Polynomial for each extension degree Let $E$ the splitting field of a polynomial in $\mathbb{Q}[x]$ of degree $3$, then $[E:\mathbb{Q}]=1,2,3,6$. 
I am asked to give an example for each case... Are the following correct?? 


*

*$[E:\mathbb{Q}]=1$ : $f(x)=(x-1)^3 $ 

*$[E:\mathbb{Q}]=2$ : $f(x)=(x-2)(x^2+x+1)$ 

*$[E:\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}]=3$ : $f(x)=x^3+x+1$ 

*$[E:\mathbb{Q}]=6$ : The splitting field is of the form $E=\mathbb{Q}(\alpha, \beta)$. That means that $f(x)=(x-a)(x^2+px+q)$, where $a \notin \mathbb{Q}$ and $p, q \notin \mathbb{Q}(a)$, right?? Then $f(x)=x^3+x^2(p-a)+x(q-ap)-aq$. But how can it be that $a \notin \mathbb{Q}$, $p \notin \mathbb{Q}(a)$ and $p-a \in \mathbb{Q}$ ??
 A: You've done a good job. 
A simple example of $[E:\mathbb{Q}] =2 $ would also be $E = Gal(x^3-1,\mathbb{Q})$. 
For $[E:\mathbb{Q}] = 3$ any irreducible polynomial of the form $f(x) = x^3 + px + q \in \mathbb{Q}[x]$ with discriminant $D = - 4p^3 - 27q^2$ (the discriminant of a third degree polynomial with real coefficients is given by $D = 18abdc - 4b^3d + 4b^2c^2 - 4ac^3 - 27a^2d^2$, now take $a = 1, b = 0, c= p$ and $d = q$).  
One example for $[E:\mathbb{Q}] = 6$ would be $E = Gal(x^3-2,\mathbb{Q})$. About the last question notice that $\sqrt{2} + 3,\sqrt{2} + 2 \notin \mathbb{Q}$ but $\sqrt{2} + 3 -(\sqrt{2} + 2) = 1 \in \mathbb{Q}$. 
A: I think there is a misunderstanding of the question posed. The splitting field of a polynomial is not the field generated (over $\mathbf Q$) by a root of this polynomial, but the field generated by all its roots. 
In particular, in the third example, $E$ is not generated  by α, though it is a simple extension, as all finite extensions of $\,\mathbf Q$. The polynomial $X^3+X+1$ has only one real root, α, and in the field $\mathbf Q(\alpha)$, it factors as
$$(X-\alpha)(X^2+\alpha X+\alpha^2-1).$$
The splitting field $F$ is obtained with the adjunction of the roots of $X^2+\alpha X+\alpha^2-1$, hence
$$[E:\mathbf Q]=[E:\mathbf Q(\alpha)][\mathbf Q(\alpha):\mathbf Q]=6.$$
An example of a splitting field of degree $3$:
Take the polynomial $f=X^3-3X^2+1$, and let $\omega$ be one of its roots. One can check that $\dfrac 1{1-\omega}$ is also a root of $f$, hence the third root is $\smash{\dfrac{\omega -1}\omega}$ (the product of the roots is $-1$).
So $f$ splits into linear factors in the field $\,\mathbf Q(\omega)$ : we've proved that $[E:\mathbf Q]=3$.
