1. NCM LECTURE NOTES ON I.N.HERSTEIN CRYPTOGRAHY
GOOD MORNING MY DEAR FRIENDS. TO DAY WE SHALL SEE ANOTHER BEAUTIFUL RESULT FROM
I. N. HERSTEIN. HAVE A GOOD DAY.
LET “ D “ BE A GIVEN INTEGRAL DOMAIN .
“ D “ IS A COMMUTATIVE RING WITH UNITY , 1 BELONGS TO “ D “ . ALSO “ D “ HAS NO ZERO
DIVISORS.
Let a , b be two given elements of “ D “ such that a . b = 0 . Then a = 0 ( OR ) b =0.
NOW WE SHALL SEE THE FOLLOWING PROBLEM :
SUPPOSE a , b belongs to “ D “ . am
= bm
AND an
= bn
for TWO RELATIVELY PRIME POSITIVE
INTEGERS “ m , n “ . G. C. D ( m , n ) = 1 . THEN “ a = b “ .
Here we should observe one thing : IN AN INTEGRAL DOMAIN , THERE IS NO GUARANTEE OF
INVERSE OF AN ELEMENT . THEREFORE NEGATIVE EXPONENT OF AN ELEMENT “ a “ or “ b “
Is undefined.
If “ a “ or “ b “ is zero , then the result is very much true . Without loss of generality , assume
That “ a “ AND “ b “ ARE NON ZERO ELEMENTS OF “ D “ .
FROM LINEAR DIOPHANTINE EQUATION THERE EXIST POSITIVE INTEGERS x , y > 0
Such that “ m . x - n . y = 1 “.
Now am
= bm
implies ( am
)x
= ( bm
)x
.
Again , a1 + n y
= b1 + n y
. This further reduces to a . an y
= b . bn y
.
Similarly , FROM an
= bn
, WE HAVE ( an
)y
= ( bn
)y
. SO an y
= bn y
.
Therefore , finally we have : ( a - b ) . an y
= 0 .
Since “ a “ is non zero , an y
can not be zero . Therefore “ a = b “
[ [ D ] ] = { ( a , a ) | “ a “ belongs to “ D “ } IS THE DIAGONAL SUBSET OF “ D X D “
NOW , LET US DEFINE A MAP “ f “ from [ [ D ] ] INTO “ D X D “

2. As f ( ( a , a ) ) = ( am
, an
) LIES IN “ D X D “ for ( m , n ) = 1.
THEN THIS MAP “ f “ IS ALWAYS INJECTIVE MAP ( one to one ).
SUPPOSE f ( ( a , a ) ) = f ( ( b , b ) ).
Then ( am
, an
) = ( bm
, bn
) . THIS IMPLIES am
= bm
and an
= bn
. BY THE PREVIOUS
OBSERVATION “ a = b “ .
NOW , LET US TAKE A JOURNEY TOWARDS “ SMART ENCRYPTION AND DECRYPTION “.
LET US TAKE “ D “ = finite integral domain with atleast “ 3 “ elements .
BY THE WELL KNOWN RESULT | D | = pn
, for some prime “ p “ and a positive integer “ n “. D IS A
FINITE FIELD.
D* = GROUP OF ALL NON ZERO ELEMENTS OF “ D “ ( CYCLIC GROUP OF pn
- 1 elements ).
D* = { a LIES IN “ D “ | “ a “ is non zero } and | D* | = | D | - 1 .
[ [ D* ] ] = { ( a , a ) | “ a “ LIES IN D* } and | [ [ D* ] ] | = | D* | = | D | - 1 .
DEFINE [ 0 , D* ] = { ( a , 0 ) , ( 0 , a ) | a lies in D* }. | [ 0 , D * ] | = 2 . | D | - 2.
S = { ( a , b ) | a , b lies in D* AND “ a “ is not equal to “ b “ }.
NOW D X D = S U [ [ D* ] ] U [ 0 , D* ] U { ( 0 , 0 ) } ( THESE SUBSETS FORMS A PARTITION
OF “ D X D “ ).
| S | = ( | D | - 1 ) . ( | D | - 2 ) .
| S U [ [ D* ]] | = ( | D | - 1 )2
.
NOW , DEFINE A MAP f : [ [ D * ] ] ………….> S U [ [ D * ] ]
DEFINED BY f ( ( a , a ) ) = ( am
, an
) LIES IN S U [ [ D * ] ] .PLEASE NOTE THAT “ a “ lies
in D* . AS | D | > 2 , THIS “ f “ is clearly INJECTIVE BUT NOT “ ONTO “.
THE VERY INTERESTING QUESTION WILL ARRISE : WHEN f ( ( a , a ) ) LIES IN “ S “
1. ( m , n ) = 1
2. ( |m – n | , | D | - 1 ) = 1
3. Here m + n = 1 ( MOD 2 ) .
FOR THE THIRD CONDITION , WE SHOULD HAVE | D | = pn
, p = odd prime