Question

C₁=5

C2 a. Describe a way to construct a field with 128 elements. (Detailed calculations are not required.) b. How many (non-isomorphic) fields are there of order C? c. Describe, using set-builder notation, the smallest field containing Z[x], the ring of polynomials with integer coefficients? Hint: look up “rational function field. (In algebraic geometry such a field is in some way | equivalent to a so-called “rational curve” e.g. a conic in the plane.)

Answer 1

128 = ph -x 27 им to the NRW Answer :- We have to construct a field with 128 element 27=ph, p → prime Then by Existence and Uniquences theorem of finite field- Any finite field with and (ph) elements is isomorphic to the splitting field of ra x* over Pp. xo, The field of 128 element is unique and isomorphic to the splitting field & -x over fe. 6 Unique fields exists of order G=5. G = s = p! p=5 prime. So, the field is upto isomorphis splitting field x5-x ② 222] be the set of all polynomial of integer co-effient. 2xt is an integrad domain So, and The associated field of fraction is the field of rational functions i.e. Q[x] in the variable a "Z. P(x) The elements of this field are 960) Whire P(x) & q(X) are polynomial of co-efficient in a with 20) 70 NEW, In particular P(x), q 12 may both be constant polynomials. so, the field of rational functions contains the field of Praction of 7. P(x) 2000) P , & €7, 2(x) = 0 9 () field of rational fraction So, Q*) is the smallest field containing 227 over /