Algebra Theory?

Question

1. Let f(x)=x^3-6x^2+x+4 and g(x)=x^5-6x+1 be polynomials in Q[x].

(a) Determine all greatest common divisors of f(x) and g(x)

(b) Let d(x) be that greatest common divisor of f(x) and g(x), in which the coefficient of the highest power of x equals 1. Determine s(x), t(x) E Q[x] such that d(x)=s(x)f(x)+t(x)g(x).

(NOTE: E means 'Element Of')

2. List all polynomials of degree 4 in Z(subscript 2)[x] and express them as products of irreducible polynomials in Z(subscript 2)[x].

3.The polynomials x^3+2x^2+2x+1 can be factored into linear factors (i.e written as a product of polynomials of degree 1) in Z(subscript 5)[x]. Find this factorisation.

If someone could please help and give full working out I would really appreciate it.

Don

with 1 i have found out:
f(x) = x^3-6x^2+x+4
g(x)= x^5-6x+1
d(x)= 85849/1864000

so if you can help with the second part of question(finding s(x) and t(x)) would be much appreciated

Answer 1

1. a) Note that 1 is a zero of f, so we get
f(x) = (x-1)(x^2-5x-4)
and the discriminant of x^2 - 5x - 4 is 5^2 - 4(1)(-4) = 41 which is prime, so the remaining roots are irrational and x^2 - 5x - 4 is irreducible in Q[x].
Try these as factors of g(x): g(1) = 1 - 6 + 1 ≠ 0, so x-1 is not a factor. (Alternatively, by division we get g(x) = (x-1) (x^4+x^3+x^2+x-5) - 4.)
By division we also get g(x) = (x^2-5x-4) (x^3+5x^2+29x+165) - 935x - 661. So the greatest and only common divisor of f(x) and g(x) is 1.

Are you certain that you have copied f(x) and g(x) correctly? This result doesn't seem to fit with the second part of the question. I've checked all my working above (though it's still possible there's an error I didn't catch).

b) No such d(x) exists, but let's find s and t such that s(x)f(x)+t(x)g(x) = 1.
x^5-6x+1 = (x^3-6x^2+x+4) . (x^2+6x+35) + (200x^2-65x-141)
x^3-6x^2+x+4 = (200x^2-65x-141) . (x/200 + 227/8000) + (5679x/1600 + 64007/8000)
200x^2-65x-141 = (5679x/1600 + 64007/8000) . (320000x/5679 - 4687064000/32251041) + 32953216400/32251041
... OK, it looks like the numbers for this are going to be really nasty, so I'll skip the second half. Just because they're rational doesn't mean they won't be horrible. ;-)

2. Linear polynomials in Z_2[x] are x and x+1; from these we can get quadratic polynomials x^2, x^2 + x and (x+1)^2 = x^2 + 1. However x^2 + x + 1 is irreducible (substitute 0 and 1 and you get 1 in both cases, so it has no linear factors). It follows that the only reducible quartic polynomial with no linear factors is (x^2+x+1)^2 = x^4+x^2+1.

Cubic polynomials in Z_2[x] must have a linear factor to be reducible, so must either have constant term 0 or constant term 1 and an odd number of nonzero x terms (so must be x^3 + 1 or x^3 + x^2 + x + 1).

Obviously the leading term must be x^4, so the possible polynomials are
x^4 = (x)^4 (obviously!)
x^4 + 1 = (x + 1)^4
x^4 + x = x (x^3+1) = x (x + 1) (x^2 + x + 1)
x^4 + x + 1 has no linear factors, and is not equal to (x^2 + x + 1)^2, so it is irreducible.
x^4 + x^2 = x^2 (x + 1)^2
x^4 + x^2 + 1 = (x^2 + x + 1)^2
x^4 + x^2 + x = x (x^3 + x + 1)
x^4 + x^2 + x + 1 = (x + 1) (x^3 + x^2 + 1)
x^4 + x^3 = x^3 (x + 1)
x^4 + x^3 + 1 has no linear factors, so is irreducible as previously
x^4 + x^3 + x = x (x^3 + x^2 + 1)
x^4 + x^3 + x + 1 = (x + 1) (x^3 + 1) = (x + 1)^2 (x^2 + x + 1)
x^4 + x^3 + x^2 = x^2 (x^2 + x + 1)
x^4 + x^3 + x^2 + 1 = (x + 1) (x^3 + x + 1)
x^4 + x^3 + x^2 + x = x (x^3 + x^2 + x + 1) = x (x+1)^3
x^4 + x^3 + x^2 + x + 1 is irreducible.

3. P(x) = x^3 + 2x^2 + 2x + 1 in Z_5[x]: check for zeros, which must exist for linear factors:
P(0) = 1, P(1) = 1, P(2) = 1, P(3) = 2, P(4) = 0.
So the only possible linear factor is (x+1), and we get
P(x) = (x + 1) (x^2 + x + 1)
However, x^2 + x + 1 is irreducible in Z_5[x], so the question is incorrect.