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How can I show that the polynomial $f(X)=X^5-2X^3+2X-2$ is irreducible over $\mathbb{Q}[\sqrt[3]{2}]$? Obviously, it is irreducible over $\mathbb{Q}$, by Eisenstein criterion, but how can I consider the extension $\mathbb{Q}[\sqrt[3]{2}]$? Thanks in advance.

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Just to remind you $k(\alpha)$ and $k[\alpha]$ are different things. –  Secret Math Jun 18 '13 at 0:26
    
Not in the case that $\alpha$ is algebraic over $k$.... –  JSchlather Jun 18 '13 at 1:03
    
@JSchlather I mean in general, that is why I use $\alpha$. I am just suggesting that we should always use better notation, i.e. when you adjoin an element to a field, it's always good to write $k(\alpha)$. –  Secret Math Jun 18 '13 at 1:08
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1 Answer

Here is what I have tried:

Let $\alpha$ be a root of $f(X)$ in $\mathbb{C}$, and denote $\beta = \sqrt[3]{2}$.

Now we consider tower of field extensions

$$ \mathbb{Q} \subset \mathbb{Q}(\alpha) \subset \mathbb{Q}(\alpha, \beta), \Rightarrow [\mathbb{Q}(\alpha, \beta): \mathbb{Q}] = [\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\alpha)][\mathbb{Q(\alpha)}:\mathbb{Q}] $$

$$ \mathbb{Q} \subset \mathbb{Q}(\beta) \subset \mathbb{Q}(\alpha, \beta), \Rightarrow [\mathbb{Q}(\alpha, \beta): \mathbb{Q}] = [\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\beta)][\mathbb{Q(\beta)}:\mathbb{Q}] $$

We know that

$$[\mathbb{Q(\alpha)}:\mathbb{Q}] = 5, [\mathbb{Q(\beta)}:\mathbb{Q}] = 3$$

$$[\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\beta)] \leq 5, [\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\alpha)] \leq 3$$

Therefore, we conclude that

$$[\mathbb{Q}(\alpha, \beta): \mathbb{Q}(\beta)] = 5$$

i.e. $f$ is irreducible in $\mathbb{Q}(\beta)$.

If this proof works well, you can generalize to two irreducible polynomials in $\mathbb{Q}[x]$ with their degree coprime.

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I lost something: why $[\mathbb{Q}(\alpha):\mathbb{Q}]=5$? I ask because $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is the degree of the minimal polynomial over $\mathbb{Q}$ with root $\alpha$. And a necessary but, not sufficient, condition to be minimal, is to be reducible... –  Edgar Almeida Jun 18 '13 at 4:16
    
@EdgarAlmeida $f(\alpha) = 0$, and $f$ is irreducible and monic, so $f = irr(\alpha, \mathbb{Q})$. –  Secret Math Jun 18 '13 at 4:25
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