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My question is about irreducible representations of groups over the field $\mathbb{Q}$.

Let $G$ be a cyclic or an abelian group. I want to check that under what conditions we have a $\mathbb{Q}G$-module that is irreducible. Because $\operatorname{char} \mathbb{Q}=0$, by Maschke 's theorem , we know that every $\mathbb{Q}$-representation of G is com­pletely reducible. Although, because $\mathbb{Q}$ is not algebraically closed, we can not say that every irreducible $\mathbb{Q}$-representations of an abelian group has degree 1.

Specially, I want to find a faithful irreducible $\mathbb{Q}$-representation of degree $\phi(n)$ (where $\phi$ is Euler's function) for a cyclic group of order $n$. I know that I should use $n^{th}$ roots of unity but I don't know how I can move this representation to $\mathbb{Q}$.

Thanks a lot.

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    $\begingroup$ You just map a generator of $C_n$ to the companion matrix of the minimal polynomial over ${\mathbb Q}$ of a primitive $n$-th root of $1$. This does indeed have degree $\phi(n)$. The result that any finite group with a faithful irreducible representation has cyclic centre follows from Schur's Lemma, and is true over any field. $\endgroup$ – Derek Holt Nov 4 '14 at 23:04
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You can classify the irreducible representations of a finite cyclic group over an arbitrary field as follows. The group algebra of the cyclic group $C_n$ of order $n$ over an arbitrary field $k$ is $k[x]/(x^n - 1)$. If $x^n - 1 = \prod_k f_k(x)^{m_k}$ is the factorization of $x^n - 1$ into irreducibles over $k$, then by the Chinese remainder theorem $k[x]$ factors (as a $k$-algebra) into a product

$$\prod_k k[x]/f_k(x)^{m_k}$$

and determining the simple modules of $k[x]/(x^n - 1)$ (which is equivalent to determining the irreducible representations of $C_n$ over $k$) from here is straightforward: they are precisely the quotients $k[x]/f_k(x)$, with a generator of $C_n$ acting as multiplication by $x$. Explicitly, $x$ acts on the basis $1, x, \dots x^{\deg f_k - 1}$ by the companion matrix of $f_k$.

Example. If $k = \mathbb{C}$, then $x^n - 1$ factors as

$$\prod_{k=0}^{n-1} (x - \zeta_n^k)$$

where $\zeta_n$ is a primitive $n^{th}$ root of unity. This recovers the familiar classification of irreducible representations of $C_n$ over the complex numbers.

Example. If $k = \mathbb{Q}$, then $x^n - 1$ factors as

$$\prod_{d | n} \Phi_d(x)$$

where $\Phi_d(x)$ is the $d^{th}$ cyclotomic polynomial. Hence there are $\sigma_0(n)$ irreducible representations, one for each divisor $d$ of $n$, which have dimension $\varphi(d)$ and are given explicitly in matrix form by the companion matrices of $\Phi_d$.

Example. Now suppose that $n = p$ is prime and that the characteristic of $k$ is also $p$. Then $x^p - 1$ factors as

$$(x - 1)^p.$$

Hence there is exactly one irreducible representation of $C_p$ over a field of characteristic $p$, namely the trivial representation.

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  • $\begingroup$ Thank you very much for your helpful answer. Would you please explain this part with an example over $\mathbb{Q}$: they are precisely the quotients $k[x]/f_k(x)$, with a generator of $C_n$ acting as multiplication by $x$. Explicitly, $x$ acts on the basis $1,x,…x^{deg \ f_k−1}$ by the companion matrix of $f_k$. $\endgroup$ – user187409 Nov 5 '14 at 7:19
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    $\begingroup$ @Alice: for example, when $n = 3$ we have $x^3 - 1 = (x - 1)(x^2 + x + 1)$, so the group algebra is $\mathbb{Q} \times \mathbb{Q}[x]/(x^2 + x + 1)$. In the basis $1, x$, the action of multiplication by $x$ is $1 \mapsto x, x^2 \mapsto -x - 1$, which has matrix $\left[ \begin{array}{cc} 1 & -1 \\ 0 & -1 \end{array} \right]$. $\endgroup$ – Qiaochu Yuan Nov 5 '14 at 8:51
  • $\begingroup$ @ Qiaochu: Thanks alot for your helpful example. I think that in your example you mean that $x\mapsto -x-1$ and the companion matrix \left [ \begin{array}{cc} 0&-1\\ 1&-1\\ \end{array} \right]. Am I right? $\endgroup$ – user187409 Nov 5 '14 at 21:35
  • $\begingroup$ @Alice: whoops, yes, right on both counts. $\endgroup$ – Qiaochu Yuan Nov 5 '14 at 21:45
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Hint: $\Bbb Q[\zeta_n]$ is a quotient ring of $\Bbb Q[C_n]$.

(All quotient rings are modules over the original ring...)

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