·
such that Rri 6= Rri+1 for all i.
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The ideal ∪iRri is principal, since R is a P.I.D.
Writing ∪iRri = Rd for some d, we see that
d ∈ Rrj for some j, But then
Rrj+1 ⊂ Rd ⊂ Rrj
a contradiction. So every r ∈ R can be written
as a finite product of irreducibles.
To prove the essential uniqueness of the factorization of r into irreducibles, suppose that
r = up1 · · · pn = vq1 · · · qm
(1)
for some integers n, m ≥ 0 and some irreducibles
pi and qj and some units u and v .
We claim Rp1 is a maximal ideal. If not, then
Rp1 is properly contained in M = Rm for some
maximal ideal M and some m ∈ R. But then
p1 = mm0 and neither m nor m0 is a unit, so
p1 is not irreducible. We’ve thus shown Rp1 is
maximal, and hence a prime ideal.
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Now (1) shows
q1 · · · qm ∈ Rr ⊂ Rp1
so one of the qi must be in Rp1. Then p1
divides qi for some i, but this implies p1 and qi
are associates since they are irreducible.
Because R is an integral domain, we can then
cancel p1 and qi from the left and right sides
of
up1 · · · pn = vq1 · · · qm
after adjusting the units u and v. We now continue by downward induction on n + m; this
proves the essential uniqueness of factorizations into irreducibles.
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