# The primitive $n^{th}$ roots of unity form basis over $\mathbb{Q}$ for the cyclotomic field of $n^{th}$ roots of unity iff $n$ is square free

Prove that the primitive $$n^{th}$$ roots of unity form a basis over $$\mathbb{Q}$$ for the cyclotomic field of $$n^{th}$$ roots of unity if and only if $$n$$ is square free

I think I have the $$(\Rightarrow)$$ direction of this, but need help with the $$(\Leftarrow)$$ direction. Just for reference, I have the $$(\Rightarrow)$$ direction displayed below:

$$(\Rightarrow)$$ (contraposition) $$n$$ is not squarefree $$\implies \exists$$ prime $$p$$ s.t. $$p^2 \mid n$$ for $$\eta_n$$ is a primitive $$n^{th}$$ root of unity. Consider $$\beta := \eta_n^{n/p}$$. Now, $$\beta^p = (\eta_n^{n/p})^p = 1 \implies \beta$$ is a primitive $$p^{th}$$ root of unity. Hence, $$\beta$$ satisfies the equation

$$$$1+ \beta + \beta^2 + \dots +\beta^{p-1}= 0$$$$

Observe also that when $$j$$ satisfies $$1 \leq j \leq p-1$$, it follows that

$$$$\eta_n \beta^j = \eta_n \cdot \eta_n^{nj/p} = \eta_n^{nj/p +1}$$$$

Since $$p^2$$ divides $$n$$, every prime $$q$$ dividing $$n$$ will also divide $$\frac{nj}{p}$$. But we can be sure that $$\gcd(\frac{nj}{p}+1, n) = 1$$. Thus, $$\eta_n^{nj/p +1}$$ is a primitive $$n^{th}$$ root of unity. Then the $$\eta_n \beta^j$$'s satisfy the equation

$$$$1+ \eta_n \beta + \eta_n \beta^2 + \dots + \eta_n\beta^{p-1} = 0$$$$

But this shows that $$\{1, \alpha, \alpha^2, \dots, \alpha^{p-1}\}$$ is linearly dependent and hence not a basis for the cyclotomic field of $$n^{th}$$ roots of unity. $$\square$$

For the $$(\Leftarrow)$$, I could pursue it in two ways: either directly ($$n$$ is square free implies the primitive roots of unity form basis) or by contraposition (the converse of the above proof). I may try the second option.

Suppose the primitive $$n^{th}$$ roots of unity, denoted $$\{1, \eta_n^{(1)}, \dots, \eta_n^{(k-1)}\}$$ do not form a basis for the cyclotomic field of $$n^{th}$$ roots of unity over $$\mathbb{Q}$$. (this could either mean the primitive roots either don't span or are not linearly independent; intuition guides me to suspect that somehow we can infer linear dependence out of this. Maybe someone can explain this to me). Then for the equation

$$1+ a_1 \eta_n^{(1)}+ a_2\eta_n^{(2)} + \dots + a_k \eta_n^{(k-1)} = 0$$

we are guaranteed that one of the $$a_i$$'s are not zero.

Let us show that in case of an $$n$$ of the shape $$n=\prod_{p\in P}p\ ,$$with a product of primes from a finite set of primes $$P$$, the primitive $$n$$.th roots of unity are generating $$F$$, the cyclotomic field of $$n$$.th roots of unity, as a vector space over $$\Bbb Q$$. The field extension $$F:\Bbb Q$$ has degree $$\phi(n)$$, exactly as many as primitive $$n$$.th roots of unity.

Fix for each $$p$$ a primitive $$p$$.th root of unity $$\zeta_p$$. Then a primitive $$n$$.th root of unity is of the shape: $$\zeta(k):=\prod_{p\in P}\zeta_p^{k(p)}\ , \qquad \text{ where k=(k(p))_{p\in P} is a multiindex }\ ,\ \bbox[yellow]{0 (Their number is thus $$\prod(p-1)=\prod p\prod\left(1-\frac 1p\right) =n\prod_{p\mid n}\left(1-\frac 1p\right)=\varphi(n)$$.)

Consider an element $$x$$ of $$F$$, and write it as a linear combination with rational scalars over all roots of order $$n$$, primitive or not (seen as "vectors"). This is possible, since $$F$$ is by definition generated by all powers of $$\zeta_n:=\exp\frac{2\pi i}n$$. Using the (multiplicative version of the) Chinese Remainder Theorem, each "vector" above is of the form $$\zeta(l):=\prod_{p\in P}\zeta_p^{l(p)}\ , \qquad \text{ where l=(l(p))_{p\in P} is a multiindex }\ ,\ \bbox[yellow]{0\le l(p)} For each $$\zeta(l)$$ which has one or more $$l$$-components equal to zero, replace in the $$\zeta(l)$$-product the corresponding factor $$1$$ by the obvious linear combination of $$\zeta_p$$-powers, extracted from $$0 = 1+(\zeta_p+\zeta_p^2+\dots+\zeta_p^{p-1})\ .$$ This leads to a wanted representation of $$x$$ in terms of $$\zeta(k)$$ primitive roots, $$k$$ having no component equal to zero.

$$\square$$

Later EDIT: Some words on the chosen notation, as mentioned in the comments.

The above may be seen explicitly using the particular case of $$n=1001=7\cdot 11\cdot 13$$. In this case the set $$P$$ of the relevant primes is $$P=\{7,11,13\}$$, and we consider it always with the natural order, so the three primes appear in this order.

We fix the primitive roots of unity of order $$7,11,13$$, and denote them by $$\tag{*} \zeta_7,\zeta_{11},\zeta_{13}\ .$$ Now we want to take each primitive root of prime order from above to some power, then multiply them. When the number of primes is small, or at least fixed, the notations are simpler. We either use powers of the shape $$a,b,c$$ here, or use only one letter and many indices. Here, as indices we may be tempted first to use one, two, three, but then there is still a translation from one, two, three to $$7,11,13$$. So we immediately take these primes, and the correspondence is easier. So far we have the chance to write such a product of powers of the units in $$(*)$$ in one of the two forms:

• $$\zeta_7^{a}\zeta_{11}^{b}\zeta_{13}^{c}$$, or
• $$\zeta_7^{k_7}\zeta_{11}^{k_{11}}\zeta_{13}^{k_{13}}$$.

So the information on the powers is covered by a finite sequence of numbers, a collection, a tuple, a multiindex of numbers, either $$(a,b,c)$$, or $$(k_7,k_{11},k_{13})$$. For known three primes, i would use $$(a,b,c)$$ for the powers, the notation is simple, simple to use, simple to read and digest. But when there are dynamically many primes, the notation may become more complex. This is the case, and in order to have one letter for the "tuple" or "list" or "collection" or "multiindex" $$(k_7,k_{11},k_{13})$$, it is a good choice to denote it by $$k$$. (So happens also with sequences in analysis, if $$(x_n)_{n\in\Bbb N}$$ is a sequence, it is often simpler called $$x$$ to have a compact name for it.) Assume now we want to build the product $$\zeta_7^4\zeta_{11}^5\zeta_{13}^6\ .$$ Then the powers are covered in $$k=(k_7,k_{11},k_{13})=(4,5,6)$$, and it is natural to write (notation) "something with $$\zeta$$ and $$k$$" for this product, so i decided to use the notation $$\zeta(k)$$ for it, so for this special value: $$k=(4,5,6)=(k_7,k_{11},k_{13})=(k_p)_{p\in\{7,11,13\}}=(k_p)_{p\in P}$$ we have $$\zeta(k)= \prod_{p\in P} \zeta_p^{k_p} = \prod_{p\in \{7,11,13\}} \zeta_p^{k_p} = \zeta_7^{k_7}\zeta_{11}^{k_{11}}\zeta_{13}^{k_{13}} = \zeta_7^4\zeta_{11}^5\zeta_{13}^6\ .$$ With this notation it is then simpler to focus on the idea of proof of the fact that primitive units generate. Some unit corresponds to some / any choice of a "power data" / "tuple of powers" / "multiindex of powers", and to have a separation in notation for the two cases (general vs. primitive) i am using the letter $$l$$ for such "power data". The components of $$l$$ have no restriction, but of course, we can take them so that $$l_p$$ is between $$0$$, including $$0$$, and $$p$$, excluding $$p$$. Such a "general unit (of order $$n$$) is always in the form $$\zeta(l)$$, with the chosen notations. A primitive unit also has such a power data, we use $$k$$ for the tuple, sequence, multiindex of powers, and the primitivity condition reads $$k_p\ne 0$$ for each $$p\in P$$. So we exclude the zero values (for any/each of the components of $$k$$), the difference is marked in yellow above. The question is now, if these $$\zeta(k)$$ primitive units generate. Yes, we can easily exhibit the $$1$$ (corresponding to $$\zeta_p^0$$, the excluded power) for each factor in the product defining $$\zeta(k)$$.

• Thanks for the response. I'm having some trouble understanding notation. Can you clarify the meaning of the following: a.) $\zeta(k)$, b.) $k=(k(p))_{p \in P}$, and c.) multiindex." Unfortunately, I am not familiar with any of this convention/terminology. Commented Apr 10 at 22:48
• The answer is now edited, did my best to explain the used notation, if something is unclear, please do not hesitate to ask, many other readers may face the same problems, so it is always a good way for me to improve the quality of the answer. The community also always profit from such a short dialog in the comments, and a resulted fine tuning. Commented Apr 11 at 10:41