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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 35 "Defining and Checking Aut
omorphisms" }}{PARA 19 "" 0 "" {TEXT -1 54 "\251Mike May, S.J., maymk@
slu.edu, Saint Louis University" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 320 "As we move th
rough Galois theory, we will be looking at the group, G, of automorphi
sms of a field extension, K/F.  Unfortunately, for students learning f
ield theory, it is often hard to identify what the allowable automorph
isms are.  This makes looking at properties of the group of automorphi
sms a near impossible task." }}{PARA 0 "" 0 "" {TEXT 256 208 "\nThis w
orksheet uses Maple to test proposed automorphisms for finite extensio
ns of the rationals.  You should probably have worked through the work
sheet on Factoring Examples before attempting this worksheet." }}}
{SECT 0 {PARA 5 "" 0 "" {TEXT -1 6 "Set up" }}{EXCHG }{EXCHG {PARA 0 "
" 0 "" {TEXT -1 129 "For this worksheet, we will restrict to the case \+
where F is the field of rationals and K is a finite extension of F.  T
hen K = F[" }{XPPEDIT 18 0 "alpha[1]" "6#&%&alphaG6#\"\"\"" }{TEXT -1 
7 ", ..., " }{XPPEDIT 18 0 "alpha[n]" "6#&%&alphaG6#%\"nG" }{TEXT -1 
7 "] with " }{XPPEDIT 18 0 "alpha[i]" "6#&%&alphaG6#%\"iG" }{TEXT -1 
38 " algebraic over F.  To each generator " }{XPPEDIT 18 0 "alpha[i]" 
"6#&%&alphaG6#%\"iG" }{TEXT -1 38 ", we associate its minimal polynomi
al " }{XPPEDIT 18 0 "f[i]" "6#&%\"fG6#%\"iG" }{TEXT -1 19 ".  An autom
orphism " }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 42 " in G can \+
be defined by giving the images " }{XPPEDIT 18 0 "sigma(alpha[1])" "6#
-%&sigmaG6#&%&alphaG6#\"\"\"" }{TEXT -1 8 " , ..., " }{XPPEDIT 18 0 "s
igma(alpha[n])" "6#-%&sigmaG6#&%&alphaG6#%\"nG" }{TEXT -1 25 " of the \+
field generators " }{XPPEDIT 18 0 "alpha[1]" "6#&%&alphaG6#\"\"\"" }
{TEXT -1 7 ", ..., " }{XPPEDIT 18 0 "alpha[n]" "6#&%&alphaG6#%\"nG" }
{TEXT -1 118 " . The question of listing automorphisms in G then reduc
es to finding the acceptable sets of images of the generators." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 90 "Our prim
e examples will look at the fields obtained by adjoining roots to the \+
polynomial  " }{XPPEDIT 18 0 "x^p - q" "6#,&)%\"xG%\"pG\"\"\"%\"qG!\"
\"" }{TEXT 258 121 ".   We will focus first on the particular case whe
n p is 3 and q is 2.  Thus we look at the field obtained by adjoining \+
 " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT 259 81 ", a primitive cu
be root of unity, and cbrt2, a cube root of 2, to the rationals. " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 179 "To check if proposed image sets d
efine an automorphism we will need the Maple commands factor, alias, R
ootOf, and subs.  You may want to look at the help files for these com
mands." }}}{EXCHG }{EXCHG }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 21 "Facto
ring and Aliases" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "We start by de
fining  " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 43 " and cbrt2
, then factoring the polynomials " }{XPPEDIT 18 0 "x^3 -1" "6#,&*$%\"x
G\"\"$\"\"\"F'!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x^3 -2" "6#,&
*$%\"xG\"\"$\"\"\"\"\"#!\"\"" }{TEXT -1 26 " over a number of fields. \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "alias(omega=RootOf(x^2 \+
+ x + 1), cbrt2=RootOf(x^3 - 2));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "factor(x^3 - 1); \nfactor(x^3 - 2);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 47 "factor(x^3 - 1, omega);\nfactor(x^3 - 2, \+
omega);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "factor(x^3 - 1, \+
cbrt2);\nfactor(x^3 - 2, cbrt2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 65 "factor(x^3 - 1, \{cbrt2, omega\});\nfactor(x^3 - 2, \+
\{cbrt2, omega\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 186 "Notice th
at Maple will factor over extension fields, using the second argument \+
of the factor command as a list of algebraic expressions to be adjoine
d to the base field of the rationals." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 39 "Substitutio
n with a factored polynomial" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "We
 start by redefining the polynomials in out example." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 37 "f[1] := x^2 + x + 1; \nf[2] := x^3 -2;" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "To use a proposed automorphism,
 we want to be able to replace a generator with an arbitrary image.  T
he command" }}{PARA 0 "" 0 "" {TEXT -1 8 "   subs(" }{XPPEDIT 18 0 "al
pha" "6#%&alphaG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " beta" "6#%%betaG
" }{TEXT -1 24 ", expression);\nreplaces " }{XPPEDIT 18 0 "alpha" "6#%
&alphaG" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "beta" "6#%%betaG" }
{TEXT -1 68 " in expression.  We are particularly interested in doing \+
this when  " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 80 " is one
 of the field generators and the expression is the minimum polynomial \+
of " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 1 "." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "g[1] := factor(f[1], \{omega, cbrt2
\});\ng[2] := factor(f[2], \{omega, cbrt2\});" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 83 "h[1] := subs(omega = -(1 + omega), g[1]);\nh[2] \+
:= subs(omega = -(1 + omega), g[2]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 51 "It is clear that the substitution process has left " }{XPPEDIT 
18 0 "g[1]" "6#&%\"gG6#\"\"\"" }{TEXT -1 13 " unchanged.  " }{TEXT 
272 106 "The substitution simply switched the factors.  Straightforwar
d algebra verifies that the same is true for " }{XPPEDIT 18 0 "g[2]" "
6#&%\"gG6#\"\"#" }{TEXT 273 114 ".  Unfortunately, it is not true that
 the algebra will always be straightforward.  Thus we want Maple to si
mplify " }{XPPEDIT 18 0 "h[2]" "6#&%\"hG6#\"\"#" }{TEXT 274 21 " and r
efactor for us." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "factor(s
implify(h[1]), \{omega, cbrt2\});\nfactor(simplify(h[2]), \{omega, cbr
t2\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 
{PARA 5 "" 0 "" {TEXT -1 35 "Choosing images for an automorphism" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "In the example above we noted tha
t the substitution left the polynomials over F unchanged.  To do that \+
we needed to send each " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT 
-1 172 " to an element of K that is a root of the same irreducible pol
ynomial.   (We factored the polynomials over K so that we could see th
e possible images for our generators.)  " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 260 83 "\nConsider the case of our favorite example. The field  \+
K is obtained by adjoining  " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }
{TEXT 261 21 " and cbrt2, roots of " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#
\"\"\"" }{TEXT 263 5 " and " }{XPPEDIT 18 0 "f[2]" "6#&%\"fG6#\"\"#" }
{TEXT 264 68 ", to the rationals.  (In fact, these polynomials split o
ver K.)  Sin" }{TEXT -1 3 "ce " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"\"
\"" }{TEXT -1 22 " has 2 roots in K and " }{XPPEDIT 18 0 "f[2]" "6#&%
\"fG6#\"\"#" }{TEXT -1 65 " has 3 roots in K,  we have 6 possible auto
morphisms of K over F." }}}{EXCHG }{SECT 0 {PARA 20 "" 0 "" {TEXT -1 
0 "" }{TEXT 275 10 "Exercises:" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "
1)  List the 6 possible automorphisms of K over Q when K = Q[" }
{XPPEDIT 18 0 "omega" "6#%&omegaG" }{TEXT -1 9 ", cbrt2]." }}}{EXCHG }
{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "2) Let K be the splitting field of
 " }{XPPEDIT 18 0 "x^5 -3" "6#,&*$%\"xG\"\"&\"\"\"\"\"$!\"\"" }{TEXT 
-1 39 ".  K is obtained by adjoining roots of " }{XPPEDIT 18 0 "x^5 - \+
3" "6#,&*$%\"xG\"\"&\"\"\"\"\"$!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 
18 0 "Phi[5](x) = x^4 + x^3 + x^2 + x + 1" "6#/-&%$PhiG6#\"\"&6#%\"xG,
,*$F*\"\"%\"\"\"*$F*\"\"$F.*$F*\"\"#F.F*F.F.F." }{TEXT -1 60 "  to Q. \+
 Describe the 20 possible automorphisms of K over Q." }}}{EXCHG }
{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "3)  In contrast, the splitting fie
ld of " }{XPPEDIT 18 0 "x^4 - 3" "6#,&*$%\"xG\"\"%\"\"\"\"\"$!\"\"" }
{TEXT -1 53 " does not have 12 automorphisms over Q.  Explain why." }}
}{EXCHG }}{EXCHG }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 33 "Working throug
h a particular case" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Let us focu
s on the potential automorphism that takes " }{XPPEDIT 18 0 "omega" "6
#%&omegaG" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "-(1 + omega)" "6#,$,&\"
\"\"F%%&omegaGF%!\"\"" }{TEXT -1 20 " and takes cbrt2 to " }{XPPEDIT 
18 0 "omega" "6#%&omegaG" }{TEXT -1 127 " cbrt2.  We want to verify th
at this substitution simply permutes the factors of the factored forms
 of the minimal polynomials." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 129 "h[1] := subs(omega = -(1 + omega), cbrt2 = cbrt2 * omega, g[1])
;\nh[2] := subs(omega = -(1 + omega), cbrt2 = cbrt2 * omega, g[2]);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "h[2] := factor(simplify(h[
2]), \{omega, cbrt2\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 173 
"rootarray := [1, omega, omega^2, cbrt2, cbrt2 * omega, cbrt2 * omega^
2];\nrootarray2 := subs(cbrt2=omega*cbrt2, omega=omega^2, rootarray);
\nrootarray3 := simplify(rootarray2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 75 "g[1] := factor(f[1], \{omega, cbrt2\});\ng[2] := fact
or(f[2], \{omega, cbrt2\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "On
ce again we need to simplify the second equation." }}}{EXCHG {PARA 0 "
" 0 "" {TEXT 290 197 "The same sequence of commands, subs followed by \+
simplify, lets us check the action of an automorphism on particular el
ements of a field.  Consider the action on rootarray, an array of the \+
roots of " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"\"\"" }{TEXT 292 5 " and
 " }{XPPEDIT 18 0 "f[2]" "6#&%\"fG6#\"\"#" }{TEXT 291 1 "." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 289 67 "To make the comparison easier, apply sim
plify to rootarray as well." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "root
array1 := simplify(rootarray);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 266 
86 "It is now clear that our action automorphism simply rearranged roo
tarray, as expected." }}}{SECT 0 {PARA 20 "" 0 "" {TEXT -1 0 "" }
{TEXT 293 10 "Exercises:" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "4)  Pi
ck another nontrivial automorphism of K = Q[" }{XPPEDIT 18 0 "omega" "
6#%&omegaG" }{TEXT -1 77 ", cbrt2] over Q.  Show that your automorphis
m simply rearranges the roots of " }{XPPEDIT 18 0 "f[1]" "6#&%\"fG6#\"
\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "f[2]" "6#&%\"fG6#\"\"#" }
{TEXT -1 1 "." }}}{EXCHG }{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "5)  Def
ine a nontrivial automorphism on the splitting field of x^5 - 3.  Veri
fy that the map permutes the roots of " }{XPPEDIT 18 0 "x^5 -3" "6#,&*
$%\"xG\"\"&\"\"\"\"\"$!\"\"" }{TEXT -1 8 " and of " }{XPPEDIT 18 0 "Ph
i[5](x)" "6#-&%$PhiG6#\"\"&6#%\"xG" }{TEXT -1 1 "." }}}{EXCHG }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 5 "" 0 "
" {TEXT 277 43 "Verifying that a map is a ring homomorphism" }{TEXT 
-1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT 265 616 "  The method we have b
een using to \"define automorphisms\", actually defines a homomorphsim
 of the additive group of the field.  (It is defined on products of th
e generators and extended by linearity.)  To check that we are definin
g a field automorphism, we need to check that the map respects multipl
ication.  (We need to check that we get the same answer whether we sim
plify a product before or after performing the map. )  Since the map i
s defined to respect addition, it suffices to check this with the prod
ucts of the powers of the generators, with the power of each generator
 less than or equal to its degree.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT 267 24 "In our case, with K = Q[" }{XPPEDIT 
18 0 "omega" "6#%&omegaG" }{TEXT 268 200 ", cbrt2], the degree of each
 generator is the same whether or not the other generators have been a
djoined to the field.  This means that we only need to consider powers
 of generators and not products." }}{PARA 0 "" 0 "" {TEXT 269 127 "\nW
e want to set up a procedure to systematically check if a proposed map
 respects multiplication and is this an automorphism.  " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 270 165 "To define the map \+
we need the generators of the field and their images.  That lets us de
fine the substitutions.  We also want a list of the powers of the gene
rators." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 173 "gens := [omega,
 cbrt2];\nimgens := [omega^2, omega*cbrt2];\nsubstitutions := \{gens[1
] = imgens[1], gens[2] = imgens[2]\};\ngenarray := [omega, omega^2, cb
rt2, cbrt2^2, cbrt2^3];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Next, \+
" }{TEXT 276 126 "we want 2 functions, one that simplifies only after \+
substituting, and one that simplifies, then substitutes, then simplifi
es. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "phi1 := (a, b) -> s
implify(subs(a, b)):\nphi2 := (a, b) -> simplify(subs(a, simplify(b)))
:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 138 "Finally we want to apply th
e substitutions to the list of elements and compare the results.  If t
he 2 lists agree we have an automorphism." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 285 "genarray1 := phi1(substitutions, genarray);\n  gen
array2 := phi2(substitutions, genarray);\n  print (`Using the substitu
tions`);\n  print (substitutions);\n  if genarray1 = genarray2 \n     \+
then print (`phi is an automorphism`);\n     else print (`phi is NOT a
n automorphism`); \n         fi:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
131 "It is useful to note that if we change the substitutions beyond t
he 6 canditdates, we see that we have not defined an automorphism." }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 404 "gens := [omega, cbrt2];\nimgens := [omega^3, omega*c
brt2];\nsubstitutions := \{gens[1] = imgens[1], gens[2] = imgens[2]\};
\ngenarray1 := phi1(substitutions, genarray);\n  genarray2 := phi2(sub
stitutions, genarray);\n  print (`Using the substitutions`);\n  print \+
(substitutions);\n  if genarray1 = genarray2 \n     then print (`phi i
s an automorphism`);\n     else print (`phi is NOT an automorphism`); \+
\n         fi:" }}}{EXCHG }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}}{SECT 0 {PARA 5 "" 0 "" {TEXT 288 33 "Non-examples, Traps, and War
nings" }{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT 271 306 "1)  We p
ointed out above that we only get an automorphism when we send roots t
o roots.  It should be noted that the subs command works with any imag
e.  If we define an the image of a root to be anything other than a ro
ot of the same minimum polynomial, the map cannot be extended to a rin
g homomorphism.  \n" }}{PARA 0 "" 0 "" {TEXT -1 272 "2) The procedure \+
defined above properly checks to see if the map defined by sending the
 generators to specified images can be extended to a homomorphism.  It
 does not check that the map is to the same field.  In particular we c
an define the obvious map from Q[cbrt2] to Q[" }{XPPEDIT 18 0 "omega" 
"6#%&omegaG" }{TEXT -1 87 " cbrt2].  It will be an isomorphism of fiel
ds but not an automorphism.  Note that (x - " }{XPPEDIT 18 0 "omega" "
6#%&omegaG" }{TEXT -1 34 " cbrt2) is not a linear factor of " }
{XPPEDIT 18 0 "x^3 -2" "6#,&*$%\"xG\"\"$\"\"\"\"\"#!\"\"" }{TEXT -1 
15 " over Q[cbrt2]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "fact
or(x^3 -2, cbrt2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 278 222 "3)  Maple
's subs command is easily confused by defining substitutions  that do \+
no use the generators, even if they define equivalent homorphisms.  Ou
r procedure will not reliably check an action on the field generated b
y  " }{TEXT 279 2 "w " }{TEXT 280 39 " if the action is defined in ter
ms of  " }{XPPEDIT 18 0 "sqrt(-3)" "6#-%%sqrtG6#,$\"\"$!\"\"" }{TEXT 
281 38 "  even though the field generated by  " }{XPPEDIT 18 0 "omega
" "6#%&omegaG" }{TEXT 286 33 " is also the field generated by  " }
{XPPEDIT 18 0 "sqrt(-3)" "6#-%%sqrtG6#,$\"\"$!\"\"" }{TEXT 282 82 " . \+
 To check that action, we need to define the field of coeficients in t
erms of  " }{XPPEDIT 18 0 "sqrt(-3)" "6#-%%sqrtG6#,$\"\"$!\"\"" }
{TEXT 283 42 " .\n\nNotice that the substitution taking   " }{XPPEDIT 
18 0 "sqrt(-3)" "6#-%%sqrtG6#,$\"\"$!\"\"" }{TEXT 284 66 "  to 12 is n
ot an isomorphism, and that, the field generated by   " }{XPPEDIT 18 
0 "sqrt(-3)" "6#-%%sqrtG6#,$\"\"$!\"\"" }{TEXT 285 42 "   is the same \+
as the field generated by  " }{XPPEDIT 18 0 "omega" "6#%&omegaG" }
{TEXT 287 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "\nf1 := x^2 + x \+
+ 1; \ng1 := factor(f1, \{omega\});\nh1 := subs(omega=12, g1);\ng2 := \+
factor(f1, \{(-3)^(1/2)\});\nh2 := subs(omega=12, g2);" }}}{EXCHG }
{EXCHG {PARA 0 "" 0 "" {TEXT -1 265 "4)  We obtained the splitting fie
ld of x^3 -2 by adjoining a root for each of 2 polynomials.  The other
 approach is to adjoin several roots to x^3-2. We can do that by adjoi
ning a root to the irreducible part left over after adjoining the firs
t part and factoring. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "
f := x^3 -2;\nalias(alpha=RootOf(f));\nfactor(f, alpha);\ng := simplif
y(f/(x-alpha));\nalias(beta = RootOf(g));\nfactor(f, [alpha, beta]);" 
}}}{EXCHG }{EXCHG }}{EXCHG }{EXCHG }}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 
1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }