# $\mathbb{Q}$-dimension of f. g. $\mathbb{Q}[\mathbb{Z}/p^l]$-modules.

My question arises from the previous question

Let $M$ be a finitely generated $\mathbb{Q}[\mathbb{Z}/p^l]$-module, where $p$ is a prime number.

Is it true that $$\dim_\mathbb{Q} M\geq \frac{p}{p-1}\dim_\mathbb{Q} M \mathbin{\mathop{\otimes}_{\mathbb{Q}[\mathbb{Z}/p^l]}}\mathbb{Q}(\zeta_{p^l})?$$

Here, $\zeta_{p^l}$ is the $p^l$th root of the unity.

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 What about $M=0$. – Martin Brandenburg Oct 25 '12 at 7:58 I edited it. I also considered the case that $\mathbb{Q}[\mathbb{Z}/p^l]$-module structure on $M$ is trivial. So I modified the question into inequality. – user9384023 Oct 25 '12 at 8:10

Your ring is the direct sum of fields $$\frac{\mathbb{Q}[x]}{x^{p^{\ell}} - 1} = \bigoplus_{i=0}^{\ell} \mathbb{Q}(\zeta_{\ell^i}),$$ and a module over your ring is just a direct sum of vector spaces over these fields; the tensor product functor you wrote just projects such a module to the $\mathbb{Q}(\zeta_{p^\ell})$-vector-space-component. The inequality is false, because you can just take $M = \mathbb{Q}(\zeta_{p^\ell})$ itself. Maybe you don't want $\mathbb{Q}$-dimension on both sides?