Degrees of field extensions Find the degrees of $\Bbb{Q(a)/Q, Q(a,b)/Q, Q(a,b,c)/Q}$ where $a,b,c$ are the roots of $x^3+x-1$ where a is real.
To solve this problem, I noticed that $x^3+x-1$ is irreducible. Then why isn’t the degree of all the field extensions $3$?
$3,9,27$ for the degrees was incorrect.
 A: The main thing to notice is that $a$ is not a rational , i.e it is not a member of the base field.
$f(x)=x^{3}+x-1$ ,has only $1$ real root and $2$ complex roots which occur as complex conjugates.
$[\Bbb{Q}(a):\Bbb{Q}]=3$ as $f(x)=x^{3}+x-1$ is irreducible.
Now the question is whether the $f(x)$ itself remains irreducible over $\Bbb{Q}(a)$ and the answer is simply no because we already have a root.
Now over $\Bbb{Q}(a)$ , $f(x)=(x-a)g(x)$ where $g(x)$ has complex roots. i.e $g(x)=(x-b)(x-c)=(x-b)(x-\bar{b})$.
Now it is clear that $g(x)$ is irreducible over $\Bbb{Q}(a)$ as those roots are complex and $g(x)$ has degree $2$.
So $[\Bbb{Q}(a,b):\Bbb{Q}(a)]=2$.
And hence $[\Bbb{Q}(a,b):\Bbb{Q}]=6$ by tower law.
And since $b,c$ are complex conjugates, $\Bbb{Q}(a,b,c)=\Bbb{Q}(a,b)$ . This is because $g(x)$ is an irreducible and separable polynomial over $\Bbb{Q}(a)$ and it gives a degree $2$ extension . And all degree $2$ extensions over a field of characteristic not $2$ coming from an irreducible polynomial(of degree $2$) contain all roots of the polynomial.
If $g(x)=x^{2}+dx+e$ then $\Bbb{Q}(a,\sqrt{d^{2}-4e})=\Bbb{Q}(a,i\sqrt{4e-d^{2}})=\Bbb{Q}(a,b,\bar{b})$ as $b=\frac{-d+i\sqrt{4e-d^{2}}}{2}$.
Otherwise you can argue that as $a+b+c=0$ , $\Bbb{Q}(a,b,c)=\Bbb{Q}(a,b)$ as the comment section indicates.
