(From Milne's field theory notes)Why does the splitting field of a degree-$3$ irreduciblle polynomial has degree $3$ or $6$ over $\Bbb Q$?

Question

There is an example in Milne's notes says:

I am not sure how to understand the example. In ths line where it says $\Bbb Q[\alpha_1,\alpha_2,\alpha_3]=\Bbb [\alpha_1,\alpha_2]$, does it use the fact that once we have $deg(f)-1$ roots in the field, we have all the roots in the field? And may I please ask when it says $[\Bbb Q[\alpha_1,\alpha_2]:\Bbb Q[\alpha_1]]$ is $1$ or $2$, in which case do we have $1$ and which case do we have $2$?

Could someone please give some explaination? Thanks so much!

Answer 1

Yes: when $n-1$ roots of a degree $n$ monic polynomial are in an extension field $F$, then also the last one is, because the quotient will belong to $F[X]$ as well. More precisely, if $f$ factors as $$ f(X)=(X-\alpha_1)\dotsm(X-\alpha_{n-1})(X-\alpha_n) $$ in $E[X]$ and $\alpha_1,\dots,\alpha_{n-1}\in F\subseteq E$, then $$ f(X)=g(X)(X-\alpha_n) $$ where $g(X)=(X-\alpha_1)\dotsm(X-\alpha_{n-1})\in F[X]$, so also $X-\alpha_n\in F[X]$.

So, the splitting field of your $f$ is $\mathbb{Q}[\alpha_1,\alpha_2]$ and $\mathbb{Q}[\alpha_1]$ has degree $3$, because $f$ is irreducible over $\mathbb{Q}$. Then $f$ factors over $\mathbb{Q}[\alpha_1]$ as $f(X)=(X-\alpha_1)g(X)$, where $g$ has degree $2$ and has $\alpha_2$ and $\alpha_3$ as roots.

There are two cases: if $g$ splits over $\mathbb{Q}[\alpha_1]$, then both $\alpha_2$ and $\alpha_3$ belong to $\mathbb{Q}[\alpha_1]$; otherwise both have degree $2$ over $\mathbb{Q}[\alpha_1]$, because $g$ is irreducible over $\mathbb{Q}[\alpha_1]$, having no root in it.

Apply the dimension formula.

Answer 2

Let $E=F(\alpha_1,\alpha_2,\alpha_3)$ be the splitting field. Then we have $$E:F=[F(\alpha_1,\alpha_2,\alpha_3):F(\alpha_1,\alpha_2,)].[F(\alpha_1,\alpha_2):F(\alpha_1)].[F(\alpha_1):F]=[F(\alpha_1,\alpha_2,\alpha_3):F(\alpha_1,\alpha_2)].[F(\alpha_1,\alpha_2):F(\alpha_1)].3$$ Now we also have $$[F(\alpha_1,\alpha_2,\alpha_3):F(\alpha_1,\alpha_2)]=1$$ since $\sum_{i=1}^3 \alpha_i \in F$ Finally we have $F(\alpha_1)(\alpha_2):F(\alpha_1)= deg(g)$ where $ g $ is the minimum polynomial of $\alpha_2$ over $F(\alpha_1)$ Now we know $\alpha_2$ satisfies $\frac{f(X)}{X-\alpha_1}$ over $F(\alpha_1)$. Therefore $g$ divides this polynomial and hence $deg(g)≤2$