ガロア第一論文及びその関連の資料スレ

[過去ﾛｸﾞ] ガロア第一論文及びその関連の資料スレ (1002ﾚｽ)
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30: １３２人目の素数さん [sage] 2021/08/29(日) 18:37:00.62 ID:7niZQGlq

>>29
つづき

He employs the notation:
NRM=1.2.3...n,=number of different values the function can take,=size of each block;
so in the foregoing example, n=3,N=3!=6,R=3,M=2. Note that R?M=N, as developed above. Let's walk through another example to exemplify the theorem, with K(x1,x2,x3,x4,x5)=x1x2+x3x4+x5. We have n=5,N=5!=120 for this function. Since the largest prime less than or equal to n is now p=5, the theorem asserts that this or any other function cannot take a total of three or four values; ie, R≠3 and R≠4 for this function. Here are a few Ks with different values:

K1=a1a2+a3a4+a5K2=a1a4+a2a3+a5K3=a1a3+a2a4+a5.
Each one is associated with a block of functions, or substitutions if you will, giving the same value. K1, for example is associated with

K′1=a4a3+a1a2+a5,
where K1 and K′1 are identified respectively with the substitutions:

(1 2 3 4)
(1 2 3 4),

(1 2 3 4)
(4 3 1 2).

I'm leaving out the 5 even though they are substitutions on 5 letters, because the 5 is always unchanged for the first block (and the other two shown for that matter). It looks like there will be another four blocks like this, where each of a4,a3,a2,a1 plays the role of that last summand sitting at the end of K; that is, it looks like R=15 and therefore M=120/R=120/15=8. If so, K certainly satisfies the theorem.
(引用終り)
以上

http://rio2016.5ch.net/test/read.cgi/math/1615510393/30

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