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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 263 38 "Valuations on C(x), Q(x) \+
and K(theta)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 42 "Let P be a complex number and let f be in " }{TEXT 256 1 
"C" }{TEXT -1 14 "(x). Then the " }{TEXT 257 5 "order" }{TEXT -1 28 " \+
of f at P (also called the " }{TEXT 258 9 "valuation" }{TEXT -1 26 " o
f f at P) is defined as:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 26 "v_P(f) = infinity when f=0" }}{PARA 0 "" 0 "" 
{TEXT -1 65 "v_P(f) = 0 when f<>0 and f does not have a root nor a pol
e at x=P" }}{PARA 0 "" 0 "" {TEXT -1 46 "v_P(f) = n when f has a root \+
at x=P of order n" }}{PARA 0 "" 0 "" {TEXT -1 48 "v_P(f) = -n when f h
as a pole at x=P of order n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 36 "If K is a field, then v is called a " }{TEXT 
259 9 "valuation" }{TEXT -1 11 " on K when:" }}{PARA 0 "" 0 "" {TEXT 
-1 22 "v(f)=infinity when f=0" }}{PARA 0 "" 0 "" {TEXT -1 36 "v(a*b) =
 v(a)+v(b)  for all a,b in K" }}{PARA 0 "" 0 "" {TEXT -1 71 "v(a+b) >=
 min(v(a),v(b)), and if v(a)<>v(b) then v(a+b)=min(v(a),v(b))." }}
{PARA 0 "" 0 "" {TEXT -1 92 "If v also has the property that v(a) is a
n integer or infinity for every a, and furthermore:" }}{PARA 0 "" 0 "
" {TEXT -1 10 "v:K --->  " }{TEXT 260 1 "Z" }{TEXT -1 17 " union \{inf
inity\}" }}{PARA 0 "" 0 "" {TEXT -1 27 "is onto then v is called a " }
{TEXT 261 19 "discrete valuation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 44 "If K is a subfield of L, then v is called
 a " }{TEXT 262 21 "valuation of L over K" }{TEXT -1 64 " when v is a \+
valuation on L, and v(a)=0 for all non-zero a in K." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "Note that for every com
plex number P, we get a discrete valuation of " }{TEXT 264 1 "C" }
{TEXT -1 9 "(x) over " }{TEXT 265 1 "C" }{TEXT -1 1 "," }}{PARA 0 "" 
0 "" {TEXT -1 5 "v_P: " }{TEXT 266 1 "C" }{TEXT -1 9 "(x) ---> " }
{TEXT 267 1 "Z" }{TEXT -1 17 " union \{infinity\}" }}{PARA 0 "" 0 "" 
{TEXT -1 63 "So every complex number P gives us a discrete valuation v
_P on " }{TEXT 268 1 "C" }{TEXT -1 198 "(x). And we have v_P(f)>0 if f
(P)=0, v_P(f)<0 if f(P)=infinity and v_P(f)=0 when f(P) is not 0 nor i
nfinity. It is easy to see that if P<>Q then v_P <> v_Q. But do we get
 all discrete valuations of " }{TEXT 269 1 "C" }{TEXT -1 9 "(x) over \+
" }{TEXT 270 1 "C" }{TEXT -1 77 " this way? The answer is no, there is
 1 more valuation, which corresponds to " }{TEXT 276 22 "the point at \+
infinity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
7 "Define " }{TEXT 271 1 "P" }{TEXT -1 3 "^1(" }{TEXT 272 1 "C" }
{TEXT -1 13 ") as the set " }{TEXT 273 1 "P" }{TEXT -1 3 "^1(" }{TEXT 
274 1 "C" }{TEXT -1 4 ") = " }{TEXT 275 1 "C" }{TEXT -1 107 " union \{
infinity\}. By stereographic projection you can make a 1-1 map from th
e points in the complex plane " }{TEXT 277 1 "C" }{TEXT -1 117 " to a \+
sphere minus 1 point. Then you can view that one point as correspondin
g to infinity. This way you can think of " }{TEXT 278 1 "P" }{TEXT -1 
3 "^1(" }{TEXT 279 1 "C" }{TEXT -1 80 ") as a sphere, called the Riema
nn sphere. If P is a point in the Riemann sphere " }{TEXT 280 1 "P" }
{TEXT -1 3 "^1(" }{TEXT 281 1 "C" }{TEXT -1 200 "), then either P is f
inite (so then P is a complex number, and we've already defined what v
_P is) or P=infinity, also called the point at infinity. We still need
 to define what v_P is when P=infinity:" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT 282 12 "Definition: " }{TEXT -1 70 "If P is
 a complex number then v_P is as above, and if P=infinity then:" }}
{PARA 0 "" 0 "" {TEXT -1 28 "  v_P(f) = infinity when f=0" }}{PARA 0 "
" 0 "" {TEXT -1 61 "  v_P(f) = degree(denom(f),x) - degree(numer(f),x)
  when f<>0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 31 "This way, for every point P in " }{TEXT 283 1 "P" }{TEXT -1 3 "
^1(" }{TEXT 284 1 "C" }{TEXT -1 35 ") we have a discrete valuation v_P
:" }{TEXT 285 1 "C" }{TEXT -1 6 "(x)-->" }{TEXT 286 1 "Z" }{TEXT -1 
57 " union \{infinity\} and you get all discrete valuations of " }
{TEXT 287 1 "C" }{TEXT -1 9 "(x) over " }{TEXT 288 1 "C" }{TEXT -1 
180 " this way.  Now if P is finite, then v_P(f) is the root order (or
 -1 times the pole order) of f at x=P. Is there a similar interpretati
on of v_P when P=infinity. The answer is yes." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 281 "Take for example:  f=(x^
4+x^3)/(x^2+5).  According to the definition, v_infinity(f)=2-4=-2. A \+
negative number, so that means a pole. Now when x goes to P=infinity, \+
then f goes to infinity, so that's exactly what a pole means. So f doe
s have a pole at x=infinity, a pole of order 2." }}{PARA 0 "" 0 "" 
{TEXT -1 180 "Likewise, f=7*x^5+2*x^3 has a pole at x=infinity of orde
r 5 which means the valuation is -5, and f=(x^2+3)/(x^5-2) has a root \+
at x=infinity of order 3, which means valuation is +3." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "If P is in " }{TEXT 
289 1 "P" }{TEXT -1 3 "^1(" }{TEXT 290 1 "C" }{TEXT -1 9 ") then a " }
{TEXT 291 16 "local parameter " }{TEXT -1 8 "at P is:" }}{PARA 0 "" 0 
"" {TEXT -1 25 "t = x-P  when P is finite" }}{PARA 0 "" 0 "" {TEXT -1 
27 "t = 1/x    when P=infinity." }}{PARA 0 "" 0 "" {TEXT -1 9 "Now f i
n " }{TEXT 292 1 "C" }{TEXT -1 72 "(x) can always be written as f=sum(
a.i * t^i, i=N..infinity) where N in " }{TEXT 293 2 "Z," }{TEXT -1 64 
" and if a.N<>0 then N is the valuation of f at x=P. For example," }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 21 "f:=(x^4+x^3)/(x^2+5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*
&,&*$)%\"xG\"\"%\"\"\"F+*$)F)\"\"$F+F+F+,&*$)F)\"\"#F+F+\"\"&F+!\"\"" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "series(f,x=infinity);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,4*$)%\"xG\"\"#\"\"\"F(F&F(\"\"&!\"\"*
&F)F(F&F*F**&\"#DF(*$F%F(F*F(*&F-F(*$)F&\"\"$F(F*F(*&\"$D\"F(*$)F&\"\"
%F(F*F**&F4F(*$)F&F)F(F*F*-%\"OG6#*&F(F(*$)F&\"\"'F(F*F(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 87 "Note that to compute a series expansion a
t x=infinity, one can simply do the following:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 18 "ff:=subs(x=1/t,f);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#ffG*&,&*&\"\"\"F(*$)%\"tG\"\"%F(!\"\"F(*&F(F(*$)F+\"
\"$F(F-F(F(,&*&F(F(*$)F+\"\"#F(F-F(\"\"&F(F-" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 15 "series(ff,t=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#+5%\"tG\"\"\"!\"#F%!\"\"!\"&\"\"!F(\"\"\"\"#D\"\"#F+\"\"$!$D\"\"\"%
F.\"\"&-%\"OG6#F%\"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "And in
deed we see a pole of order 2 when t->0 (i.e. when x=1/t  goes to infi
nity), so a pole of order 2 at infinity." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "Now we list the essential properti
es of differentiation in " }{TEXT 294 1 "C" }{TEXT -1 389 "(x), the on
es that we really need in the integration algorithm. We will see later
 that in the logarithmic case, we have almost the same properties. Tha
t is why integration in the logarithmic case was so similar to integra
ting rational functions. However, in the exponential case we will see \+
a bigger difference, and although the difference is small it will comp
licate matters considerably." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT 322 13 "Proposition 1" }{TEXT -1 25 ": Let P be an ele
ment of " }{TEXT 295 1 "P" }{TEXT -1 3 "^1(" }{TEXT 296 1 "C" }{TEXT 
-1 34 "). Let f be a non-zero element of " }{TEXT 297 1 "C" }{TEXT -1 
5 "(x). " }}{PARA 0 "" 0 "" {TEXT -1 9 "P finite:" }}{PARA 0 "" 0 "" 
{TEXT -1 74 "   If v_P(f)=0 then v_P(f') can be any element of \{0,1,2
,...\} or infinity." }}{PARA 0 "" 0 "" {TEXT -1 39 "   If v_P(f)<>0 th
en v_P(f') = v_P(f)-1" }}{PARA 0 "" 0 "" {TEXT -1 11 "P infinity:" }}
{PARA 0 "" 0 "" {TEXT -1 74 "   If v_P(f)=0 then v_P(f') can be any el
ement of \{2,3,4,...\} or infinity." }}{PARA 0 "" 0 "" {TEXT -1 39 "  \+
 If v_P(f)<>0 then v_P(f') = v_P(f)+1" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 179 "When P is finite we already knew thi
s. If P=infinity, then just express things in terms of the local param
eter t=1/x. However, realise that the differentiation ' = d/dx = d/d(1
/t)" }}{PARA 0 "" 0 "" {TEXT -1 27 "Now d(a(t))=diff(a(t),t)*dt" }}
{PARA 0 "" 0 "" {TEXT -1 180 "So d/dx=d/d(1/t) = d/(-1/t^2 * dt) = -t^
2 * dt.  Now this factor -t^2 has valuation +2, which explains why at \+
x=infinity the valuations all turn out 2 higher than at finite points.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 "We a
lso notice that the behavior is the same at all but a finite number of
 points. Those finitely many exception points will be called " }{TEXT 
298 14 "special points" }{TEXT -1 18 ". We see that for " }{TEXT 299 
1 "C" }{TEXT -1 202 "(x), there is only 1 special point, the point at \+
infinity. In the logarithmic case, infinity is also the only special p
oint. In the exponential case, there are two special points, namely 0 \+
and infinity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 335 "The reason we will only consider exponential extensions \+
and logarithmic extensions in this course is that, for the algebraic e
xtensions, to study these valuations requires some algebraic geometry,
 namely the study of algebraic curves, which is outside of the scope o
f this course. For example, the valuations on the differential field \+
" }{TEXT 300 1 "C" }{TEXT -1 114 "(x, sqrt(x^3+x)) will correspond to \+
points on the algebraic curve y^2=x^3+x, and just like with the valuat
ions on " }{TEXT 301 1 "C" }{TEXT -1 251 "(x), there will also be poin
t(s) at infinity. Furthermore, the integration of algebraic functions \+
leads to a difficult problem in algebraic geometry, a complete procedu
re for elementary integration could only be given when that problem ha
d a solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "We will now consider all \+
valuations of " }{TEXT 302 1 "Q" }{TEXT -1 9 "(x) over " }{TEXT 303 1 
"Q" }{TEXT -1 11 ".  If f in " }{TEXT 304 1 "Q" }{TEXT -1 134 "(x), in
 other words f is a rational function and its coefficients are rationa
l numbers, then f can only have a root or a pole at P in " }{TEXT 305 
1 "P" }{TEXT -1 3 "^1(" }{TEXT 306 1 "C" }{TEXT -1 78 ") when either P
 is infinity or P is an algebraic number. So each element P of " }
{TEXT 307 1 "P" }{TEXT -1 3 "^1(" }{TEXT 308 1 "Q" }{TEXT -1 6 "_bar)=
" }{TEXT 309 1 "Q" }{TEXT -1 50 "_bar union \{infinity\} gives us a va
luation v_P on " }{TEXT 310 1 "Q" }{TEXT -1 34 "(x), and we get all va
luations on " }{TEXT 311 1 "Q" }{TEXT -1 22 "(x) in that way. Here " }
{TEXT 312 1 "Q" }{TEXT -1 102 "_bar stands for the set of all algebrai
c numbers, the set of all roots of all non-zero polynomials in " }
{TEXT 313 1 "Q" }{TEXT -1 122 "[T] where T is a variable. However, we \+
note that if P=sqrt(2) or P=-sqrt(2), then v_P is the same valuation, \+
because f in " }{TEXT 314 1 "Q" }{TEXT -1 166 "(x) can only have a roo
t (or pole) at x=sqrt(2) if it has a root (or pole) at x=-sqrt(2). In \+
fact, whenever P and Q are roots of the same irreducible polynomial d \+
in " }{TEXT 315 1 "Q" }{TEXT -1 219 "[T], then v_P=v_Q. Because in suc
h situation, v_P(f)=3 just means that d^3 divides numer(f) and d^4 doe
s not divide numer(f), but v_Q(f)=3 means the same thing. And v_P(f)=-
2 means that d^2 but not d^3 divides denom(f)." }}{PARA 0 "" 0 "" 
{TEXT 317 11 "Definition:" }{TEXT -1 63 " If K is a subfield of L, and
 if P,Q in L, then P,Q are called " }{TEXT 316 17 "conjugated over K" 
}{TEXT -1 67 " when P and Q are roots of the same irreducible polynomi
al in K[T]." }}{PARA 0 "" 0 "" {TEXT -1 104 "So if P and Q are two alg
ebraic numbers, conjugated over Q, then v_P and v_Q are the same valua
tions on " }{TEXT 318 1 "Q" }{TEXT -1 35 "(x). In other words: valuati
ons on " }{TEXT 319 1 "Q" }{TEXT -1 82 "(x) correspond to the set of a
ll monic (i.e. lcoeff=1) irreducible polynomials in " }{TEXT 320 1 "Q
" }{TEXT -1 169 "[T], union \{infinity\}. The valuation at infinity is
 like before, and the valuation at an irreducible polynomial d just me
asures how often d divides numer(f) or denom(f)." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 13 "d:=x^4+3*x+3;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"dG,(*$)%\"xG\"\"%\"\"\"F**&\"\"$F*F(F*F*F,F*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "factor(d);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,(*$)%\"xG\"\"%\"\"\"F(*&\"\"$F(F&F(F(F*F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "alpha:=RootOf(d);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%&alphaG-%'RootOfG6#,(*$)%#_ZG\"\"%\"\"\"F-*&\"\"$F
-F+F-F-F/F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "f:=(x^8+6*x^
5+8*x^4+9*x^2+24*x+16)/(x^11+6*x^8+6*x^7+9*x^5+18*x^4+9*x^3);\n" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&,.*$)%\"xG\"\")\"\"\"F+*&\"\"'
F+)F)\"\"&F+F+*&F*F+)F)\"\"%F+F+*&\"\"*F+)F)\"\"#F+F+*&\"#CF+F)F+F+\"#
;F+F+,.*$)F)\"#6F+F+*&F-F+F(F+F+*&F-F+)F)\"\"(F+F+*&F4F+F.F+F+*&\"#=F+
F1F+F+*&F4F+)F)\"\"$F+F+!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "
Now the valuation of f at infinity is:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "degree(denom(f),x)-degree(numer(f),x);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Wh
ich is positive, which means that f has a root at x=infinity." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "The valua
tion of f at x=alpha is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 
"factors(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"\"7%7$,(*$)%\"xG
\"\"%F$F$*&\"\"$F$F*F$F$F+F$\"\"#7$,(F(F$*&F-F$F*F$F$F-F$!\"#7$F*!\"$
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "v_alpha(f) = -2 because that \+
is the multiplicity of d in f  (d^2 appears in denom(f))." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 235 "We will now consider the valuations on K(theta) over K
, where theta is logarithmic over K. Remember that v is a valuation on
 K(theta) over K means that v(a)=0 for all non-zero a in K. Also remem
ber that theta is logarithmic means that:" }}{PARA 0 "" 0 "" {TEXT -1 
30 "*) theta'=a'/a for some a in K" }}{PARA 0 "" 0 "" {TEXT -1 91 "*) \+
there is no b in K for which b'=a'/a, which implies that theta is tran
scendental over K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 239 "If theta is transcendental over K, then we can view it a
s a variable over K, which we effectively did in previous worksheets b
y replacing the logarithmic function theta by a variable Theta, as thi
s was necessary in some Maple computations." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 535 "For every monic irreducible po
lynomial d in K[T] we have one corresponding valuation on K(theta), an
d there is the valuation for the point at infinity. Lets see how they \+
behave under differentiation. First the finite \"points\". Let alpha b
e a root of the polynomial d in K[T]. Then at theta=alpha, we have the
 local parameter: t-alpha. Now if f in K(theta), we can do a series ex
pansion of f at t=alpha  (to do that in Maple, first replace the logar
ithm theta by a variable Theta, because Maple's series command will ex
pect a variable)." }}{PARA 0 "" 0 "" {TEXT -1 67 "f = sum(a.i * t^i, i
=N..infinity)  with a.N<>0. Then differentiate:" }}{PARA 0 "" 0 "" 
{TEXT -1 55 "f' = sum(a.i' * t^i  + a.i*i*t^(i-1)*t', i=N..infinity)" 
}}{PARA 0 "" 0 "" {TEXT -1 655 "Now t'=(theta-alpha)'=a'/a-alpha' whic
h contains no theta, and can not be zero because we've seen that there
 exists no algebraic alpha with alpha'=a'/a unless K itself has an ele
ment alpha with alpha'=a'/a. The term with the lowest power in t is th
en: a.N*N*t^(N-1)*a'/a which has valuation N-1, unless when N=0. When \+
N=v_alpha(f)=0 then we do not have terms with negative powers in t, so
 the valuation of f' can be any non-negative integer or can also be in
finity when f was constant. When N<>0, then a.N*N*t^(N-1)*(a'/a-alpha'
) is the term with the lowest power of t, so that's the only term that
 determines the valuation of f', so then v_alpha(f')=N-1." }}{PARA 0 "
" 0 "" {TEXT -1 77 "Now the point at infinity, which means that the lo
cal parameter is t=1/theta." }}{PARA 0 "" 0 "" {TEXT -1 67 "f = sum(a.
i * t^i, i=N..infinity)  with a.N<>0. Then differentiate:" }}{PARA 0 "
" 0 "" {TEXT -1 55 "f' = sum(a.i' * t^i  + a.i*i*t^(i-1)*t', i=N..infi
nity)" }}{PARA 0 "" 0 "" {TEXT -1 873 "but now t'=(1/theta)' = (-1/the
ta^2)*theta'=(-1/theta^2)*a'/a=-t^2*a'/a and this has valuation 2. So \+
what is the valuation, i.e. what is the lowest power in t that occurs?
 Well, that depends. We have a.N'*t^N + a.N*N*t^(N-1)*(-t^2)*a'/a whic
h has valuation N, except when a.N is constant, because then we have v
aluation N-1 + 2 =N+1 (assuming that N<>0 because when N=0 that term v
anishes since it has a factor N in it). We've seen this behavior in di
fferentiation of polynomials in theta (note that the term of highest d
egree corresponds to the term with lowest valuation at infinity); if y
ou differentiate a polynomial of degree -N in theta, with N<>0, then t
he degree drops by 1 (i.e. the valuation N increases by 1) if and only
 if lcoeff(f,theta) is a constant, and if it is not a constant then th
e degree stays the same (N stays the same). We can summarize as follow
s:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 323 14 "Pr
oposition 2:" }{TEXT -1 121 " Let theta be logarithmic over K. Let f b
e a non-zero element of K(theta), and let v_P be a valuation on K(thet
a) over K." }}{PARA 0 "" 0 "" {TEXT -1 9 "P finite:" }}{PARA 0 "" 0 "
" {TEXT -1 33 "   If v_P(f)=0 then v_P(f') >= 0." }}{PARA 0 "" 0 "" 
{TEXT -1 39 "   If v_P(f)<>0 then v_P(f') = v_P(f)-1" }}{PARA 0 "" 0 "
" {TEXT -1 11 "P=infinity:" }}{PARA 0 "" 0 "" {TEXT -1 75 "   If a.N i
s not constant then v_P(f')=v_P(f).\n   If a.N is constant, then:" }}
{PARA 0 "" 0 "" {TEXT -1 35 "        If v_P(f)=0 then v_P(f')>0." }}
{PARA 0 "" 0 "" {TEXT -1 43 "        If v_P(f)<>0 then v_P(f')=v_P(f)+
1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "No
tice that just like for rational functions, v_P(f') is never -1 at any
 finite point P. Furthermore, there is only 1 special point, namely th
eta=infinity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "In a similar way you can \+
prove in the exponential case that (try this yourself!):" }}{PARA 0 "
" 0 "" {TEXT 324 14 "Proposition 3:" }{TEXT -1 121 " Let theta be expo
nential over K. Let f be a non-zero element of K(theta), and let v_P b
e a valuation on K(theta) over K." }}{PARA 0 "" 0 "" {TEXT -1 18 "P fi
nite, but P<>0" }}{PARA 0 "" 0 "" {TEXT -1 33 "   If v_P(f)=0 then v_P
(f') >= 0." }}{PARA 0 "" 0 "" {TEXT -1 39 "   If v_P(f)<>0 then v_P(f'
) = v_P(f)-1" }}{PARA 0 "" 0 "" {TEXT -1 3 "P=0" }}{PARA 0 "" 0 "" 
{TEXT -1 33 "   If v_P(f)=0 then v_P(f') >= 0." }}{PARA 0 "" 0 "" 
{TEXT -1 37 "   If v_P(f)<>0 then v_P(f') = v_P(f)" }}{PARA 0 "" 0 "" 
{TEXT -1 11 "P=infinity:" }}{PARA 0 "" 0 "" {TEXT -1 32 "  If v_P(f)=0
 then v_P(f') >= 0." }}{PARA 0 "" 0 "" {TEXT -1 36 "  If v_P(f)<>0 the
n v_P(f') = v_P(f)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 537 "So there are two special points, P=0 and P=infinity. Bot
h are particularly nasty in the sense that if you have a non-zero valu
ation, the valuation always stays the same under differentiation. That
 makes it hard to \"reduce pole orders\" of \"rational\" functions or \+
\"reduce the degree\" of \"polynomial\" functions (note that reducing \+
pole orders means: increasing the valuation at finite points to make t
hese valuations non-negative, and reducing the degree of \"polynomials
\" means increasing the valuation at infinity to make it non-negative)
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 321 8 "Exam
ple:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "theta:=exp(3*x^2+2)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG-%$expG6#,&*$)%\"xG\"\"#
\"\"\"\"\"$F,F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "f:=norma
l(theta^2*(theta-3)*(theta^2-x)^(-2));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"fG*&*&)-%$expG6#,&*$)%\"xG\"\"#\"\"\"\"\"$F/F0F/F0,&F(F0F1!
\"\"F0F0*$),&*$F'F0F3F.F0F/F0F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 28 "factor(subs(theta=Theta,f));" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#*&*&)%&ThetaG\"\"#\"\"\",&F&F(\"\"$!\"\"F(F(*$),&*$F%F(F+%\"xGF(F'F(
F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "f_prime:=normal(diff(
f,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(f_primeG,$*&*&)-%$expG6#,
&*$)%\"xG\"\"#\"\"\"\"\"$F0F1F0F1,.*&)F)F2F1F/F1F2*(\"\"*F1F)F1F.F1F1*
(\"#=F1F(F1F/F1!\"\"*&F9F1F.F1F:F)F:F2F1F1F1*$),&*$F(F1F:F/F1F2F1F:F0
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "factor(subs(theta=Theta
,f_prime));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&)%&ThetaG\"\"#\"
\"\",.*&)F'\"\"$F)%\"xGF)F-*(\"\"*F)F'F))F.F(F)F)*(\"#=F)F&F)F.F)!\"\"
*&F3F)F1F)F4F'F4F-F)F)F)*$),&*$F&F)F4F.F)F-F)F4F(" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 15 "As you can see:" }}{PARA 0 "" 0 "" {TEXT -1 66 "v_
infinity(f)=degree(denom(f),Theta)-degree(numer(f),Theta)=4-3=1." }}
{PARA 0 "" 0 "" {TEXT -1 66 "v_infinity(f')=6-5=1, it stayed the same,
 like proposition 3 says." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 86 "Furthermore, at the point P is a root of the irred
ucible polynomial Theta^2-x we have:" }}{PARA 0 "" 0 "" {TEXT -1 22 "v
_P(f)=-2, and we saw:" }}{PARA 0 "" 0 "" {TEXT -1 115 "v_P(f')=-3, whi
ch is what it should be according to proposition 3, we had a non-zero \+
valuation, which dropped by 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 142 "Likewise, at the irreducible polynomial \+
Theta-3  (i.e. at the point P=3) we had valuation 1, and after differe
ntiation this dropped by 1 to 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 150 "However, at P=0 we had valuation 2 (a fa
ctor Theta^2 in the numerator) and after differentiation the valuation
 was unchanged, like proposition 3 says." }}}}{MARK "0 0 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 }