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We define $K:=\mathbb{Q}[\sqrt{-3}]$, in particular $e^{2\pi i/3} \in K$.

If $f \in \mathbb{Q}[X]$ is a monic, irreducible polynomial with $\text{deg}(f)=3$, why is $f$ also irreducible over $K$?

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up vote 3 down vote accepted

Hint: If $f$ were reducible over $K$, $K$ would contain an element of degree $3$ over $\Bbb Q$. Is there such an element?

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I don't know if it's the best way to proof that such an element doesn't exist. But what I know is that the minimal polynomial of $x_1=\sqrt{-3}$ is $f_{x_1}= X^2+3$ and the minimal polynomial of $x_2=e^{2\pi i/3}$ is $f_{x_2}= X^2+X+1$. But maybe the fact that $[K \colon \mathbb{Q}] =2$ shows that there is no element of degree $\geq 3$. – cem_astor May 18 '14 at 10:47
@cem_astor Elements in $K$ have the form $a + b \sqrt{-3}$ for $a, b \in \Bbb Q$. Such elements have degree $2$ over $\Bbb Q$ via polynomials similar to $X^2 + 3$. – Ayman Hourieh May 18 '14 at 11:01
I think it also follows from the fact that $[K \colon \mathbb{Q}]=2 < 3$. – cem_astor May 18 '14 at 11:06
@cem_astor Yes. You can do that. – Ayman Hourieh May 18 '14 at 11:06

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