# Linear Dependence of Primitive Roots of Unity

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!

• 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.
– KCd
Commented May 29 at 17:07

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.
• 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. Commented May 29 at 17:17
• 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 $\mathbf Q$.