# Abstract Algebra Homework

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ABSTRACT ALGEBRA HOMEWORK

AbstractAlgebra Homework

AbstractAlgebra Homework

Question1

UsingSchonemann and Einstein’s criterion,

P(x) = anxn+an-1xn-1+……. + anx+ ao:

Ifthere exists a prime number p such that:

• P divides each ai for i6 =n

• P does not an, and

• P2 does not divide a0,

ThenP(X) is irreducible over the rationals.

Giventhe polynomial

P(X)= P(X1,…………,Xn) and by the fact that its Irreducible its also equal toP(S1,………..,Sn)the two polynomials are equal to zero.

Notealso that we can use Cohn’s irreducibility coefficient

Forexample for the polynomial P(X) = X7+ 5X4+ 35 can be shown to be irreducible.

Takingthe coefficient of X7,which is 1, this is not divisible 5 (obtained because it’s a factorof 5 and 35). Dividing this factor by 5 and 35, we find that itsdivisible hence using Einstein’s criterion, the polynomial isirreducible and equal to zero.

Fromthe remainder theorem, P(x) = Q(x)D(x) + R(x)

P(s)= Q(s)D(s) + R(s)

Sinceboth P(x) = o and P(s) = 0 0 = Q(x)D(x) + R(x)

O= Q(s)D(s) + R(s)

Sincep(x) =P(s), P(x,……..xn)= 0

Question3

2(a) From Perron’s Criterion

GivenP(X) = Xn+ an-1Xn-1+…….+ ao

Polynomialsmod P. An integer polynomial

P(X)= anXn+ an-1Xn-1+ …………..+ aomodP

=(anmodP)Xn+ (an-1modP)Xn-1+ …….. +aomodP, this implies that:

Q(,):Q= [)2+ ,)2+(]= 129

b)Q(+)= Q(+)2= 6 +2

Question4

Thefield extension Gal[Q(,j)/Q] is Galois of degree 4, hence its Galois group which has order 4.

Theelements of the Galois group are determined by the values of  and j. The Q conjugates of and j are and j.Sowe get four possible automrpisms in the Galois group.

Sincethe Galois group has order 4, there are 4 possible assignment ofvalues to and all exist

 J -j – J – -j

Eachnonidentity automorphism in the above table has order 2 sinceGal[Q(,j)/Q] contains 4 elements of order 2 , Gal[Q(,j)/Q] has 4 subfields Kisuch that [Gal[Q(,j):ki]= 2 or equivalently

[Ki:Q]=4/2 =2 and that completes the list. Below is a diagram of allsubfields

Q(,j)

Q() Q(j)Q(j)

Q

In the table above, the subgroup fixing Q() is the first and second row, the subgroup fixing Q(j) is thefirst and third row and the subgroup fixing Q(j)is the first and fourth row.