How to calculate the degree of the splitting field $X^3+2$ over $\mathbb{Q}$.
I know that I need to first find the minimal polynomial which will give me what the splitting field is equal to but I don't know how to factorise it.
Degree of a splitting field over $\mathbb{Q}$
abstract-algebraminimal-polynomialssplitting-field
Best Answer
The splitting field is $\mathbb{Q}(r_1,r_2,r_3)$, where $r_1,r_2,r_3$ are the three roots of the polynomial $X^3+2$.
This is just the definition, but it's the starting point anyway. How do we know that the polynomial has distinct roots? The derivative test, of course: the derivative is $3X^2$ which has no nonconstant common factor with $X^3+2$.
Now let's look at the polynomial again: it has no rational roots, because $\pm1$ and $\pm2$ aren't roots. Since it has degree three, it is irreducible over $\mathbb{Q}$, but this is not that important in order to find the splitting field.
Suppose we are in an extension where all roots exist. Thus we can factor $X^3+2=(X-r_1)(X-r_2)(X-r_3)$; however, $r_1^3=-2$, so we also get the factorization $$ (X-r_1)(X^2+r_1X+r_1^2) $$ What does this entail? That $(X-r_2)(X-r_3)=X^2+r_1X+r_1^2$ and therefore $$ r_2+r_3=-r_1,\quad r_2r_3=r_1^2 $$ This is the same as saying that $a+b=-1$, $ab=1$, where $a=r_2/r_1$, $b=r_3/r_1$. In particular $a$ and $b$ are roots of $X^2+X+1$, so they're the nonreal cube roots of $1$. If we call $\omega$ one of them, the other one is $\omega^2$.
Hence $r_2=\omega r_1$ and $r_3=\omega^2 r_1$. Now we can make a choice: take $r_1=r=-\sqrt[3]{2}$ (the real cube root). Then the splitting field is $$ \mathbb{Q}(r,\omega r,\omega^2r)=\mathbb{Q}(r,\omega) $$ You know that $[\mathbb{Q}(r):\mathbb{Q}]=3$ and $[\mathbb{Q}(\omega):\mathbb{Q}]=2$.
Can you compute $[\mathbb{Q}(r,\omega):\mathbb{Q}(r)]$?
You're starting with the wrong foot. The given polynomial is irreducible, so it is the minimal polynomial of any of its roots.
You also seem to believe that if $f(X)$ is the minimal polynomial for some algebraic number, then $\mathbb{Q}[X]/(f(X))$ is the splitting field for $f(X)$. This is not true, in general, and the example shows it: the minimal polynomial for $r=-\sqrt[3]{2}$ is $X^3+2$, but $\mathbb{Q}[X]/(X^3+2)\cong\mathbb{Q}(r)$ is not the splitting field because it doesn't contain $\omega$.