# Minimal polynomials

• Apr 5th 2011, 12:10 PM
mathematic
Minimal polynomials
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
• Apr 5th 2011, 12:30 PM
TheChaz
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?
• Apr 5th 2011, 12:39 PM
mathematic
(a) u = a^5 + 3a^3 − a + 1
a^3a^2+3a^3-a+1=5a^2+3(5)-a+1=5a^2-a+16
b) for this one i"m unsure on finding the inverse
• Apr 5th 2011, 12:48 PM
TheChaz
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!
• Apr 5th 2011, 12:56 PM
mathematic
Ok that makes sense
• Apr 5th 2011, 01:20 PM
mathematic
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
• Apr 5th 2011, 04:37 PM
mathematic
II. Do I look at 1=u*m(x)?

III. We have degree=m<infinity
I'm tempted to look at something like this:
[M:K]=[M:L]*[L:K]. but we only have [F : K].
Like we could have [F:K]=[F:L]*[L:K] if K was a subfield of l, but I'm not seeing that
• Apr 5th 2011, 10:37 PM
zoek
Quote:

Originally Posted by mathematic
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.

Since $\displaystyle u$ be a nonzero element of $\displaystyle F$ that is algebraic over $\displaystyle K$ with minimal polynomial $\displaystyle m(x)$ means that $\displaystyle m(u)=0$.

So you have $\displaystyle u^n+a_{n-1}u^{n-1}+...+a_1u+a_0 = 0$

$\displaystyle \overset{\cdot u^{-n}}{\Longrightarrow } 1+a_{n-1}u^{-1}+...+a_1u^{-(n-1)}+a_0u^{-n}=0\Rightarrow$

$\displaystyle \Rightarrow 1+a_{n-1}u^{-1}+...+a_1\left( u^{-1}\right)^{(n-1)}+a_0\left( u^{-1}\right)^n=0\Rightarrow$

$\displaystyle \Rightarrow p\left( u^{-1}\right)=0$ where $\displaystyle p\left(x \right)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n$.
• Apr 5th 2011, 11:15 PM
zoek
Quote:

Originally Posted by mathematic
III. We have degree=m<infinity
I'm tempted to look at something like this:
[M:K]=[M:L]*[L:K]. but we only have [F : K].
Like we could have [F:K]=[F:L]*[L:K] if K was a subfield of l, but I'm not seeing that

If $\displaystyle p(x)$ has a root $\displaystyle y\in F$ then you can take $\displaystyle L = K(y)$.
You know that $\displaystyle [K(y):K] = deg(p) = n$.