If we have an algebraic number $\alpha$ with (complex) absolute value $1$, it does not follow that $\alpha$ is a root of unity (i.e., that $\alpha^n = 1$ for some $n$). For example, $(3/5 + 4/5 i)$ is not a root of unity.

But if we assume that $\alpha$ is an algebraic integer with absolute value $1$, does it follow that $\alpha$ is a root of unity?

I know that if all conjugates of $\alpha$ have absolute value $1$, then $\alpha$ is a root of unity by the argument below:

The minimal polynomial of $\alpha$ over $\mathbb{Z}$ is $\prod_{i=1}^d (x-\alpha_i)$, where the $\alpha_i$ are just the conjugates of $\alpha$. Then $\prod_{i=1}^d (x-\alpha_i^n)$ is a polynomial over $\mathbb{Z}$ with $\alpha^n$ as a root. It also has degree $d$, and all roots have absolute value $1$. But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that $\alpha^n=\sigma(\alpha)$ for some Galois conjugation $\sigma$. If $\sigma^m(\alpha) = \alpha$, then $\alpha^{n^m} = \alpha$.

Thus $\alpha^{n^m - 1} = 1$.

  • $\begingroup$ Consider (3+4i)/5. $\endgroup$ Sep 9, 2010 at 7:10
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    $\begingroup$ @Pierre-Yves Gaillard: but that is not an algebraic integer. $\endgroup$ Sep 9, 2010 at 7:36
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    $\begingroup$ @Qiaochu Yuan : You're right! Thanks! Shame on me! Moreover, this is explicitly mentioned in the question. I have no excuse! [I often forget the very first rule: read carefully the question!] $\endgroup$ Sep 9, 2010 at 7:54
  • $\begingroup$ @Jonas Kibelbek : It seems to me you can also phrase your very nice argument as follows. Let $K\subset\mathbb C$ be Galois of finite degree over $\mathbb Q$. Then the algebraic integers of $K$ which lie, together with their conjugates, on the unit circle form a finite (multiplicative) group. $\endgroup$ Sep 9, 2010 at 11:09
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    $\begingroup$ @Jonas Kibelbek - Here is a formulation of the argument by Kevin Buzzard: mathoverflow.net/questions/10911/… - This link is supposed to lead to KB's answer, but seems to lead to the question itself... At the time of writing, KB's answer is the first one. $\endgroup$ Sep 17, 2010 at 4:35

6 Answers 6


Let $x$ be an algebraic number with absolute value $1$. Then $x$ and its complex conjugate $\overline{x} = 1/x$ have the same minimal polynomial. Writing $f(T)$ for the minimal polynomial of $x$ over $\mathbb{Q}$, with degree $n$, the polynomials $T^nf(1/T)$ and $f(T)$ are irreducible over $\mathbb{Q}$ with root $\overline{x}$, so the polynomials are equal up to a scaling factor: $$T^nf(1/T) = cf(T).$$ Setting $T = 1$, $f(1) = cf(1)$.

Assuming $x$ is not rational (i.e., $x$ is not $1$ or $-1$), $f$ has degree greater than $1$, so $f(1)$ is nonzero and thus $c = 1$. Therefore $$T^nf(1/T) = f(T),$$ so $f(T)$ has symmetric coefficients. In particular, its constant term is $1$. Moreover, the roots of $f(T)$ come in reciprocal pairs (since $1$ and $-1$ are not roots), so $n$ is even.

Partial conclusion: an algebraic number other than $1$ or $-1$ which has absolute value $1$ has even degree over $\mathbb{Q}$ and its minimal polynomial has constant term $1$. In particular, if $x$ is an algebraic integer then it must be a unit.

There are no examples of algebraic integers with degree $2$ and absolute value $1$ that are not roots of unity, since a real quadratic field has no elements on the unit circle besides $1$ and $-1$ and the units in an imaginary quadratic field are all roots of unity (and actually are only $1$ and $-1$ except for $\mathbb{Q}(i)$ and $\mathbb{Q}(\omega)$). Thus the smallest degree $x$ could have over $\mathbb{Q}$ is $4$ and there are examples with degree $4$: the polynomial $$x^4 - 2x^3 - 2x + 1$$ has two roots on the unit circle and two real roots (one between $0$ and $1$ and the other greater than $1$).

  • $\begingroup$ Thank you; I appreciate the thorough explanation. In my comment above, I was simply wondering about the extension of least degree, curious if degree 4 was enough-- as you've shown it is. Checking various symmetric polynomials in Wolfram Alpha, I see there are many such unimodular units. x^4-3x^3+3x^2-3x+1 has discriminant just -275. $\endgroup$ Sep 10, 2010 at 2:09
  • $\begingroup$ How to see that a root in the explicit polynomial you mentioned is of modulus 1? $\endgroup$
    – Student
    Sep 24, 2019 at 12:51
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    $\begingroup$ @Student see the first paragraph of kconrad.math.uconn.edu/blurbs/galoistheory/numbersoncircle.pdf. $\endgroup$
    – KCd
    Sep 24, 2019 at 18:16
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    $\begingroup$ +1! A beautiful paper with beautiful pictures. $\endgroup$
    – Student
    Sep 24, 2019 at 18:24

No. There are algebraic integers on the unit circle which aren't roots of unity.

This paper by Ryan Daileda provides some useful information and references.

Also see Salem numbers.

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    $\begingroup$ Thank you! That's an excellent reference and paper (with the added benefit of being very clear and just 3 pages long)! I'm curious now what is the smallest extension containing such a unit. $\endgroup$ Sep 9, 2010 at 8:13
  • $\begingroup$ What do you mean by smallest extension? Extension of least degree? Least discriminant (in abs. value)? $\endgroup$
    – KCd
    Sep 9, 2010 at 12:01
  • $\begingroup$ @Robin In the paper by Ryan Daileda, he propesed at very first that the unit group $U_k$ of a number field $k$ is a finitely generated group. May I ask why? $\endgroup$ Dec 6, 2018 at 19:02
  • $\begingroup$ @WembleyInter the fact that $U_k$ is finitely generated is part of Dirichlet's unit theorem. It is part of the easier direction. Dirichlet's unit theorem says $U_k$ has rank $r(k) + s(k) - 1$, where $r(k)$ and $s(k)$ are defined in Daileda's paper (number of real embeddings and number of pairs of complex embeddings, respectively). To show $U_k$ has rank at most $r(k) + s(k) - 1$ is much easier than showing the rank is exactly $r(k) + s(k) - 1$, and getting that finite upper bound is enough to see that $U_k$ is finitely generated. $\endgroup$
    – KCd
    Nov 14, 2021 at 20:22

The answer to your question is no - not all algebraic integers with absolute value 1 are roots of unity - as you may have learned by now from the 2005 paper of Daileda that was referenced in the answer supplied by Robin Chapman. However, you should be aware that Daileda has in fact rediscovered these simple folklore results on unimodular units. These results are probably at least a half-century old - if not much older. Indeed, I recall reading similar results in papers published by Iwasawa in the fifties or sixties. I don't have the time now to locate said Iwasawa reference, but here is another reference that suffices to prove my point. This 1979 paper of Nakahata includes the same results and the same simple proofs found in Daileda's paper. Moreover, Nakahata places these results naturally into a more general context. Thus I highly recommend that you consult Nakahata's paper in addition to Daileda's. I've appended its Zbl review below.

alt text

  • $\begingroup$ Thank you, it's nice to have another reference and to see a more general result. $\endgroup$ Sep 17, 2010 at 3:35

Let me first mention an example in Character Theory. Let $G$ be a finite group of order $n$ and assume $\rho$ is a representation with character $\chi:=\chi_\rho$ which is defined by $\chi(g)=Tr(\rho(g))$. Since $G$ is a finite group then, by invoking facts from linear algebra, one can show $\chi(g)\in\mathbb{Z}[\zeta_n]$. For abelian groups, it is easy to see $\chi(g)$ is a root of unity, when $\chi$ is irreducible, but what about non-abelian groups? In other words let $|\chi(g)|=1$, what can we say about $\chi(g)$?

This relates to your question. Let assume $K/\mathbb{Q}$ be an abelian Galois extension inside $\mathbb{C}$, and take an algebraic integer $\alpha\in\mathcal{O}_K$ such that $|\alpha|=1$, then for any $\sigma\in Gal(K/\mathbb{Q})$ we have $$ |\sigma(\alpha)|^2=\sigma(\alpha)\overline{\sigma(\alpha)} $$ Since $K/\mathbb{Q}$ is abelian then $\overline{\sigma(\alpha)}=\sigma(\overline{\alpha})$ so $$ |\sigma(\alpha)|^2=\sigma(|\alpha|)=1 $$ Then norm of all its conjugate is one so it must be a root of unity. This answer to the question was posed, therefore if $|\chi(g)|=1$ then $\chi(g)$ is root of unity.


Notice however that if all roots of a polynomial with integer coefficients lie on the unit complex circle (i.e., all roots z have |z|=1), then, by a theorem of Kronecker, these roots are indeed roots of unit.

  • $\begingroup$ It's a good comment, but appears to be a comment rather than an answer. $\endgroup$
    – 6005
    Jan 14, 2015 at 5:50
  • $\begingroup$ Agree with Goos. This is a very good argument. But you should follow up on it by giving an example. Given that the question is over 4 years old, I don't think that a hint is any longer appropriate. $\endgroup$ Jan 14, 2015 at 6:00
  • $\begingroup$ Can you state the theorem, please? $\endgroup$
    – Ninja
    Oct 23, 2017 at 1:09
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    $\begingroup$ @Ninja You can find it in a really short way here (it is kindof included in the proof): crypto.stanford.edu/pbc/notes/numberfield/kummerlemma.html (all the roots are conjugate to each other). $\endgroup$
    – ctst
    Oct 27, 2017 at 10:09
  • $\begingroup$ I am checking it, thank you! $\endgroup$
    – Ninja
    Oct 27, 2017 at 23:07

As pointed out above, an algebraic integer with absolute value equal to 1 does not have to be a root of unity. But if all the conjugates of the algebraic integer have absolute value 1, then it is indeed the case.

Let $\alpha$ be an algebraic integer with minimum polynomial $f\in \mathbb{Z}[X]$, which is monic and say of degree $n$. We can assume $n \geq 2$ otherwise $\alpha \in \{-1,1\}$. Assume all the roots of $f$ have absolute value 1. Let us concentrate on the coefficient of $X^i$: the sum of the roots taking $i$ of them each time is in absolute value bounded by ${n}\choose{i}$ by the triangle inequality. Thus the coefficient of $X^i$ is in absolute value bounded by ${n}\choose{i}$, hence for any $n$ there are only finitely many algebraic integers of degree $n$ such that all conjugates have absolute value $1$, since there are only finitely many polynomials in $\mathbb{Z}[X]$ with given bounded coefficients.
Next, consider the powers of $\alpha$. They are all algebraic integers of degree at most $n$, and furthermore all their conjugates also have absolute value $1$ since the Galois actions map powers of $\alpha$ to powers of its conjugates. Thus, the powers of $\alpha$ are elements of a finite set. This implies $\alpha$ must be a root of unity.

In the case of a finite group $G$, $g \in G$ and complex (not necessarily irreducible) character $\chi$ with $|\chi(g)|=1$ all the Galois conjugates of the algebraic number $\chi(g)$ also have absolute value $1$. Let $n=o(g)$ and $K=\Omega_{\mathbb{Q}}^{X^n-1} \subseteq \mathbb{C}$ the splitting field. Let $\mathfrak{G}=Gal(K/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^*$. If $\sigma \in \mathfrak{G}$ and $\varepsilon$ is an $n$th-root of unity, then $\sigma(\varepsilon)=\varepsilon^m$, for some $m \in \mathbb{Z}$, with gcd$(m,n)=1$. Now, $\chi(g)=\varepsilon_1 + \cdots + \varepsilon_{\chi(1)}$, a sum of $n$th-roots of unity. Hence, $\sigma(\chi(g))=\varepsilon_1^m + \cdots + \varepsilon_{\chi(1)}^m=\chi(g^m)$. Note that $\mathfrak{G}$ is abelian and that the restriction of complex conjugation to $K$ induces an element of $\mathfrak{G}$ of order $2$. Hence, $|\sigma(\chi(g))|^2=\sigma(\chi(g)) \cdot \overline{\sigma(\chi(g))}=\sigma(\chi(g)) \cdot \sigma(\overline{\chi(g)})=\sigma(|\chi(g)|^2)=\sigma(1)=1$, which yields $|\chi(g^m)|=1$.


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