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I am attempting to solve the following problem for Artin's Algebra textbook. The problem is from a section about degrees of field extensions, so I assume that will be critical to the proof.

Prove that the polynomial $f(x)=x^4+3x+3$ is irreducible over the field $\mathbb{Q}[2^{1/3}]$

My attempt

Let $F=\mathbb{Q}$. Let $α\in \mathbb{C}$ such that $f(α)=0$. By Eisenstein, I see that $f$ is irreducible in $\mathbb{Q}[x]$, so it is the irreducible polynomial for $α$. Thus, $[F(α):F]=4$. It is easy to see that $x^3-2$ is the irredubile polynomial for $2^{1/3}$ and $[F(2^{1/3}):F]=3$ by similar logic.

Now, $[F(α,2^{1/3}):F]=[F(α,2^{1/3}):F(α)][F(α):F]=[F(α,2^{1/3}):F(2^{1/3})][F(2^{1/3}),F]$.

So $[F(α,2^{1/3}):F]=4[F(α,2^{1/3}):F(α)]=3[F(α,2^{1/3}):F(2^{1/3})]$.

So $[F(α,2^{1/3}):F]$ is a multiple of $12$.

Now, it seems to me that $[F(α,2^{1/3}):F]=12$, although I have not proven this yet. But I see that, if this is true, then we obtain $[F(α,2^{1/3}):F(2^{1/3})]=4$ and consequentially that $α$ has degree 4 over $F(2^{1/3})$ and thus $f(x)$ is actually irreducible over $F(2^{1/3})$.

If I can prove $[F(α,2^{1/3}):F]=12$, I will be done. How can I do this?

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Hint : you already have that $[F(α,2^{1/3}):F] \geq 12$, so it will suffice to show that $[F(α,2^{1/3}):F] \leq 12$. Now, notice that $[F(α,2^{1/3}):F(α)]\leq 3$ because $2^{1/3}$ is the zero of some polynomial of degree $3$ with coefficents in $F(α)$ (guess which one).

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  • $\begingroup$ If I'm not mistaken, $[F(α)](2^{1/3})=F(α,2^{1/3})$, right? $\endgroup$ – Pascal's Wager Apr 5 '18 at 22:33
  • $\begingroup$ @Pascal'sWager Yes. $\endgroup$ – Ewan Delanoy Apr 6 '18 at 6:08
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Let $[F(\alpha):F]=4$ and $[F(2^{\frac{1}{3}}):F]=3$ as both divide $[F(\alpha, 2^{\frac{1}{3}}):F]$ we have $12|[F(\alpha, 2^{\frac{1}{3}}):F]$.

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