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Consider the cyclotomic field $\mathbb{Q}(\zeta_n)$. We know that the set of primitive roots $\Pi_n=\{\zeta_n^m:(m,n)=1\}$ generates $\mathbb{Q}(\zeta_n)$ as a field. However, what happens when we consider what it generates as a vector space?, i.e. the $\mathbb{Q}$-vector space $$\langle\Pi_n\rangle_{\mathbb{Q}}=\left\{\sum_{(m,n)=1}\lambda_m\zeta_n^m\ \Bigg\vert\ \lambda_m\in\mathbb{Q}\right\}$$ More precisely, my question is: What is the dimension of ${\langle\Pi_n\rangle_{\mathbb{Q}}}$? We know that $\varphi^*(n):=\text{dim}_{\mathbb{Q}}(\langle\Pi_n\rangle_\mathbb{Q})\leq\text{dim}_{\mathbb{Q}}(\mathbb{Q}(\Pi_n))=\varphi(n)$. Note that $\Pi_n$ can be linearly dependent. For instance, if $n$ is a multiple of $4$, we have $\varphi^*(n)\leq\varphi(n)-2$. To see this, write $n=4k$ and note that $$(2k\pm1,n)=(2k\pm1,4k)=(2k\pm1,\mp2)=(\pm1,\mp2)=1$$ so that $\zeta^{2k\pm1}_n=-\zeta_n^{\pm1}\in\Pi_n\Rightarrow \zeta^{2k\pm1}_n$ and $\zeta_n^{\pm1}$ $\mathbb{Q}$-linearly dependent. As an example, $$\Pi_{12}=\{\zeta_{12},\zeta_{12}^5,\zeta_{12}^7,\zeta_{12}^{11}\}=\{\zeta_{12},\zeta_{12}^5,-\zeta_{12},-\zeta_{12}^{5}\}$$ and $\zeta_{12}$ and $\zeta_{12}^5$ are linearly independent since $\zeta_{12}^5/\zeta_{12}=\zeta_{12}^4=\zeta_3\not\in\mathbb{Q}$ which implies that $\varphi^*(12)=2$. Thanks in advance!

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  • $\begingroup$ I think your last point would be made more clear by giving a specific example: $i$ and $-i$ are primitive 4th roots of unity and are $\mathbf Q$-linearly dependent. $\endgroup$
    – KCd
    Commented May 29 at 17:07

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The number of primitive $n$th roots of unity equals $[\mathbf Q(\zeta_n):\mathbf Q]$ and are a full set of $\mathbf Q$-conjugates by ${\rm Gal}(\mathbf Q(\zeta_n)/\mathbf Q)$. So having $\{\zeta_n^m : (m,n) = 1\}$ be $\mathbf Q$-linearly independent means $\{\zeta_n^m :(m,n) = 1\}$ is a normal basis of $\mathbf Q(\zeta_n)/\mathbf Q$. That happens if and only if $n$ is squarefree. This was the subject of an earlier MSE question here.

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  • $\begingroup$ Interesting! Is it possible to express the dimension $\varphi^*(n)$ of $\langle\zeta_n^m:(m,n)=1\rangle_\mathbb{Q}$ in the general case? With this, we know that $\varphi^*(n)=\varphi(n)\iff n$ is square-free. $\endgroup$
    – wakewi
    Commented May 29 at 17:17
  • $\begingroup$ If there is a linear relation $\sum_{(m,n)=1} a_m\zeta_n^m =0$ with rational coefficients then $\sum_{(m,n)=1} a_mx^m$ has $\zeta_n$ as a root, so it must be divisible by the $n$th cyclotomic polynomial. For example, $i$ and $-i = i^3$ are linearly dependent and the relation $0 = i+i^3$ is due to $x+x^3 = x(1+x^2)$, where $1+x^2$ is the minimal polynomial of $i$ over $\mathbf Q$. $\endgroup$
    – KCd
    Commented May 29 at 20:18
  • $\begingroup$ Maybe you will find something useful in Lenstra's paper on vanishing sums of roots of unity: math.leidenuniv.nl/~hwl/PUBLICATIONS/1979d/art.pdf. Look especially at section 2 on primitive relations on roots of unity. $\endgroup$
    – KCd
    Commented May 29 at 20:20

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