12-240/Fields' Further proof: Difference between revisions
Michael.Wang (talk | contribs) No edit summary |
Michael.Wang (talk | contribs) No edit summary |
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see the (a+1)th row |
see the (a+1)th row |
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There must be a "1" in this row, each row precisely to meet the rule. ( The rule... you know, I cannot find some notations.) |
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If F(n) is a field, then |
If F(n) is a field, then |
Revision as of 09:51, 12 September 2012
In the first class, Professor says something about particular fields. Forgive me, because I am an international student, if I can not express information precisely.
About: F(n) F(1) F(2) and F(3) are a field, but F(4) is not. Professor said that any number N which is not a prime number can not "form" a field. We should ask why...
Here is the proof.
If we have a field F(n), and n=a*b (a,b,n∈ N*, a,b≠1)
IN defination of multiplication
- 0 1 2 3 .......... b.......n-1
0 . . . . ......................
1 . . . . ......................
2 . . . . ......................
3 . . . . .......................
4 . . . . .......................
.
a 0 (a) (2a) (3a)...........(a*b).....(n-1)*a ( in this row, every element mod n)
.
.
.
.
n-1..................................
see the (a+1)th row
There must be a "1" in this row, each row precisely to meet the rule. ( The rule... you know, I cannot find some notations.)
If F(n) is a field, then
1.m*a=k*n+1 (k,m∈N*, m<n) 2.n=ab
==>>m=kb+1/a
unless a=1
m will not exist. so F(n), when n is not a prime number, is not a field.
There is a large need for me to improve my format. Editing is welcomed.
PS: But till now, there are still some questions existing. How can we prove that a prime number N can absolutely form a field? Is there any exception? I am still working on it. ----Michael.Wang