step in the field synthesis problem is to assume the existence of a field GF(q), and attempt to construct a larger field containing it. Emulating Appendix 5A. Finite Field Arithmetic the structures of the previous section, let P(z) be a polynomial of degree dm over GF(q) in an indeterminate Z. It is easily demonstrated that the collec- tion of pseudopolynomials in X of degree dm - 1 or less forms a commuta- tive ring under multiplication modulo P(x). Again, the question of field existence resolves to determining when multiplicative inverses exist. THEOREM 5A.7. If P(z) is an irreducible polynomial over GF(q), having x as a root, then the nonzero pseudopolynomials in X over GF(q) of degree less than the degree of P(z) all have multiplicative inverses under multipli- cation mod P(x). Proof. Let J(z) be a polynomial in Z, 1 <deg J(z) < deg P(z). Since P(z) is irreducible, the greatest common divisor of J(z) and P(z) is 1. Hence, there exist polynomials N1(z) and N2(z) such that 1 = gcd(J(z), P(z)) = N1(z)J(z) + N2(z)P(z) (5A.38) Reducing this equation mod P(z) gives 1 = ((N1(z) mod P(z)).((z)) mod P(z) (5A.39) Hence, by setting the indeterminate Z equal to X, it follows that N1(x) mod P(x) is the inverse of J(x) under multiplication mod P(x). By now the reader should recognize the similarity of the roles played by the prime numbers within the ring of integers and the irreducible polyno- mials within the ring of polynomials. The proof of the following theorem fol- lows exactly the proof of the equivalent theorem for the integers. THEOREM 5A.8. If P(z) is not irreducible over the field GF(q) then at least one non-zero pseudopolynomial in X over GF(q) with degree less than the degree of P(x) does not have an inverse under multiplication mod P(x). With the background developed thus far, it is possible to conclude the following: THEOREM 5A.9. Let GF(q) be a known field. A larger field GF(qm) con- taining GF(q) exists if, and only if, a polynomial P(z) of degree m over GF(q), irreducible over GF(q), exists. The distinct m