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Computational Number Theory: Homomorphisms, Isomorphisms, and Chinese Remainder Theorem, Slides of Cryptography and System Security

The concepts of homomorphisms and isomorphisms in the context of computational number theory. It covers the chinese remainder theorem, which provides a solution to simultaneous congruences when the moduli are pairwise relatively prime. The document also discusses algebraic implications, including the existence of isomorphisms on multiplicative groups and the relationship between linear transformations and invertibility modulo each prime. Additionally, it touches upon the topic of finding square roots modulo composite numbers.

Typology: Slides

2012/2013

Uploaded on 04/27/2013

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Computational
Number Theory 2
2
Homomorphisms
DEF: A function between rings is
called a ring homomorphism if for all x, y
f ( x + y ) = f ( x ) + f ( y )
f ( x ! y ) = f ( x ) ! f ( y )
and f (1) = 1
Note: it follows that f (0) = 0, f (-x) = x, and
for invertible elements
Example: For M divisible by m
defined by f (n) = n mod m is homomorphism.
f:R1R2
f(x1) = f(x)1
f:ZMZm
3
Isomorphisms
DEF: A homomorphism that is bijective is a
called an isomorphism.
Example: Index theorem says that the
exponential function defined
by is an isomorphism if x is
primitive.
NOTE: is viewed as a ring if re-interpret
multiplication as addition, exponentiation by
index as multiplication, 1 as 0, and x as 1.
f:Zp1Z
p
Z
p
f(i)=ximod p
4
Chinese Remainder
Theorem
Suppose . There is a
homomorphism
defined by
NOTE: domain and codomain have same size
THM: If all the are pairwise relatively
prime, then f is an isomorphism. Furthermore,
the inverse is given by a linear function
with
mi
f(n)=(nmod m1,...,nmod mr)
M=m1·m2···mr
g(n1,n2,...,nr)=(c1n1+c2n2+...+crnr)mod M
ci=!M
mi"·#!M
mi"1
mod mi$
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Computational

Number Theory 2

2

Homomorphisms

DEF: A function between rings is called a ring homomorphism if for all x, y

• f^ (^ x^ +^ y^ ) =^ f^ (^ x^ ) +^ f^ (^ y^ )

• f^ (^ x^!^ y^ ) =^ f^ (^ x^ )^!^ f^ (^ y^ )

• and^ f^ (^1 ) =^1

Note: it follows that f (0) = 0, f (- x ) = x , and for invertible elements

Example: For M divisible by m defined by f ( n ) = n mod m is homomorphism.

f : R 1 → R 2

f ( x −^1 ) = f ( x )−^1

f : Z M → Z m

3

Isomorphisms

DEF: A homomorphism that is bijective is a called an isomorphism.

Example: Index theorem says that the exponential function defined by is an isomorphism if x is primitive.

NOTE: is viewed as a ring if re-interpret multiplication as addition, exponentiation by index as multiplication, 1 as 0, and x as 1.

f : Z p − 1 → Z∗ p

Z∗ p

f ( i ) = x i^ mod p

4

Chinese Remainder

Theorem

Suppose. There is a homomorphism defined by NOTE: domain and codomain have same size THM: If all the are pairwise relatively prime, then f is an isomorphism. Furthermore, the inverse is given by a linear function

with

f : Z M → Z m 1 × Z m 2 × · · · × Z mr

mi

f ( n ) = ( n mod m 1 ,... , n mod mr )

M = m 1 · m 2 · · · m r

g ( n 1 , n 2 ,... , n r ) = ( c 1 n 1 + c 2 n 2 +... + cr nr ) mod M

ci =

M

mi

[(

M

mi

mod mi

]

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5

Algebraic Implications

Assuming with all pairwise relatively prime.

LEMMA 1 : There is an isomorphism on multiplicative groups

COR:

LEMMA2: A linear transformation on the space (i.e. a k by k square matrix) is invertible iff it is invertible modulo each.

COR:

N = n 1 · n 2 · · · nr ni

Z∗ N ≈ Z∗ n 1 × Z∗ n 2 × · · · × Z∗ nr

!( N ) = !( n 1 ) · !( n 2 ) · · · !( nr )

Z k N

ni

Mk ( N )∗^ ≈ Mk ( n 1 )∗^ × · · · × Mk ( nr )∗

6

Square Roots mod- pq

For Prime Factors p, q

LEMMA: Let n = pq with p, q different odd primes. For each quadratic residue s mod n there are exactly four square roots of s. Furthermore, if , are the square roots of s respectively mod p and mod q , then the square roots of s mod n are all the sums:

THM: Factoring n and taking square roots mod n are equivalent in the class BPP.

± rp ± rq

[± q ( q −^1 mod p ) rp + ± p ( p −^1 mod q ) rq ] mod n

7

Taking e ’th Roots and

Factoring

Recall: for , such that the exponent e is relatively prime to , “ e ’th root” of b calculated by:

RESULT: If factorization of n is known, taking e ’th roots mod n is tractable.

FACT: For n = pq, knowing gives p, q.

PARTIAL CONVERSE: If know e ’th root exponent d then can factor n.

FULL CONVERSE? - Open problem

a ∈ Z∗ n

!( n )

b = ae^ mod n

a = be

− (^1) mod !( n )

mod n

!( n )

8

Miller-Rabin Primality

Let n be an odd number. Let q be the odd part of n -1, so n -1= , and b be any integer in. DEF: n is a strong pseudoprime relative to b if , or for some i < k. THM: For any odd composite n and random b Pr( n is strong pseudoprime rel. b )! ". NOTE: Non-prime pseudoprimes much rarer in practice. Worst case probability for n = 9. Miller-Rabin-Primality-Test(positive integer n ) if ( n == 1 OR n is even) return “NO” choose at random if ( gcd( b,n ) > 1 ) return “NO” return TestIfStrongPseudoPrime( n,b )

2 kq Z n

b ∈ [ 2 , n − 2 ]

b^2

iq

bq^ ≡ n 1 ≡ n − 1

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