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Wednesday, 10 April 2019

Fields and Modular Arithmetic

I only recently became aware of the fact that a set of elements like {0, 1, 2, 3, 4, 5, 6} forms a field under modular arithmetic with modulus 7 (the number of elements in the set). The condition is that there are a prime number of elements in the set. Let's firstly define this idea of modular arithmetic:

Figure 1: extract from
https://crypto.stanford.edu/pbc/notes/numbertheory/arith.html

Now what is a field? Here's a definition from http://www.applet-magic.com/field.htm:
A field F consists of a set S of elements and two binary functions f and g defined on S and whose values are in S. Both binary functions are associative, f(a,f(b,c) is equal to f(f(a,b),c) for all a, b and c in S and likewise for g. The binary function f, called addition is necessarily commutative; i.e., f(a,b) is equal to f(b,a) for all a and b in S. The other function g, called multiplication, is not necessarily commutative. There is an identity for each functions; i.e. there exist e such that f(e,a)=a for all a in S and an l such that g(l,a)=a for all a in S. There exist additive inverses for all elements of S and multiplicative inverses for all elements of S except the additive identity. This means that for any element a in S there exist a and b such that f(a,b)=e. Such a b is denoted as a. For any element a in S that is not e there exists a and b such that g(a,b) is equal to l. The element b is usually denoted as a1. Furthermore there is distributivity of multiplication with respect to addition; i.e. g(a,f(b,c)) is equal to f((g(a,b),g(a,c)) for all a,, b and c in S.
The set of n x n matrices is an example of a field in which multiplication is not commutative. Putting the whole business less formally, we can say that a set of elements form a field if the set is closed under the operations of addition, subtraction, multiplication and division. In other words, these operations when applied to two elements in the set always produce a third element that is also in the set. An example is {0, 1, 2, 3, 4, 5, 6} and it is easy to set up tables for all four operations verifying the set is indeed closed, as it is under all sets with a prime number of elements. Figure 2 shows what happens to the elements of the set under the operation of multiplication by 2:

Figure 2: mapping of elements of mod 7 under multiplication by 2

Figure 3 shows the full multiplication table:

Figure 3: multiplication table for mod 7

Of course, as stated above there is no multiplicative inverse for the additive identity element. In the case of the mod 7, this element is 0 and it means we can't divide by this element just as in regular arithmetic. Figure 4 shows the full division table:

Figure 4: division table for mod 7 with case of 5 divided by 3 highlighted

Figure 5 explains the condition for division to be possible in a modular system:

Figure 4: extract from
https://crypto.stanford.edu/pbc/notes/numbertheory/arith.html

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