For GF(2^4) arithmetic there are two
irreducible polynomials p(x)=x^4+x+1  & r(x)=x^4+x^3+1
we will use p(x)

addition: a+b -> same as a xor b
multiplication: first shift-add to find a(x)*b(x)
example: a(x)=x^3+x and b(x)=x^2+x+1
a(x)*b(x):    a: 1010
              b: 0111
---------------------
                 1010
                1010
               1010
              0000
------------------------ xor addition
              0110110
           = x^5+x^4+x^3+x^2

Now divide this polynomial by p(x) and find the remainder
fast remaindering algorithm: 
replace  x^4 by x + 1 
         x^5 by x^2 + x 
         x^6 by x^3 + x^2

reduction of x^5+x^4+x^3+x^2 = (x^2+x)+(x+1)+x^3+x^2 
= x^3+1
Therefore, a(x)*b(x) mod p(x) is found as x^3+1

GF(17) is arithmetic is mod 17 arithmetic
Numbers are 0,1,2, .., 16
they are 4-bit numbers except 16 which is a 5-bit number

a+b mod n example: 7+12=19 mod 17
                       =2

a*b mod n example: 7*12=84 mod 17
                       =-1 or +16

We can create tables for multiplication for both GF(17) &
GF(2^4) arithmetic.