# Thread: Irreducible, real root, extension, linear combination

1. ## Irreducible, real root, extension, linear combination

a) Prove that the polynomial $\displaystyle f(X)=X^3-6X^2+9X+3$ is irreducible over $\displaystyle Q$

b) Consider the extension $\displaystyle Q \subset Q(u)$ where $\displaystyle u$ is a real root of $\displaystyle f$. Express $\displaystyle u^4$ and $\displaystyle (u+1)^{-1}$ like a linear combination of elements from the basis $\displaystyle \{1,u,u^2\}$.

Thanks a lot!

2. Originally Posted by roporte
a) Prove that the polynomial $\displaystyle f(X)=X^3-6X^2+9X+3$ is irreducible over $\displaystyle Q$

Thanks a lot!
if a polynomial is of degree 3 and is reducible, then the polynomial can be factored into either a linear and a quadratic or 3 linear factors..

in either case, it need a linear factor,, so if you want to prove it to be reducible, all you have to do is try to find a linear factor. if you can't then it is reducible..

so for your problem, try showing that it has no linear factor.
you can use the rational root test (which is the easiest), or mod-p test, or other test you know..

3. Originally Posted by roporte
a) Prove that the polynomial $\displaystyle f(X)=X^3-6X^2+9X+3$ is irreducible over [tex]Q
apply Eisenstein's criterion with p = 3.E
isenstein criterion
b) Consider the extension $\displaystyle Q \subset Q(u)$ where $\displaystyle u$ is a real root of $\displaystyle f$. Express $\displaystyle u^4$ and $\displaystyle (u+1)^{-1}$ like a linear combination of elements from the basis $\displaystyle \{1,u,u^2\}$.
$\displaystyle u^3=6u^2 - 9u - 3 \Longrightarrow u^4=6u^3 - 9u^2-3u=6(6u^2 - 9u - 3) - 9u^2 - 3u = 27u^2-57u-18.$

$\displaystyle (u+1)^{-1}=au^2+bu+c$ if and only if $\displaystyle (au^2+bu+c)(u+1)=1.$ now multiply out and use the fact that $\displaystyle u^3=6u^2-9u-3$ to find the coefficients $\displaystyle a,b,c.$