from a = 5^(1/3), we get a^3 = 5.
So rewrite a^5 as (a^3)(a^2) = 5(a^2)
etc.
Can you take it from here?
I. Let a = 5^(1/3) in R, algebraic over Q with minimal polynomial m(x) = x^3 −5. Write the
following elements of Q(a) as Q-linear combinations of {1, a, a^2}:
(a) u = a^5 + 3a^3 − a + 1
(b) v = (a^2 − 2a + 4)^−1.
My thought is to plug in a and see what u and v equal then.
II. Let F be an extension field of K and let u be a nonzero element of F that is algebraic
over K with minimal polynomial m(x) = x^n + a_(n−1)x^n−1 + · · · + a_1x + a_0. Show that
u^−1 is algebraic over K by finding a polynomial p(x) in K[x] such that p(u^−1) = 0.
Well I know a number u is algebraic if p(u)=0 for a plynomial p(x)
III. Let F be an extension field of K with [F : K] = m < infinity, and let p(x) in K[x] be a
polynomial of degree n that is irreducible over K. Show that if n does not divide m,
then p(x) has no roots in F.
n does not divide m, so we can't have m=nq
I'm still thinking about this one, but I'm 95% sure it has something to do with the "difference of cubes" factorization,
since "a" is a cube root, and because (a^2 - 2a + 4), when multiplied by (a - 2), = (a^3 - 8).
So let's see...
v = (.......)^-1
Multiply both sides by 1/(a - 2) to get
v = 1/(a^3 -8) = 1/(5 - 8) = -1/3
Double check that!
II. Let F be an extension field of K and let u be a nonzero element of F that is algebraic
over K with minimal polynomial m(x) = x^n + a_(n−1)x^n−1 + · · · + a_1x + a_0. Show that
u^−1 is algebraic over K by finding a polynomial p(x) in K[x] such that p(u^−1) = 0.
Well I know a number u is algebraic if p(u)=0 for a plynomial p(x)
Would I set m(x)=0
Then x^n + a_(n−1)x^n−1 + · · · + a_1x + a_0=0