12-240/Fields' Further proof: Difference between revisions

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[[Image:12-240-Splash.png]]
[[12-240]][[Image:12-240-Splash.png]]

[[12-240/Classnotes for Thursday September 13]]
[[12-240/Classnotes for Tuesday September 11]]
[[12-240/Classnotes for Tuesday September 11]]


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.
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)
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.

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 F(n)


If you do not understand what the F(n) means, you can look through the file "12-240/Classnotes for Tuesday September 11"
If you do not understand what the F(n) means, you can look through the file "12-240/Classnotes for Tuesday September 11"
'''Why all the numbers which are not prime numbers can not form a field F(n)?'''

Here is the proof.
Here is the proof.


If we have a field F(n), and n=a*b (a,b,n∈ N*, a,b≠1)
If we have a field F(n), and n=a*b (a,b,n∈ N*, a,b≠1) , which means n is not a prime number.


IN defination of multiplication
IN defination of multiplication


* 0 1 2 3 .......... b.......n-1
* 0 1 2 3 .......... b.......n-1


0 . . . . ......................
0 0 0 0 0 ...........0........0


1 . . . . ......................
1 0 . . . ......................


2 . . . . ......................
2 0 . . . ......................


3 . . . . .......................
3 0 . . . ......................


4 . . . . .......................
4 0 . . . ......................


. . . . . .......................
.. 0 . . . ......................


a 0 (a) (2a) (3a)...........(a*b).....(n-1)*a ( in this row, every element mod n)
a 0 (a) (2a) (3a).........(a*b)....(n-1)*a ''' ( in this row, every element mod n)'''


....................................
....0.....................................


....................................
....0.....................................


......................................
....0.....................................


.....................................
....0.....................................
n-1..................................
(n-1) 0...................................


see the (a+1)th row
see the (a+1)th row


There must be a "1" in this row, actually each row or column, to meet the rule :Existence of Negatives/Inverses.
There must be a "1" in this row, actually each row or column, to meet the rule :Existence of Negatives/Inverses. (About the rule, seen in the file "12-240/Classnotes for Tuesday September 11")






So if F(n) is a field, then
So if F(n) is a field, then
1.m*a=k*n+1 (k,m∈N*, m<n) there must exist k,m.
there must exist k,m ∈N*, m<n
to meet the equation:
2.n=ab
m*a=k*n+1

And we know that n=ab


==>>m*a=k*a*b+1 (a≠1)
So m*a=k*a*b+1 (a≠1)


==>>m=k*b+1/a
Hence m=k*b+1/a


unless a=1
unless a=1

Latest revision as of 16:13, 16 September 2012

12-24012-240-Splash.png

12-240/Classnotes for Tuesday September 11

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 F(n)

If you do not understand what the F(n) means, you can look through the file "12-240/Classnotes for Tuesday September 11"

Why all the numbers which are not prime numbers can not form a field F(n)?

Here is the proof.

If we have a field F(n), and n=a*b (a,b,n∈ N*, a,b≠1) , which means n is not a prime number.

IN defination of multiplication

  • 0 1 2 3 .......... b.......n-1

0 0 0 0 0 ...........0........0

1 0 . . . ......................

2 0 . . . ......................

3 0 . . . ......................

4 0 . . . ......................

.. 0 . . . ......................

a 0 (a) (2a) (3a).........(a*b)....(n-1)*a ( in this row, every element mod n)

....0.....................................

....0.....................................

....0.....................................

....0.....................................

(n-1) 0................................... see the (a+1)th row

There must be a "1" in this row, actually each row or column, to meet the rule :Existence of Negatives/Inverses. (About the rule, seen in the file "12-240/Classnotes for Tuesday September 11")


So if F(n) is a field, then

    there must exist k,m ∈N*, m<n
         to meet the equation:
                 m*a=k*n+1
 
  And we know that n=ab

    So m*a=k*a*b+1 (a≠1)
    Hence m=k*b+1/a 

unless a=1

     m will not exist, because m should be an integer.
   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