Algebraic function fields
Places
Definition An algebraic function field F/K of one variable over K is an
extension field F ⊇ K such that F is a finite algebraic extension of K(x)
for some element x ∈ F which is transcendental over K.
K = {z ∈ F : z is algebraic over K}
K is called the field of constants of F/K.
K is algebraically closed in F if K = K.
We have
1. K ⊆ K ⊂ F
2. F/K is an algebraic function field over K (why?)
1
Definition F/K is called rational, if F = K(x) for some x ∈ F transcendental over K.
Definition A valuation ring of the function field F/K is a ring O ⊆ F
with the following properties.
1. K ⊂ O ⊂ F
2. For any z ∈ F , z ∈ O or z −1 ∈ O
Example Let p(x) ∈ K[x] be an irreducible polynomial. Then
f (x)
: f (x), g(x) ∈ K[x] and p(x) 6 | g(x)
Op(x) =
g(x)
is a valuation ring of K(x)/K.
2
Proposition Let O be a valuation ring of the function field F/K. Then
(a) O is a local ring, i.e., O has a unique maximal ideal P = O \ O ∗ ,
where O ∗ is the group of units of O.
(b) For 0 6= x ∈ F , x ∈ P if and only if x−1 6∈ O.
(c) K ⊂ O and K ∩ P = {0}.
Proof
(a) It suffices to show that P = O \ O ∗ is an ideal of O.
• Let x ∈ P and let z ∈ O. Then xz 6∈ O ∗ (why?), implying xz ∈ P .
• Let x, y ∈ P . We may assume that x/y ∈ O. Then 1 + x/y ∈ O,
implying x + y = y(1 + x/y) ∈ P .
(b) ...
3
(c) Let z ∈ K and assume to the contrary that z 6∈ O. Then z −1 ∈ O.
Since z −1 is algebraic over K, there are elements a1 , . . . , ar ∈ K such
that
ar (z −1 )r + · · · + a1 z −1 + 1 = 0,
implying
z −1 (ar (z −1 )r−1 + · · · + a1 ) = −1.
Thus,
z = −(ar (z −1 )r−1 + · · · + a1 ) ∈ K[z −1 ] ⊆ O
which contradicts our assumption.
2
4
Theorem Let O be a valuation ring of the function field F/K and let
P be its unique maximal ideal. Then
(a) P is a principal ideal.
(b) If P = tO, then any 0 6= z ∈ F has a unique representation of the
form z = tn u for some n ∈ Z and u ∈ O ∗ .
(c) O is a principal ideal domain, and if P = tO and {0} 6= I ⊆ O is
an is an ideal, then I = tn O for some n ∈ N.
Definition A ring possessing the above properties is called a discrete
valuation ring.
Lemma Let O be a valuation ring of the algebraic function field F/K,
P be its maximal ideal, and let 0 6= x ∈ P . Let x1 , . . . , xn ∈ P be such
that x1 = x and xi ∈ xi+1 P , i = 1, . . . , n − 1. Then n ≤ [F : K(x)] < ∞.
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Proof The inequality [F : K(x)] < ∞ is immediate (why?) and for the
proof of n ≤ [F : K(x)] it suffices to show that x1 , . . . , xn are linearly
independent over K(x).
Assume to the contrary that
n
P
ϕi xi = 0, where ϕi ∈ K(x), not all equal
i=1
zero. We may assume that all ϕi ∈ K[x] and that x does not divide all
of them (why).
Let j = max{i : x 6 | ϕi }. Then for all i > j there exists a polynomial
n
P
ϕi xi = 0 that
gi ∈ K[x] such that ϕi = xgi , and it follows from
i=1
P
xi ϕi . Therefore,
−xj ϕj =
X xi
X x
i6=j
ϕi +
xi g i ,
−ϕj =
x
x
i<j j
i>j j
implying ϕj ∈ P (why?).
However, since ϕj = ϕj (0)+xgj with gj ∈ K[x] ⊆ O, ϕj (0) = ϕj −xgj ∈
2
P ∩ K (why?), which is impossible, because ϕj (0) 6= 0 (why?).
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Corollary If t ∈ P , then
∞
T
tn O = {0}.
n=1
Proof Assume to the contrary that for some z 6= 0, z ∈
n−1
∞
T
n=1
n−2
tn O. Then
for all n ∈ N, the sequence x1 = z, x2 = t
, x3 = t
, . . . , xn = t
satisfies the lemma prerequisite xi ∈ xi+1 P , i = 1 . . . , n − 1. However,
this is impossible for n > [F : K(x)].
2
7
Proof of the theorem
(a) Assume to the contrary that P is not principal and let 0 6= x1 ∈ P .
Since P 6= x1 O, there is x2 ∈ P \ x1 O, implying x2 x−1
1 6∈ O. Therefore,
x1 x−1
2 ∈ P and x1 ∈ x2 P .
We proceed by induction and obtain an infinite sequence x1 , x2 , . . . ∈ P
such that xi ∈ xi+1 P for all i = 1, 2, . . .. However, this contradicts the
lemma.
(b) The uniqueness of the representation is immediate (why?), and for
the proof of the existence we may assume that z ∈ O (why?).
If z is a unit, then z = t0 z, i.e., u = z. Otherwise, z ∈ P (why?) and,
by the corollary to the lemma, there is a maximal non-negative integer
n such that z ∈ tn O. Therefore, z = tn u, where u is a unit (why?).
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(c) Let {0} 6= I ⊆ O be an ideal an let A = {r ∈ N : tr ∈ I}. Then
A 6= ∅ (why?) and let n = min(A).
Since tn O ⊆ I, the proof of part (c) of the theorem will be complete if
we show the converse inclusion.
Let 0 6= x ∈ I. Since x = ts u for some u ∈ O ∗ and s ≥ n, ts ∈ I as well.
Therefore, x = tn ts−n u ∈ tn O.
2
Definition
(a) A place P of the function field F/K is the maximal ideal of some
valuation ring O of F/K. Any element t ∈ P such that P = tO
is called a prime element for P (or local parameter or uniformizing
variable).
(b) PF = {P : P is a place of F/K}.
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Remark If O is a valuation ring of F/K and P is its maximal ideal,
then O is uniquely determined by P :
O = {z ∈ F : z −1 6∈ P }.
We call Op = O the valuation ring of the place P .
Definition (The arithmetic of Z ∪ {∞}) For all n ∈ Z,
• ∞+∞=n+∞=∞+n
• ∞>n
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Definition A discrete valuation of F/K is a function v : F → Z ∪ {∞}
with the following properties.
(1) v(x) = ∞ if an only if x = 0.
(2) v(xy) = v(x) + v(y) for all x, y ∈ F .
(3) v(x + y) ≥ min{v(x), v(y)} for all x, y ∈ F .
(4) There exists an element z ∈ F with v(z) = 1.
(5) v(a) = 0 for all 0 6= a ∈ K.
It follows from (2) and (4) that v : F → Z ∪ {∞} is surjective (how?).
Also it follows from the definition that v(x) = v(−x) for all x ∈ F
(how?).
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Lemma (Strict Triangle Inequality) Let v be a discrete valuation of
F/K and let x, y ∈ F be such that v(x) 6= v(y). Then v(x + y) =
min{v(x), v(y)}.
Proof Since v(x) 6= v(y), we may assume that v(x) < v(y). Now, assume
to the contrary that v(x + y) 6= min{v(x), v(y)}. Then v(x + y) > v(x)
(why?), implying a contradiction
v(x) = v((x + y) − y) ≥ min{v(x + y), v(y)} > v(x).
2
Definition To a place P ∈ PF we associate a function vP : F → Z∪{∞}
such that vP (0) = ∞ and for 0 6= z ∈ F vP (z) is defined as follows.
Let t be a prime element for P . Then for some u ∈ O ∗ and some
n ∈ Z, z = tn u and this representation of z is unique (for a given t). We
put vP (z) = n.
This function vP is well-defined (why?).
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Theorem Let F/K be a function field.
(a) For any place P ∈ PF , the function vP is a discrete valuation of
F/K. In addition,
OP
OP∗
= {z ∈ F : vp (z) ≥ 0},
= {z ∈ F : vp (z) = 0},
P
= {z ∈ F : vp (z) > 0}.
An element z of F is a prime element for P if and only if vP (z) =
1.
(b) Conversely, if v is a discrete valuation of F/K, then
• Pv = {z ∈ F : v(z) > 0} is a place of F/K, and
• OPv = {z ∈ F : v(z) ≥ 0} is the corresponding valuation ring.
(c) Any valuation ring O of F/K is a maximal proper subring of F .
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Proof
(a) We shall prove the triangle inequality only. Let x, y ∈ F , vP (x) = n,
and vP (y) = m, where n ≤ m < ∞. Then x = tn u1 and y = tm u2 ,
where u1 , u2 ∈ OP∗ and
x + y = tn (u1 + tm−n u2 ) = tn z,
where z ∈ OP . If z = 0, then
vP (x + y) = ∞ > min{n, m}.
Otherwise, z = tk u, where k ≥ 0 and u ∈ OP∗ , implying
vP (x + y) = vP (tn+k u) = n + k ≥ n = min{vP (x), vP (y)}.
The rest of (a) ...
(b) ...
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(c) Let O be a valuation ring of F/K and let z ∈ F \ O. We shall prove
that O[z] = F . Let y ∈ F and let P be the maximal ideal of O. Then, for
sufficiently large k, vP (yz −k ) ≥ 0 (why?). Consequently, w = yz −k ∈ O,
implying y = wz k ∈ O[z].
2
Since P is a maximal ideal of OP , OP /P is a field, and for z ∈ OP we
define z(P ) to be the residue class of z modulo P . If z ∈ F \ OP , we
put z(P ) = ∞.1 Sometimes we shall also write z + P instead of z(P ) for
z ∈ OP .
Since K ⊂ OP (K ⊂ OP ) (why?) and K∩P = {0} (K∩P = {0}) (why?),
the residue class map OP → OP /P induces a canonical embedding of K
(K) into OP /P .
We shall always consider K (K) as a subfield of OP /P via this embedding.
1 This
∞ is not the same as in discrete valuations.
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Definition Let P ∈ PF .
(a) FP = OP /P is called the residue class field of P and the map
x 7→ x(P ) from F to FP ∪ {∞} is called the residue class map with
respect to P .
(b) deg P = [FP : K] is called the degree of P .
What happens when deg P = 1?
Proposition Let P be a place of F/K and let 0 6= x ∈ P . Then
deg P ≤ [F : K(x)] < ∞.
Proof We have already seen that [F : K(x)] < ∞ (where?), and for
the proof of deg P ≤ [F : K(x)] it suffices to show that any elements
z1 , . . . , zn of OP , whose residue classes z1 (P ), . . . , zn (P ) are linearly independent over K, are linearly independent over K(x).
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So, let z1 (P ), . . . , zn (P ) be linearly independent over K, and assume to
n
P
ϕi z i = 0
the contrary that there is a non-trivial linear combination
i=1
with ϕi ∈ K(x).
We may assume that the ϕi are polynomials in x and not all of them
are divisible by x (why?). That is, ϕ = ai + xgi , where ai ∈ K and
gi ∈ K[x], and not all ai = 0.
Applying the residue class map to
n
P
ϕi zi = 0 we obtain
i=1
0 = 0(P ) =
n
X
ϕi (P )zi (P ) =
i=1
n
X
ai zi (P ),
i=1
which contradicts the linear independence of z1 (P ), . . . , zn (P ) over K.
2
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Definition Let z ∈ F and let P ∈ Pf . We say that P is a zero (respectively, a pole) of z if and only if vP (z) > 0 (respectively, vP (z) < 0).
If vP (z) = m > 0 (respectively, vP (m) = −m < 0), then P is a zero
(respectively, pole) of P of order m.
Theorem Let F/K be a function field, K ⊂ R be a subring of F , and
let I 6= {0} be a proper ideal of R. Then there is a place P ∈ PF such
that I ⊆ P and R ⊆ OP .
Proof Let
F = {S : S is a subring of F with R ⊆ S and IS 6= S}.2
F 6= ∅, because R ∈ F, and we shall prove that F is inductively ordered
by inclusion (what does it mean?).
2 IS
is the ideal of S generated by I.
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S
Indeed, if H ⊆ F is a totally ordered subset of F, then T = {S : S ∈
H} is a subring of S with R ⊆ T . We have to show that IT 6= T .
Pn
For this, assume to the contrary that i=1 ai si = 1, with ai ∈ I and si ∈
T . Since H is totally
Pn ordered, there is an S ∈ H such that s1 , . . . , sn ∈ S.
Therefore, 1 = i=1 ai si ∈ S as well, in contradiction with the definition
of F (which contradiction?).
By Zorn’s lemma, F contains a maximal element O that, as we prove
below, is a valuation ring of F/K.
We have K ⊂ O ⊂ F (why?).
Assume to the contrary that there exists an element z of F with z 6∈
O and z −1 6∈ O. Then IO[z] = O[z] and IO[z −1 ] = O[z −1 ] (why?).
Consequently, there are a0 , . . . , an , b0 , . . . , bm ∈ IO such that
1 = a 0 + a1 z + · · · + a n z n
1 = b0 + b1 z −1 + · · · + an z −m
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It follows that m, n ≥ 1 (how?) and we may assume that m and n are
chosen minimally and m ≤ n (why?). Multiplying the first equality by
1 − b0 and the second by an z n we obtain
1 − b0 = (1 − b0 )a0 + (1 − b0 )a1 z + · · · + (1 − b0 )an z n
0 = (b0 − 1)an z n + b1 an z n−1 + · · · + bm an z n−m
Adding these equalities yields
1 = c0 + c1 z + · · · + cn z n−1
with coefficients ci in IO (why?). This contradicts the minimality of n.
That is, we have shown that for any z ∈ F , z ∈ O or z −1 ∈ O.
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2
Corollary Let F/K be a function field and let z ∈ F be transcendental
over K. Then z has at least one zero and one pole.3
Proof Let R = K[z]. Then I = zK[z] is a proper ideal of R. Thus,
there is a place P ∈ PF with z ∈ P (hence P is a zero of z).
To prove that z has a pole we start with ...
2
Corollary The field K of constants of F/K is a finite extension of K.
Proof Let P ∈ PF . Since K embeds into FP via the residue class map
OP → FP , it follows that
[K : K] ≤ [FP : K] (< ∞).
2
3 In
particular, PF 6= ∅.
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