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Some consequences of the Riemann-Roch theorem
Proposition Let g0 ∈ Z and W0 ∈ DF be such that for all A ∈ DF ,
dim A = deg A + 1 − g0 + dim(W0 − A).
Then g0 = g and W0 is a canonical divisor.
Proof We have already seen (where?) that the prerequisites of the theorem
imply dim W0 = g0 and deg W0 = 2g − 2.
Let A ∈ DF be such that deg A ≥ max{2g − 2, 2g0 − 2}. Then dim A =
deg A + 1 − g and dim A = deg A + 1 − g0 , implying g0 = g.
Finally, let W be a canonical divisor. Then
dim W = g = (2g − 2) + 1 − g + dim(W0 − W ).
Therefore, dim(W0 − W ) = 1, which, together with deg(W0 − W ) = 0, implies
that W0 − W is principal. Therefore, W0 ∼ W , i.e., W0 id also canonical. 2
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Proposition A divisor B is canonical if and only if deg B = 2g − 2 and
dim B ≥ g.
Proof Let deg B = 2g − 2 and dim B ≥ g and let W be a canonical divisor.
Then
g ≤ dim B = deg B + 1 − g + dim(W − B) = g − 1 + dim(W − B)
(why?). Therefore, dim(W − B) ≥ 1 and, since deg(W − B) = 0, W − B is
principal...
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Proposition An algebraic function field F/K is rational if and only if g = 0
and there is a divisor A ∈ DF such that deg A = 1.
Proof We have already seen the “only if” direction of the proposition.
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For the proof of the “if” direction, let A ∈ DF be a divisor of degree 1. Since
deg A = 1 ≥ 2g − 1 = −1,
dim A = deg A + 1 − g = 2.
Therefore, there is an integral divisor A0 ∈ [A].
Since dim A0 = 2, there is a non-zero element x ∈ L(A0 ) \ K such that
(x) + A0 ≥ 0,
which is possible if and only if A0 = (x)∞ , because deg A0 ≥ 0 and deg A0 = 1.
Therefore,
[F : K(x)] = deg(x)∞ = deg A0 = 1.
That is, F = K(x).
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Theorem (Strong Approximation Theorem) Let S be a proper subset of PF
and let P1 , . . . , Pr ∈ S. Let x1 , . . . , xr ∈ F and let n1 , . . . , nr ∈ Z. Then there
exists an element x of F such that
• vPi (x − xi ) = ni , for all i = 1, . . . , r, and
• vP (x) ≥ 0, for all P ∈ S \ {P1 , . . . , Pr }.
Proof Let the adele α = (αP )P ∈PF ∈ AF be defined by
xi if P = Pi , i = 1, . . . , r
.
αP =
0 otherwise
Let Q ∈ PF \ S. For sufficiently large m ∈ N,
AF = A F
mQ −
r
X
i=1
(why?).
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(ni + 1)Pi
!
+F
Therefore, there is an element z of F such that
z − α ∈ AF
mQ −
r
X
(ni + 1)Pi
i=1
!
.
In particular,
• vPi (z − xi ) > ni , for all i = 1, . . . , r, and
• vP (z) ≥ 0, for all P ∈ S \ {P1 , . . . , Pr }.
Let y1 , . . . , yr ∈ F be such that vPi (yi ) = ni , i = 1, . . . , r, and let y ∈ F be
such that
• vPi (y − yi ) > ni , for all i = 1, . . . , r, and
• vP (y) ≥ 0, for all P ∈ S \ {P1 , . . . , Pr }.
(Why there is such a y?)
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Then
vPi (y) = vPi ((y − yi ) + yi ) = ni , i = 1, . . . , r
(why?) and, for x = y + z,
vPi (x − xi ) = vPi (y + (z − xi )) = ni , i = 1, . . . , r
(why?) and, for P ∈ S \ {P1 , . . . , Pr },
vP (x) = vP (y + z) ≥ 0
(why?).
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Proposition Let P ∈ PF . Then, for any n ≥ 2g, there is an element x of F
such that (x)∞ = nP .
Proof Since
dim(n − 1)P = (n − 1) deg P + 1 − g
(why?), and
dim nP = n deg P + 1 − g
(why?), L((n − 1)P ) is a proper subspace of L(nP ). Therefore, for all x ∈
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L(nP ) \ L((n − 1)P ), (x)∞ = nP (why?).
Definition Let P ∈ PF . A non-negative integer n is called a pole number of
P if there is an element x of F such that (x)∞ = nP . Otherwise, n is called
a gap number of P .
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Remark
• An non-negative integer n is a pole number of P if and only if
dim(nP ) > dim((n − 1)P ).
• An non-negative integer n is a gap number of P if and only if
L(nP ) = L((n − 1)P ).
• If n1 and n2 are pole numbers of P , then n1 + n2 is also a pole number
of P .
Theorem (Weierstrass Gap Theorem) Let F/K be of positive genus g and
let P be a place of degree one. Then there are exactly g gap numbers
1 = i1 < · · · < ig ≤ 2g − 1
of P .
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Proof Consider the following sequence of vector spaces
K = L(0) ⊆ L(P ) ⊆ L(2P ) ⊆ · · · ⊆ L((2g − 1)P ).
Since dim L(0) = 1, dim L((2g − 1)P ) = g (why?), and for all i,
dim L(iP ) ≤ dim L((i − 1)P ) + 1
(why?), there are exactly g − 1 numbers between 1 and 2g − 1 such that L(iP )
is a proper subspace of L((i−1)P ). The remaining g numbers are gap numbers
of P .
It remains to show that 1 is a gap number of P . Were 1 a pole number of
P , there would not be gap numbers of P at all (why?), which is impossible,
because g > 0 (why?).
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Definition A divisor A ∈ DF is called non-special if i(A) = 0. Otherwise A
is called special.
Remarks
(a) A is non-special if and only if dim A = deg A + 1 − g.
(b) If deg A > 2g − 2, then A is non-special.
(c) The “specialty” of A depends only on the class [A] of A in the divisor
class group CF .
(d) Canonical divisors are special.
(e) Any divisor A with dim A > 0 and deg A < g is special.
(f) If A is non-special and B ≥ A, then B is non-special.
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Lemma Let P1 , . . . , Pg ∈ PF be pairwise different places of degree one and
let A ∈ DF , A ≥ 0, be such that dim A = 1 and deg A ≤ g − 1. Then for some
j = 1, . . . , g, dim(A + Pj ) = 1.
Proof Assume to the contrary that dim(A + Pj ) > 1 for all j = 1, . . . , g, and
let zj ∈ L(A + Pj ) \ L(A), j = 1, . . . , g. Then
vPj (zj ) = −vPj (A) − 1 and vPi (zj ) ≥ −vPi (A) for i 6= j,
and, by Strict Triangle Inequality, the g + 1 elements 1, z1 , . . . , zg of F are
linearly independent over K.
Let D ≥ A + P1 + · · · + Pg be of degree 2g − 1. Then 1, z1 , . . . , zg ∈ L(D),
implying dim D ≥ g + 1. However, by the Riemann-Roch theorem,
dim D = deg D + 1 − g ≤ g.
Therefore, our assumption was wrong.
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Proposition If there are g pairwise different places P1 , . . . , Pg ∈ PF of degree
one, then there exists a non-special divisor B ≥ 0 such that deg B = g and
supp B ⊆ {P1 , . . . , Pg }.
Proof By the lemma, there is a sequence of divisors
0 < Pi1 < Pi1 + Pi2 < · · · < Pi1 + Pi2 + · · · + Pig = B,
{i1 , . . . , ig } ⊆ {1, . . . , g}, such that
dim(Pi1 + Pi2 + · · · + Pij ) = 1,
j = 1, . . . , g. In particular, dim B = 1, implying
deg B + 1 − g = g + 1 − g = 1 = dim B.
That is, B is non-special.
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Lemma Let A, B ∈ DF be such that dim A, dim B > 0. Then
dim A + dim B ≤ 1 + dim(A + B).
In this course we make the additional assumption that K is infinite.
Proof of the lemma Since dim A, dim B > 0, there are positive divisors A0
and B0 such that A0 ∼ A and B0 ∼ B.
Let
X = {D ∈ DF : D ≤ A0 and L(D) = L(A0 )}.
Since, by definition, A0 ∈ X, X 6= ∅.
Let D0 be an element of X of minimal degree (why there is such an element?).
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Let supp B0 = {P1 , . . . , Pr }. Since L(D0 − Pi ) is a proper subspace of L(D0 )
(why?), i = 1, . . . , r, and a vector space over an infinite field is not a union of
finitely many proper subspaces (why?), for some z ∈ F ,
z ∈ L(D0 ) \
r
[
L(D0 − Pi ).
i=1
Let ϕ : L(B0 ) → L(D0 + B0 )/L(D0 ) be the (K-linear) map defined by ϕ(x) =
zx + L(D0 ). We contend that Ker φ = K.
The inclusion K ⊆ Ker φ is obvious (why?), and, for the converse inclusion,
let x ∈ L(B0 )\K. Then each pole of x is in supp B0 and x has a pole. That is,
for some i = 1, . . . , r, vPi (x) < 0. Therefore, since vPi (z) = −vPi (D0 ) (why?),
vPi (zx) = vPi (z) + vPi (x) = −vPi (D0 ) + vPi (x) < −vPi (D0 ),
implying ϕ(x) 6∈ L(D0 )). That is, ϕ(x) 6= 0.
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Therefore,
dim B0 − 1 ≤ dim(D0 + B0 ) − dim D0 ,
implying
dim D0 + dim B0 ≤ 1 + dim(D0 + B0 ).
Finally,
dim A + dim B = dim A0 + dim B0 = dim D0 + dim B0
≤ 1 + dim(D0 + B0 ) ≤ 1 + dim(A0 + B0 ) = 1 + dim(A + B).
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Theorem (Clifford’s Theorem) Let A ∈ DF be such that 0 ≤ deg A ≤ 2g − 2.
Then
1
dim A ≤ 1 + deg A.
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Proof The case of dim A = 0 is trivial, and if dim(W − A) = 0, where W is
canonical, then
1
1
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deg A + (deg A − 2g) ≤ 1 + deg A.
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If both dim A and dim(W − A) are positive, then, by the lemma,
dim A = deg A + 1 − g = 1 +
dim A + dim(W − A) ≤ 1 + dim W = 1 + g
and, by the Riemann-Roch theorem,
dim A − dim(W − A) = deg A + 1 − g
Adding the last two (in)equalities and dividing by 2 yields the desired result.
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