Question

Let F49 be the field of 49 elements constructed in class. The definition of this field is F19={la(x)]F: a(r) e Z,a}} where Z7]is the ring of polynomials in r with coefficients in the field Z7 and a(x)p = {a(x)+ (1]zz + [4],)5(x) : 5(#) e Z7(a]} and addition is given by [a(r)]F+ [b(r)]F = [a(r) + b(2)]F and multiplication is given by [a(r)]F[b(x)]F = [a(z)b(1)]p. 1. Let Fa9t represent the ring of polynomials with coefficients in F9 (a) Show that for each [k]7 € Z7 there exists a(x) E Z7[r] such that (b) Let f(t)E F19[] be given by f(t) = [1];]pt? + [[2};]pt + [[2]};]p. Find both roots of f(t) in F49 (e) Show that f(t)= [[1];}]pt + [[6]z + [G] 7 a]p is irreducible in Fs9[¢]. 2. Let F25 be the name of a field of 25 elements. (a) Construct Fs, similar to the construction of F given in class. (b) Find the inverse of every non-zero element in F25 (c) Using left-regular representation, find an isomorphismo between F25 and a certain subset of 2 x 2 matrix with coefficients in Z3. Prove that o satisfies the two properties (a+B) (a)+ø(8) and (aB) = (a)6(B) for all a and B in F25 (d) Find a E F5 such that F = (a).

Answer 1

solution de fotela 1) a) To show that for each (kin ez, there exists a6) €2, (c) such that [a (x)} - {[k]] So, all pasible values of [k], are 0 1,3,5,7,131 (3[:] Now [o]} =[0]?. [1? =[]), [2]*-[4], , [3], [2],, [+]:[], [5,7 [4], [:] [:], So, for [k], = [o], we can the a (x)=[o], E 2, [x] [k], - []we can the a (x) = [], € 2} (x). Similarly for [k], [2], and [n], we can take a (x) = [3] and [2], respectively, het [k], [3]then take a(z) = [1],* €2, [x] Then (a (x)] = [[+], x) F - {[1],?-[], -[17] = [[n];} = [[3.]]] For [k), = [57, take a few] = {2],x € 3, G] Then [as]² = [w] x² - [u]; [[13x°[-]]io] - [], [61],x? +(w),]-[2]7 - [ato ] f *[-[13], = [52]

And for [k], = [6], , we can take a 67 = (3], 2 E 2, (c) . Then , {a(x)}2 = [],x? = {4, ([.],x+[u]n] - 36], iarad [-Low),] -- [1] = []; ..[4(x)] = [63] .6) given's RCD E Faq [t]. fle): 107,]f&%+[2],]+6+ [[2],]F. then by Sridham, formula roots of f(t) =o. { [[2]]F + V[CZ]2-4 [_!_]e[[2]] 2. [[37]F [[53]+ Yea]}e – [CZ]E = G), = -[-2.] [cz]]c. [[:],]pV[]]€ 1 [u],= (), [[2]] Now [497=(x+(47; -[u]n]e. • [[6]a] F = [[3),] .. V[032] F = [2] [[_32F in Flig is [[w), ] [-]][73] [ceb] [121] - [8], 5L!!1] [ab]e • [[n]c

+ [Cu]y][c].]e + [] = [[277] f + [[mx].] - [[67]e + [[u];x] ..there roots of f (t) are, ([621] F = [[m], ] in Fua c) tet lide show that it (t) = 169,76€ +[16]3+ [6];2]F Than {[[@]z + [3], x]?) - [[6)7]F => [a] + [b],)} + [[6], + [6],x], + (60716 => [(a+],+ [zab].* + [b*), 3*]e +[Co],+C4)x] e [ch] =>/14), ** + [aab+6],x + [a2+6]]= [colade =>(3), [[1]px?+ [c]]+ [2ab +6], x + [a’+6-452]] -0.7.] => [2ab +6), x + [a+6–4597]. = [[a]e Then, [2ab +6], = [], .... 0 [a76-46?), = [o], ... ☺ I implies [ab+3], -(0)2 ---(3) From (2], [a2-4b, = [0-6), = [4] - - [a+2b], [a-ab]z = [1],

NOW Now in 21 [ ]n can be written in the following a ways anly (+)2 = (+]; [+], + [6],[6]7 $[2], [u], - (33.[5]; thise cases one by one. (i)r. Let [a+26]2= ['11 = (a -26] Then [za] = [a+2b], +[a-2b]; = ['Js+ [], - [2], => [a], -[1], [1+2b] = [] i => [b], - [0] 2. . But then [ab+3] = [o+3] = [3],+ [o], so (2) is not