In my field theory class, I had to prove that $[E:F]=3$ or $6$ if $E$ is a splitting field of an irreducible polynomial $f(x)$ (of degree $3$) over $F$. I see three separate cases:
$f(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$: $f(x)$ is separable, so I can create a tower of simple field extensions. $F(\alpha_1)$ is a degree $3$ extension over $F$, and $F(\alpha_1,\alpha_2)$ is a degree $2$ extension over $F(\alpha_1)$. I got the degrees from the degree of the irreducible polynomial for each field, and by the tower law, $[F(\alpha_1,\alpha_2):F]=3\cdot 2$.
Why shouldn't there be another extension s.t. $E=F(\alpha_1,\alpha_2,\alpha_3)$, rather than $E=F(\alpha_1,\alpha_2)$? Viewing the degree of the field extension through the degree of the irreducible polynomial, it's clear to me that $irr(\alpha_3,F(\alpha_1,\alpha_2)=x-\alpha_3$, which is degree $1$; $[F(\alpha_1,\alpha_2,\alpha_3):F(\alpha_1,\alpha_2)]=1$, so they're the same field (*). I understand this explanation, but I can't really visualize it. I thought of the example of the splitting field of $x^3-2$ over $\mathbb Q$, but I don't see how this applies with $\alpha_3$ – the zeros of $x^3-2$ are $\{\sqrt [3]2,\omega\sqrt [3]2,\omega^2\sqrt [3]2\}$ and the splitting field is $\mathbb Q(\sqrt [3]2,\omega)$, but what if $\alpha_3$ can't be represented as the sum/product of $\alpha_1,\alpha_2?$
$f(x)=(x-\alpha_1)^3$: A simple extension will give the splitting field, and $[F(\alpha):F]=3$.
$f(x)=(x-\alpha_1)^2(x-\alpha_2):$ How would the splitting field be of degree $3$ or $6$, not degree $2$? $irr(\alpha_1,F)$ is of degree 2, so $[F(\alpha_1:F]=2$. Following the logic of (*) in my first question, the total degree of the splitting field should be $2\cdot 1$.
Thank you in advance for reading and responding!