A Course in Number Theory and Cryptography (2nd Edition) by Neal Koblitz

By Neal Koblitz

It is a considerably revised and up-to-date advent to mathematics issues, either historic and sleek, which have been on the centre of curiosity in functions of quantity concept, relatively in cryptography. As such, no heritage in algebra or quantity idea is believed, and the booklet starts off with a dialogue of the elemental quantity conception that's wanted. The method taken is algorithmic, emphasising estimates of the potency of the innovations that come up from the idea, and one unique characteristic is the inclusion of modern functions of the idea of elliptic curves. large routines and cautious solutions are a vital part all the chapters.

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Extra resources for A Course in Number Theory and Cryptography (2nd Edition) (Graduate Texts in Mathematics, Volume 114)

Example text

The word "isomorphic" means that we have a 1-to-1 correspondence that preserves addition and multiplication. In some cases the fields F( a) and F( a') are the same, in which case we obtain an automorphism of the field. For example, v'2 has one conjugate, namely -v'2, over Q, and the map a+bv'2 1--+ a-bv'2 is an automorphism of the field Q( v'2) (which consists of all real numbers of the form a + bv'2 with a and b rational). If all of the conjugates of a are in the field F(a), then F(a) is called a Galois extension of F.

Since the union of all the cosets exhausts F~, this means that F~ is a disjoint union of d-element sets; hence dl(q - 1). Second proof. First we show that aq - l = 1. To see this, write the product of all nonzero elements in F q. There are q - 1 of them. If we multiply each of them by a, we get a rearrangement of the same elements (since any two distinct elements remain distinct after multiplication by a). Thus, the product is not affected. But we have multiplied this product by aq - l . Hence aq - l = 1.

A second method of solution would be first to compute 21000000 mod 7 (since 1000000 = 6· 166666 + 4, this is 24 == 2) and also 21000000 mod 11 (since 1000000 is divisible by 11-1, this is 1), and then use the Chinese Remainder Theorem to find an x between 0 and 76 which is == 2 mod 7 and == 1 mod 11. Modular exponentiation by the repeated squaring method. , finding the least nonnegative residue) when both m and n are very large. There is a clever way of doing this that is much quicker than repeated multiplication of b by itself.

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