$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Gal{Gal}$Let $ L $ be a cyclic Galois extension of $ \mathbb{Q} $ of degree $ 6 $. So $ G = \Gal(L/\mathbb{Q}) $ is a cyclic group of order $ 6 $. Then we have a homomorphism $ \phi : G \rightarrow \GL (L) $ defined by $ \phi(\sigma)(g) = \sigma(g) $ for all $ \sigma \in G $. Thus by representation theory we can consider $ L $ as a $\mathbb{Q}G $ module. Then $ L $ can be written as a direct-sum decomposition of $ r $ distinct irreducible submodules. What is the value of $ r $? And what are the irreducible components?
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By the normal basis theorem, $L$ is the regular representation of $G$, i.e., $r = 6$ and the components are the $6$ distinct characters of $G$.
EDIT: As pointed out in the comments, particularly by @FrançoisBrunault, that is $L \otimes_{\mathbb Q} \mathbb C$ as a $\mathbb C[G]$-module; $L$ itself as a $\mathbb Q[G]$-module has $4$ summands, corresponding to the trivial and sign characters, and the two other pairs of complex conjugate characters.
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2$\begingroup$ I don't think that there will be 6 summands over ${\mathbb Q}$; it seems that $r=4$ in this case since $x^6-1$ decomposes into 4 irreducible factors over ${\mathbb Q}$. $\endgroup$ Commented Dec 17, 2021 at 18:25
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$\begingroup$ @VictorOstrik, I'm not sure I follow why the number of irreducible factors of $x^6 - 1$ controls the number of summands in a decomposition of a cyclic extension of degree $6$. $\endgroup$– LSpiceCommented Dec 17, 2021 at 18:27
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3$\begingroup$ The group algebra $\mathbb{Q}G$ is isomorphic to $\mathbb{Q} \times \mathbb{Q} \times \mathbb{Q}[x]/(x^2+x+1) \times \mathbb{Q}[x]/(x^2-x+1)$ and this is the decomposition as a $G$-module for the left multiplication by $G$. Over $\mathbb{C}$ your answer is correct of course. $\endgroup$ Commented Dec 17, 2021 at 19:52
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$\begingroup$ @FrançoisBrunault, ah, yes, I see, I was indeed unthinkingly working with the $\mathbb C$-algebra. $\endgroup$– LSpiceCommented Dec 17, 2021 at 20:12