5
ABSTRACT ALGEBRA HOMEWORK
AbstractAlgebra Homework
AbstractAlgebra Homework
Question1
UsingSchonemann and Einstein’s criterion,
P(x) = a_{n}x_{n}+a_{n1}x_{n1}+……. + a_{n}x+ a_{o}:
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(X_{1},…………,X_{n}) and by the fact that its Irreducible its also equal toP(S_{1},………..,S_{n})the two polynomials are equal to zero.
Notealso that we can use Cohn’s irreducibility coefficient
Forexample for the polynomial P(X) = X_{7}+ 5X_{4}+ 35 can be shown to be irreducible.
Takingthe coefficient of X_{7},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,……..x_{n})= 0
Question3
2(a) From Perron’s Criterion
GivenP(X) = X_{n}+ a_{n1}X^{n1}+…….+ a_{o}
Polynomialsmod P. An integer polynomial
P(X)= a_{n}X^{n}+ a_{n1}X^{n1}+ …………..+ a_{o}modP
=(a_{n}modP)X^{n}+ (a_{n1}modP)X^{n1}+ …….. +a_{o}modP, 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 K_{i}such that [Gal[Q(_{},j):k_{i}]= 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.