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 , a 1 + n y = b 1 + n y. This further reduces to a . a n y = b . bn y .

Similarly , FROM a n = b n , WE HAVE ( a n ) y = ( b n ) y . SO a n 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 “

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