= = Hence, a2+a+1 is the lowest degree pseudopolynomial in a which is zero, and the minimum polynomial of a is m,(z) = z2 + z z 1. The smallest subfield containing a is GF(4), whose elements are represented element a-representation and whose addition and multiplication tables are characterized completely by the GF(2) addition and multiplication tables and the fact that a2 = a + 1 = 0. The element C of Figure 5A.1 is not in GF(4). To determine the minimum polynomial of C over GF(4), evaluate the monic pseudopolynomials in C over GF(4). c+1=f,ca=d,c+b=j,c2d. Note that cadc>c2+ca=0 = (Addition and subtraction are identical operations in fields containing GF(2).) Hence, mcz)z2+za(over GF(4)). If instead the minimum polynomial of C over GF(2) had been determined, it would be seen that GF(2)). Representations of GF(16) elements as pseudopolynomials in C over GF(4) and GF(2) are shown in Table 5A.1. It is obvious from the above pseudopoly- nomial representations that the elements of GF(16) may be viewed as a 2- dimensional vector space over GF(4) or a 4-dimensional vector space over GF(2). In either case, vector addition corresponds to field element addition. For example, the field element h can be viewed as the vector (a, b) of pseudopolynomial coefficients over GF(4), or as (1,0,1,1) over GF(2). Using the c-representations over GF(4) of GF(16) elements, only the GF(4) arithmetic tables and the fact that C2 +++++ are necessary to do calculations. For example: Appendix 5A. Finite Field Arithmetic e(m+k) = (bc+a)((bc) +(ac +a)) (representation) (bc + a)(c + a) (addition in GF(4)) bc2+(a+ab)c+ a2 (distributive law) = b(c+a)+(a+ab)c+a2 (c2 = c + a) (GF(4) arithmetic) The same calculations using c-representations over GF(2) along with the fact that A+c+1=0is: e(m+k) = c((c3+c2+c)+(c3+c) (representation) (GF(2) arithmetic) c(c + 1) (c4 = c + 1) c2 + c (distributive 1 law) (representation) The above agree with the direct calculation e(m+k)=ed=a = from Figure 5A.1. Clearly the cumbersome tables of Figure 5A.1 can be totally el