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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 33 "The extended Euclidean Al
gorithm." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
128 "If a and b are positive integers, and d is the gcd, then we have \+
that the ideal (d) equals the ideal (a,b), so: Z*d = Z*a + Z*b." }}
{PARA 0 "" 0 "" {TEXT -1 120 "In other words: a and b are Z-linear com
binations of d (i.e. multiples of d) and d is a Z-linear combination o
f a and b." }}{PARA 0 "" 0 "" {TEXT -1 112 "So d = s*a + t*b for some \+
integers s and t. These s and t can be computed with the Extended Eucl
idean Algorithm." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "a:=72;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG\"#s" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 6 "b:=50;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG
\"#]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "d:=igcdex(a,b,'s','
t');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG\"\"#" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 10 "s*a + t*b;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "s;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 2 "t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#8" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 113 "Note that s and t are not uniquely determined. If
 you replace [s,t] by [s + b, t - a] you'll get the same result." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "s, t := s+b, t-a;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>6$%\"sG%\"tG6$\"#T!#f" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 10 "s*a + t*b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 233 "Of course, by adding o
r substracting a suitable multiple of [b,-a] to [s,t] we can get s in \+
the range 0..b-1, and then s,t will be uniquely determined. Or we coul
d bring t in the range -(a-1) .. 0, that would give us the same result
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "For \+
polynomials this all works very similar." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "a:=x^3-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG,&*
$)%\"xG\"\"$\"\"\"\"\"\"!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 9 "b:=x^4-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG,&*$)%\"xG\"
\"%\"\"\"\"\"\"!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "d
:=gcdex(a,b,x,'s','t');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,&!\"
\"\"\"\"%\"xGF'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s*a+t*b;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&%\"xG\"\"\",&*$)F%\"\"$\"\"\"F
&!\"\"F&F&F,*$)F%\"\"%F+F&F,F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 10 "expand(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&!\"\"\"\"\"%\"
xGF%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "If R is the ring Q[x], t
he ideal (a,b)=R*a+R*b equals the ideal (d)=R*d, so d is an R-linear c
ombination of a and b." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 12 "d = s*a+t*b." }}{PARA 0 "" 0 "" {TEXT -1 86 "This eq
uation will still hold if we replace [s,t] by [s,t] + some multiple of
 [b, -a]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "expand(s*a+t*b
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&!\"\"\"\"\"%\"xGF%" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ra:=randpoly(x,degree=2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#raG,(*$)%\"xG\"\"#\"\"\"!#&)F(!#b!#P\"\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "s, t := expand(s+ra*b
), expand(t-ra*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%\"sG%\"tG6$,
.%\"xG\"#a*$)F)\"\"'\"\"\"!#&)*$)F)\"\"#F.\"#&)*$)F)\"\"&F.!#b*$)F)\"
\"%F.!#P\"#P\"\"\",.!#OF=F4F3F0F/F8\"#bF)F7*$)F)\"\"$F.F<" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "expand(s*a+t*b);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,&!\"\"\"\"\"%\"xGF%" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 425 "So if ra is some arbitrary polynomial, we can add ra*b t
o s (and substract ra*a from t) and the equation s*a+t*b=d remains val
id. If s is an element of Q[x], we have seen that we can always find a
 unique polynomial q such that s-q*b will have degree < degree(b). Nam
ely, q is the quotient of the division of s by b. So we can make the s
olution [s,t] of the equation s*a+t*b=d unique by requiring that degre
e(s,x)<degree(b,x)." }}{PARA 0 "" 0 "" {TEXT -1 110 "If you compare de
grees in this equation you will see that this condition implies that d
egree(t,x)<degree(a,x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 106 "Concluding: If a and b are polynomials, and d is th
e gcd, then there are unique polynomials s,t such that:" }}{PARA 0 "" 
0 "" {TEXT -1 14 "1) d = s*a+t*b" }}{PARA 0 "" 0 "" {TEXT -1 26 "2) de
gree(s,x)<degree(b,x)" }}{PARA 0 "" 0 "" {TEXT -1 27 "3) degree(t,x)<d
egree(a,x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 131 "Now we will study how to compute these s and t, in the case of
 polynomials. The case of integers (the igcdex) works very similarly.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 118 "In t
he Euclidean Algorithm we reduce the ideal (a,b) to the ideal (a',b) w
here a'=rem(a,b,x)=a-q*b where q=quo(a,b,x)." }}{PARA 0 "" 0 "" {TEXT 
-1 47 "Then, by recursion we can find s',t' such that:" }}{PARA 0 "" 
0 "" {TEXT -1 36 "s'*a' + t'*b = d where d is the gcd." }}{PARA 0 "" 
0 "" {TEXT -1 25 "So s'*(a-q*b) + t'*b = d." }}{PARA 0 "" 0 "" {TEXT 
-1 27 "So s'*a  + (t'-s'*q)*b = d." }}{PARA 0 "" 0 "" {TEXT -1 60 "So \+
we should take s,t = s', t'-s'*q.       (formula for s,t)" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "Let's implement th
is." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1045 "GCDEX:=proc(a,b,x)
\n  # Produce d,s,t such that:\n  # d = gcd(a,b)\n  # d = s*a+t*b\n  #
 degree(s,x)<degree(b,x)\n  # degree(t,x)<degree(a,x)\n  local c,r,q,d
,s,t;\n  if b=0 then\n      c:=content(a,x);\n      # The following li
ne avoids that it gives gcd = -1.\n      if a/c = -1 then c:=-c fi;\n
\n      # Since b=0 we see that a is the gcd. We divide a by the\n    \+
  # content c because that makes a smaller. For example,\n      # if a
 was something like a=2*x+2 then c would be 2\n      # and then a/c wo
uld be x+1.\n      # So then the gcd is a/c = 1/c * a + 0 * b,\n      \+
# so s, t = 1/c, 0.\n      a/c, 1/c, 0\n  else\n      # The following \+
computes the remainder r as wel as\n      # the quotient q:\n      r:=
rem(a,b,x,'q');\n      # Interchange s,t because a (is now: r) and b a
re being\n      # interchanged by calling the procedure with b,r inste
ad\n      # of r,b.\n      d, t, s := procname(b,r,x);\n      # Now we
've computed the gcd d by recursion. Now apply\n      # the formula fo
r s,t given above and return the answer:\n      d, s, expand(t-s*q)\n \+
 fi\nend;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&GCDEXGR6%%\"aG%\"bG%\"
xG6(%\"cG%\"rG%\"qG%\"dG%\"sG%\"tG6\"F1@%/9%\"\"!C%>8$-%(contentG6$9$9
&@$/*&F<\"\"\"F8!\"\"!\"\">F8,$F8FC6%F@*&FAFAF8FBF5C%>8%-%$remG6&F<F4F
=.8&>6%8'8)8(-9!6%F4FJF=6%FRFT-%'expandG6#,&FS\"\"\"*&FTFgnFOFgnFCF1F1
F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "a;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,&*$)%\"xG\"\"$\"\"\"\"\"\"!\"\"F)" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 2 "b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$
)%\"xG\"\"%\"\"\"\"\"\"!\"\"F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "d,s,t:=GCDEX(a,b,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6%%\"
dG%\"sG%\"tG6%,&!\"\"\"\"\"%\"xGF+,$F,F*F+" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 16 "expand(s*a+t*b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#,&!\"\"\"\"\"%\"xGF%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "a:
=x^50-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG,&*$)%\"xG\"#]\"\"\"
\"\"\"!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "b:=x^32-1;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG,&*$)%\"xG\"#K\"\"\"\"\"\"!
\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "d,s,t:=GCDEX(a,b,
x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6%%\"dG%\"sG%\"tG6%,&!\"\"\"\"
\"*$)%\"xG\"\"#\"\"\"F+,0*$)F.\"#5F0F**$)F.\"\"'F0F*F,F**$)F.\"#9F0F**
$)F.\"#GF0F**$)F.\"#CF0F**$)F.\"#?F0F*,8F+F+F8F+F2F+F5F+F;F+F>F+FAF+*$
)F.\"#KF0F+*$)F.\"#YF0F+*$)F.\"#UF0F+*$)F.\"#QF0F+" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 16 "expand(s*a+t*b);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&!\"\"\"\"\"*$)%\"xG\"\"#\"\"\"F%" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 2 "d;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&!\"\"
\"\"\"*$)%\"xG\"\"#\"\"\"F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 210 "You can verify (for a pro
of you need to use the principle of mathematical induction because the
 algorithm uses recursion) that the conditions on the degrees of s and
 t will hold when we use our algorithm GCDEX." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 413 "Previous time we looked \+
at the ring Q[x]/(f) where f is an element of Q[x]. Every element in t
his ring can be represented uniquely by a polynomial in x of degree sm
aller than degree(f,x). We can add and multiply in this ring, when we \+
multiply and the result would get result >= degree(f,x) then you just \+
reduce modulo f, i.e. you replace your result by rem(result,f,x) and i
t will have degree < degree(f,x) again." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 127 "So we know how to add and multiply \+
in the ring Q[x]/(f). Basically all we need for that are the Maple fun
ctions expand and rem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 478 "Now we want to divide as well. Suppose that g in Q[
x]/(f), so we can represent g uniquely by a polynomial of degree < deg
ree(f,x). We want to do divisions, so we want to compute 1/g if it exi
sts. In other words, we want to find (if it exists) an element t of Q[
x]/(f) such that t*g = 1. In the ring Q[x]/(f) a polynomial equals 1 w
hen its remainder modulo f is 1. So this means that when g is given by
 a polynomial in Q[x], that we want to find a polynomial t in Q[x] suc
h that:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
23 "  rem( t*g, f, x)  = 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 75 "In other words: t*g = q*f + r  where q = quo(s*
g,f,x) and r=rem(s*g,f,x)=1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 33 "So:  s*f + t*g = 1  where t = -q." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "This means: 1 is i
n the ideal (f,g), in other words gcd(f,g)=1." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Conclusion: There exists
 a polynomial t in Q[x] such that t*g=1 in Q[x]/(f) if and only if gcd
(f,g)=1." }}{PARA 0 "" 0 "" {TEXT -1 63 "Furthermore: we can find t by
 the Extended Euclidean Algorithm." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "f:=x^3-2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
13 "g:=x^2-5*x+3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "d, s, \+
t := GCDEX(f,g,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "t;" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "So the gcd is 1. And t*g should n
ow be 1 modulo f. Let's check:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 4 "g*t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "expand(%);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "rem(%,f,x);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 60 "One more example. Take g = x. So we're go
ing to compute 1/x." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 121 "Note that Q[x]/(x^3-2) is isomorphic to Q[ 2^(1/3) ], \+
so computing 1/x is just like computing 1/2^(1/3). And we know that" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "1 / 2^(1
/3) =  1/2 * ( 2^(1/3) )^2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
2 "f;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "g:=x;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "d,s,t := GCDEX(f,g,x);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "t;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 125 "Every element g in Q[x]/(f) is represented by a polynomial of \+
degree < degree(f,x). Now 1/g exists if and only if gcd(f,g)=1." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 131 "Suppose \+
now that f is an irreducible element of Q[x], i.e. there do not exist \+
polynomials f1, f2 of lower degree such that f1*f2=f." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 275 "Then you will see tha
t every non-zero polynomial g of degree < degree(f,x) will have no fac
tors in common with f, because f simply doesn't have any non-trivial f
actors of degree < degree(f,x). In other words: every non-zero g in Q[
x]/(f) will have an inverse 1/g in Q[x]/(f)." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 194 "Q[x]/(f) is a commutative ring
, i.e. we can add and multiply and multipication is commutative. If f \+
is irreducible then we can also divide by any non-zero element. Such a
 ring is called a field." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 158 "Note that if f is reducible, suppose f = f1*f2 wi
th degrees f1,f2 smaller than degree of f, then you could not divide b
y f1 in the ring Q[x]/(f). We conclude:" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 71 "Q[x]/(f) is a field if and only if f
 is an irreducible element of Q[x]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 749 "When f \+
is an irreducible polynomial we can multiply in the field Q[x]/(f) by \+
expand and rem. We can divide by the Extended Euclidean Algorithm (cal
led gcdex in Maple, for the help page type ?gcdex ). It would be easie
r to be able to just write * and / for products and quotients, and to \+
have only one Maple command that will evaluate expressions in the ring
 Q[x]/(f). This is done by the command evala. The element x modulo f i
n Q[x]/(f) should then be denoted by: RootOf(f,x). This RootOf should \+
be interpreted as: \"some unspecified root of the polynomial f\". So i
f alpha := RootOf(x^3-2,x) then you can not say that alpha equals 2^(1
/3), you should think of alpha as an unspecified root of x^3-2, and no
t just one of the three roots in particular." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 2 "f;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "
A:=(x^4+2*x)/(x+1)+x^5/(x-2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "n:=normal(A);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "n,d:=numer(A), denom(A);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "gcdex(f,d,1,x,'s','t');" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "t;" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 4 "t*n;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "
A_simplified := rem(%,f,x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "No
w we'll simplify A modulo f using Maple's evala and RootOf." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "alpha:=RootOf(f,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Ar:=subs(x=alpha,A);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Ar:=evala(Ar);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 115 "The name of the command evala stand for \+
EVALuate Algebraic. This is Maple's command for handling algebraic num
bers." }}}}{MARK "33 2 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }