{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Geneva" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 } 1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Geneva" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 4 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 3" -1 5 1 {CSTYLE "" -1 -1 "Geneva" 1 12 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "List Item " -1 14 1 {CSTYLE "" -1 -1 "Geneva" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 } 1 1 0 0 3 3 1 0 1 0 2 2 14 5 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Geneva" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 6 4 1 0 1 0 2 2 19 1 }{PSTYLE "Author" -1 19 1 {CSTYLE "" -1 -1 "Geneva" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 4 4 1 0 1 0 2 2 0 1 }{PSTYLE "R3 Font 0" -1 256 1 {CSTYLE "" -1 -1 "Monaco" 1 9 0 0 255 1 2 2 2 2 2 2 1 1 1 1 } 1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "R3 Font 0" -1 257 1 {CSTYLE "" -1 -1 "Geneva" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "R3 Font 2" -1 258 1 {CSTYLE "" -1 -1 "Geneva" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 28 "Galois Groups of Polynomi als" }}{PARA 19 "" 0 "" {TEXT -1 54 "\251Mike May, S.J., maymk@slu.edu , Saint Louis University" }}{PARA 0 "" 0 "" {TEXT -1 73 "A worksheet t o look at how Maple finds the Galois group of a polynomial.\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 8 "Outline:" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "M aple has a command \"galois(f(x));\" that computes the Galois group of f(x), an \nirreducible polynomial of degree at most 8 over the ration als." }}{PARA 0 "" 0 "" {TEXT -1 35 "In this worksheet we would like t o:" }}{PARA 14 "" 0 "" {TEXT -1 107 "1) Learn the syntax of the Maple command and how to get more information on it from Maple's help facil ity;" }}{PARA 14 "" 0 "" {TEXT -1 70 "2) Work through the computation s to see how to compute galois groups;" }}{PARA 14 "" 0 "" {TEXT -1 96 "3) Explore determining the degree of the splitting field to find \+ the order of the Galois group." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 43 "Understanding how to use the galois command" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "The ba sic sytax of the command is galois(f); where f is a polynomial of degr ee 8 or less in a single variable." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "galois(x^6 + 2*x + 2);" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 198 "The result gives us the name of the group from two diffe rent lists, whether the group is odd or even, its order, and a set of \+ generators for the group expressed as a subgroup of the Symmetric grou p " }{XPPEDIT 18 0 "S[n]" "6#&%\"SG6#%\"nG" }{TEXT -1 171 ", where n i s the degree of the polynomial.\nThe help command will give us more in formation on the galois command and the naming convensions used in the second set of names." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "?ga lois" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "?`group/transnames` " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "For this class, we are inter ested not only on the result, but also in following the argument to se e how Maple produces its answer." }}{PARA 257 "" 0 "" {TEXT -1 152 "\n We can get more information on how Maple is doing the computations by \+ changing\nthe value of \"infolevel[galois]\" from its default value o f 0 to 1 or 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "infolevel [galois] := 1;\ngalois(x^6 +3*x +3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "When the inforlevel is 1, we get the discriminant Maple compute s in finding the Galois group." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "infolevel[galois] := 2;\ngalois(x^6 + 5*x + 5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "When the infolevel is 2 we get a flurry \+ of information that gives the results of intermediate computations." } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 223 "[Note that Maple version 6 has \+ an error in the table here. The group with 720 elements and the given generators is S(6), not A(6). Other groups seem to be correctly iden tified. The error is corrected in Maple version 7.]" }}}{EXCHG {PARA 5 "" 0 "" {TEXT -1 9 "Exercise:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 85 "1) Use the help pages to describe the tw o Galois groups found above as named groups." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 41 "Maple's method of computing Galois groups " }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 59 "Step 1 - Set up the polynomia l and look at possible groups." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 261 "The information we got from running galois with infolevel at 2 is a b it overwhelming and needs a lot of explaining. We will step through t he process that Maple uses to find the galois group of a polynomial. \+ We start by defining the polynomial we will look at." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "f := x^6 + 3*x + 3;" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 75 "The next thing to do is to check that the polynom ial is in fact irreducible" }{TEXT 256 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "factor(f);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 172 "If the polynomial factors, Maple will give an error message that \+ it does not know how to compute the Galois group. Since our polynomia l is irreducible Maple will continue." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "degf := degree(f);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 419 "The galois function in Maple is table driven. Maple stores in formation \nabout all the transative subgroups of Sn for n <8 in a tab le called `galois/groups`. \nIt will check information about the grou ps, then look up in the table to see what groups still qualify. When \+ it gets down to a single group it quits. It is instructive to look at part of the table even if the meaning of the entries are not yet self -evident." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "grps:= `galois /groups`:\nprint(grps[6,1]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 271 " The table above names the groups as dTn where d is the degree and n is the number of the group in a list of transative groups on d elements. We would like access to more descriptive names. The following proce dure gets names of transative groups on sets of size up to 9." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 191 "groupnamer := proc(x)\nloca l deg, numb:\ndeg := substring(x,1):\nnumb := parse(substring(x,3..-1) ):\n`group/transgroup/names`||deg[numb]:\nend proc:\ngroupnamer(\"6T9 \");\ngroupnamer(grps[6,1][1][1]);" }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 9 "Exercise:" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "2) Give examp les to show that the Galois group of a reducible polynomial cannot be \+ determined by knowing the Galois groups of the factors. (Hint, it suf fices to look at 4th degree polynomials that factor as the product of \+ 2 quadratic polynomials.)" }}}{EXCHG }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 43 "Step 2 - Se e if the discriminant is square." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 390 "The group array has 24 entries indexed by ordered pairs (n,d) wit h n between 1 and 12 for the degree of the polynomial and d either 0 o r 1 determined by whether or not the group is contained in the alterna ting group. (The entries for degrees 9 through 12 are rather incomple te.) Recall that the Galois group is contained in the alternating gro up if and only if the discriminant is a squar" }{TEXT 257 2 "e." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "We check \+ this with our test polynomial." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "disc:= discrim(f, x);\nissqr(disc);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 154 "Since the discriminant is not a square we want the (6, 1) entry of the table.\nTo make this easier to read, we just want to \+ look at the names of the groups" }{TEXT 258 1 "." }}{PARA 0 "" 0 "" {TEXT -1 46 "We will do that under both naming conventions." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 178 "possible := grps[6,1];\npos sgrps := `galois/short`(possible);\nprint(`The possible groups are:`); \nfor i from 1 to nops(possgrps) do\nprint(possgrps[i] =groupnamer(pos sgrps[i]));\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 317 "Thus we have 11 possible groups \+ at this point. The first set of names used, the names listed in possg rps, follows Butler and McKay, Comm. Alg. 11(1983) pp. 863-911. The s econd set of name was explained in the help page called earlier, but s hould be more familiar anyway. You should recognize the symmetric gro ups " }{XPPEDIT 18 0 "S(6);" "6#-%\"SG6#\"\"'" }{TEXT -1 4 "and " } {XPPEDIT 18 0 "S(3);" "6#-%\"SG6#\"\"$" }{TEXT -1 3 ". " }{XPPEDIT 18 0 "PGL(2,5);" "6#-%$PGLG6$\"\"#\"\"&" }{TEXT -1 98 " is a projectiv e general linear group. It is obtained by taking invertible 2 by 2 m atrices over " }{XPPEDIT 18 0 "Z[5]" "6#&%\"ZG6#\"\"&" }{TEXT -1 201 " , then modding out by the normal subgroup of scalar matrices. The cyc lic groups are denoted by C(n). The group [S(3)^2]2 is the semidirect product of C(2) over the direct product of 2 copies of S(3)." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 5 "" 0 " " {TEXT -1 70 "Step 3 - Use primes to check for cycle types of element s of the group." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 183 "The next techn ique used is to factor f(x) mod p for a variety of primes. We have s een in class that if f(x) mod p has distinct roots and factors into i rreducible factors of length " }{XPPEDIT 18 0 "d[1]" "6#&%\"dG6#\"\"\" " }{TEXT -1 8 ", ... , " }{XPPEDIT 18 0 "d[s]" "6#&%\"dG6#%\"sG" } {TEXT -1 6 " over " }{XPPEDIT 18 0 "Z[p]" "6#&%\"ZG6#%\"pG" }{TEXT -1 117 ", then the Galois group G of f(x) contains an element that can be written as a product of disjiont cycles of length " }{XPPEDIT 18 0 " d[1]" "6#&%\"dG6#\"\"\"" }{TEXT -1 8 ", ... , " }{XPPEDIT 18 0 "d[s]" "6#&%\"dG6#%\"sG" }{TEXT -1 239 ". This can be used to eliminate grou ps that do not have elements with the right cycle length.\n\nThe fouth element of information on each group above is a vector that encodes i nformation about the cycle structure of elements of the group. " } {XPPEDIT 18 0 "S[6]" "6#&%\"SG6#\"\"'" }{TEXT -1 55 " obviously has e lements of all possible cycle types. " }{XPPEDIT 18 0 "Z[6]" "6#&%\"Z G6#\"\"'" }{TEXT -1 278 " on the other hand only has elements of type \+ 2,2,2, type 3,3, and type 6. If we order cycle types lexicographicall y with 1,1,1,1,1,1 omitted from the front as trivial, followed by 2,1 ,1,1,1 and ending with 6, we have 11 possible cycle structures. Of th e 10 nontrivial types, " }{XPPEDIT 18 0 " Z[6]" "6#&%\"ZG6#\"\"'" } {TEXT -1 51 " has elements of the third, sixth, and tenth type.\n" }} {PARA 0 "" 0 "" {TEXT -1 299 "For f(x) to have distinct roots mod p we need the leading coefficient of f(x) and the discriminant of f(x) to \+ both be nonzero mod p. Since we started with a monic polynomial, we o nly need to check the discriminant. We see that it is zero mod 3 and \+ 17 but not mod p for any other prime less thn 40." }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 81 "print(disc mod 2, disc mod 3, disc mod 7, di sc mod 11, disc mod 13, disc mod 17);" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 50 "We are ready to start considering specific primes." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g = Factor(f) mod 2;" }}} {EXCHG {PARA 257 "" 0 "" {TEXT -1 223 "Since this polynomial is irredu cible, G must contain an element that is a single 6-cycle. The comma nds cylcepattern and encode can be used to tell us that this is the 10 th nontrivial partition of 6. That lets us remove " }{XPPEDIT 18 0 " S[3] " "6#&%\"SG6#\"\"$" }{TEXT -1 4 "and " }{XPPEDIT 18 0 "S[4]" "6#& %\"SG6#\"\"%" }{TEXT -1 1 "/" }{XPPEDIT 18 0 "Z[4]" "6#&%\"ZG6#\"\"%" }{TEXT -1 33 " from the list of possible groups" }{TEXT 259 1 "." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 414 "shape := `galois/cyclepatte rn`(f, x, 2);\n e := `galois/encode`(`galois/initpart`(6), shape); \n rm := \{\}:\n rmgrp := \{\}:\n for i to nops(possible) do\n g rp := possible[i];\n if grp[4][e] = 0 then \n rm := rm uni on \{grp\};\n rmgrp := rmgrp union \{[grp[1],groupnamer(grp[1]) ]\}; fi\n od:\n print(`groups removed `=rmgrp);\n possible := p ossible minus rm:\n grpleft := `galois/short`(possible);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Factor(f) mod 5; Factor(f) mod 7;" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 210 "Using p = 5, the polynomial \+ factored the same way, so it did not give us any new information, but \+ using p = 7 give a shape of 3, 2, 1, the fifth nontrivial partition o f 6. This lets us eliminate 7 more groups" }{TEXT 260 1 "." } {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 512 "shape \+ := `galois/cyclepattern`(f, x, 7);\n e := `galois/encode`(`galois/ini tpart`(6), shape); \n rm := \{\}:\n rmgrp := \{\}:\n for i to nops( possible) do\n grp := possible[i];\n if grp[4][e] = 0 then \+ \n rm := rm union \{grp\};\n rmgrp := rmgrp union \{[grp [1],groupnamer(grp[1])]\}; fi\n od:\n print(`groups removed `=rm grp);\n possible := possible minus rm:\n print(\"The groups left are :\");\n for i from 1 to nops(possible) do\n print([possible[i] [1],groupnamer(possible[i][1])]);\n od;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 194 "Maple next tries the next 7 primes without being able to eliminate either of the remaining two groups. Furthermore, we hav e a group that has to be eliminated to eliminate any other groups. " }}{PARA 0 "" 0 "" {TEXT -1 322 "(The group 6T13 is [S(3)^2]2, the semi direct product of the cyclic group C(2) acting on S(3)xS(3). This gro up is minimal in the set of groups still allowed under the ordering im posed by the shapes.) These two facts makes us suspicious that the mi nimal group is the Galois group. We need another way to eliminate gro ups." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 56 "Step 4 - Look at the factoring of resolvant pol ynomials." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 220 "The other approach w e use is to look at how some related polynomials factor. (We have alr eady used this method with the discriminant. We implicitly checked to see if the polynomial x^2 - disc(f(x)) factored in step 2.)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Now we will look at resolvant pol ynomials. Corresponding to any nth degree polynomial p(x) with roots \{" }{XPPEDIT 18 0 "r[1]" "6#&%\"rG6#\"\"\"" }{TEXT -1 2 ", " } {XPPEDIT 18 0 "r[2]" "6#&%\"rG6#\"\"#" }{TEXT -1 8 ", ... , " } {XPPEDIT 18 0 "r[n]" "6#&%\"rG6#%\"nG" }{TEXT -1 161 "\} is a polynomi al of degree (n, s) with roots that are products of s roots of p(x). \+ Thus our 6th degree polynomial has a 15th degree resolvant with roots like " }{XPPEDIT 18 0 "r[1]*r[2]" "6#*&&%\"rG6#\"\"\"F'&F%6#\"\"#F'" }{TEXT -1 45 " and a 20th degree resolvant with roots like " } {XPPEDIT 18 0 "r[1]*r[2]*r[3]" "6#*(&%\"rG6#\"\"\"F'&F%6#\"\"#F'&F%6# \"\"$F'" }{TEXT -1 113 ". The fifth and sixth entries of the group en tries looks at how these polynomials factor. We see that if G is " } {XPPEDIT 18 0 "S[6]" "6#&%\"SG6#\"\"'" }{TEXT -1 137 ", both of these \+ polynomials are irreducible while they factor in the other case. Let' s look at the resolvant using products of two roots" }{TEXT 261 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 620 "resolv := `galois/rsetpo l`(f, x, 2);\n factrsetpoly := factor(resolv);\n shapersetpoly := [` galois/degrees`(factrsetpoly,x)];\n rm := \{\}:\n for i to nops(poss ible) do\n grp := possible[i];\n if grp[5] <> shapersetp oly then \n rm := rm union \{grp\};\n print(`We remov e the group `,[grp[1],groupnamer(grp[1])]);\n print(`whose res olvant poly has factor shape `, grp[5]);fi\n od:\npossible := pos sible minus rm:\nprint(`The groups left are:`);\nfor i from 1 to nops( possible) do\n print([possible[i][1],groupnamer(possible[i][1])],\n \+ ` with res poly factor shape `, possible[i][5]);\n od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Since the resolvant polynomial factors, we can remove " } {XPPEDIT 18 0 "S[6]" "6#&%\"SG6#\"\"'" }{TEXT -1 37 " and are left wit h our single answer." }}}{EXCHG {PARA 5 "" 0 "" {TEXT -1 12 "Exercises : \n" }}{PARA 0 "" 0 "" {TEXT -1 59 "3) Write out a clear argument fo r why the Galois group of " }{XPPEDIT 18 0 "x^6 + 3 x + 3" "6#,(*$%\"x G\"\"'\"\"\"*&\"\"$F'F%F'F'F)F'" }{TEXT -1 116 " is [S(3)^2]2. To use a Maple computation, you have to be able to justify the how it helps \+ you find a Galois group." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "4) Find the Galois group of \+ " }{XPPEDIT 18 0 "x^5 +3*x + 3" "6#,(*$%\"xG\"\"&\"\"\"*&\"\"$F'F%F'F' F)F'" }{TEXT -1 49 ". Write a clear argument justifying your answer. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 0 {PARA 4 " " 0 "" {TEXT -1 51 "Another approach - Computing the size of the group ." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "If K is the splitting field over F of f(x), then K = F[" } {XPPEDIT 18 0 "alpha[1]" "6#&%&alphaG6#\"\"\"" }{TEXT -1 7 ", ..., " } {XPPEDIT 18 0 "alpha[n]" "6#&%&alphaG6#%\"nG" }{TEXT -1 12 "] where th e " }{XPPEDIT 18 0 "alpha[i]" "6#&%&alphaG6#%\"iG" }{TEXT -1 29 " are \+ the roots of f(x). Let " }{XPPEDIT 18 0 "F[i]" "6#&%\"FG6#%\"iG" } {TEXT -1 5 " = F[" }{XPPEDIT 18 0 "alpha[1]" "6#&%&alphaG6#\"\"\"" } {TEXT -1 7 ", ..., " }{XPPEDIT 18 0 "alpha[n]" "6#&%&alphaG6#%\"nG" } {TEXT -1 11 "]. Define " }{XPPEDIT 18 0 "f[i](x)" "6#-&%\"fG6#%\"iG6# %\"xG" }{TEXT -1 38 " to be the irreducible polynomial of " } {XPPEDIT 18 0 "alpha[i]" "6#&%&alphaG6#%\"iG" }{TEXT -1 7 " over " } {XPPEDIT 18 0 "F[i-1]" "6#&%\"FG6#,&%\"iG\"\"\"F(!\"\"" }{TEXT -1 25 " , a polynomial of degree " }{XPPEDIT 18 0 "s[i]" "6#&%\"sG6#%\"iG" } {TEXT -1 53 ". Then the degree of K over F is the product of the " } {XPPEDIT 18 0 "s[i]" "6#&%\"sG6#%\"iG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "We use this approach t o find the order of the Galois groups of " }{XPPEDIT 18 0 "x^4 + p*x + p" "6#,(*$%\"xG\"\"%\"\"\"*&%\"pGF'F%F'F'F)F'" }{TEXT -1 24 " when p is 2, 3, and 5." }}{PARA 0 "" 0 "" {TEXT -1 20 "We start with p = 5. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "f := x^4+5*x+5: alias(a lpha = RootOf(f)): factor(f, alpha);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Since f factors completely over " }{XPPEDIT 18 0 "F[1]" "6#&%\" FG6#\"\"\"" }{TEXT -1 82 ", the extension is of degree 4. Since the g roup must be a transitive subgroup of " }{XPPEDIT 18 0 "S[4]" "6#&%\"S G6#\"\"%" }{TEXT -1 20 ", the group must be " }{XPPEDIT 18 0 "Z[4]" "6 #&%\"ZG6#\"\"%" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "f[1] := x^4+3*x+3; alias(alpha[1] = RootOf(f[1])): \ng := fact or(f[1], alpha[1]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 253 "This has \+ a quadratic factor and 2 linear factors. Adjoining a root to the quad ratic factor splits the polynomial completely. (Counting the constant term in front, the quadratic factor is the second factor of g. We ca n recover it with the op command.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "f[2] := op(2,g); alias(alpha[2] = RootOf(f[2])):\nh : = factor(f[1], \{alpha[1], alpha[2]\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Thus we see that the degree of the extension must be 8 wh en p is 3." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "When p is 2 we hav e to work harder. We first note that adjoining a root of the polynomi al leaves a cubic factor." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 129 "f[1] := x^4+2*x+2; alias(alpha[3] = RootOf(f[1])): \ng := factor( f[1], alpha[3]);\nf[2] := op(2,g); alias(alpha[4] = RootOf(f[2])):" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "Thus the degree of the splitting field is at least 12. We now adjoin a root of that polynomial and fa ctor again." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "h := factor( f[1], \{alpha[3], alpha[4]\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "We still have an irreducible quadratic factor. Thus the splittin g field must be a degree 24 extension. This means it must be " } {XPPEDIT 18 0 "S[4]" "6#&%\"SG6#\"\"%" }{TEXT -1 1 "." }}}{EXCHG {PARA 5 "" 0 "" {TEXT -1 10 "Exercises:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "5) Show that the order of the Galoi s group of " }{XPPEDIT 18 0 "x^8 - 2" "6#,&*$%\"xG\"\")\"\"\"\"\"#!\" \"" }{TEXT -1 9 " is 16. " }}}{EXCHG }{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "6) Show that the order of the Galois group of " }{XPPEDIT 18 0 "x^8 - 3" "6#,&*$%\"xG\"\")\"\"\"\"\"$!\"\"" }{TEXT -1 7 " is 32." } }}{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 }