Construct the Normal Closure for this extension

Question

I am trying to find the normal closure for the following extension

$\Bbb Q\subset\Bbb Q(t)$, where "$t$" is a zero of $x^3-3x^2+3$ and $\Bbb Q$ are the rational numbers.

I know the normal closure should have all of the roots of the the function in it (including $b=e^{i2\pi/3}$ since the function has degree $3$). However, I am having a hard time to find the zeros for this function.

I really hope you guys can help. Thank you very much.

Answer 1

Extended hints:

• The polynomial is irreducible, because Eisenstein with $p=$___ says so.
• The discriminant of this polynomial is $81$, which is a ________. Therefore the Galois group is __________.
• The first step in solving a cubic is to get rid of the quadratic term with the aid of a linear substitution. Here this amounts to calculating that $$ f(x+1)=x^3-3x+1. $$
• We could follow up on Cardano's method, locally annotated e.g. here. Leaving that to you, because I happened to recognize this polynomial. What follows is thus more than a bit ad hoc - sorry about that.

Let $\zeta=e^{2\pi i/9}=\cos(2\pi/9)+i\sin(2\pi/9)$. Then $t=\zeta+\overline{\zeta}=2\cos(2\pi/9)$, and using $\overline{\zeta}=1/\zeta$ we get $$ t^3=(\zeta+\frac1\zeta)^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}=3t+(\zeta^3+\zeta^{-3}). $$ Here $$ \zeta^3+\zeta^{-3}=e^{2\pi i/3}+e^{-2\pi i/3}=2\cos(2\pi/3)=-1, $$ so $t$ satisfies the equation $$ t^3=3t-1. $$ Leaving the rest to you.