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Let $ \alpha $ be the root of some polynomial with integer coefficients. Can $ \alpha $ always be written as an algebraic expression using only rational number and roots of unity?

This is equivalent to asking if $$ \mathbb{Q}(\zeta_3,\zeta_4,\zeta_5, \dots ) $$ the infinite dimensional extension constructed by adjoining the $ n $th root of unity for every $ n $, is equal to the algebraic closure of $ \mathbb{Q} $.

If this is not true then what is an example of an algebraic number that cannot be expressed in terms of roots of unity?

Ian Gershon Teixeira
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No. If $\zeta$ is a root of unity, then the Galois group of $\zeta$ over $\mathbb{Q}$ is abelian (since any automorphism must send $\zeta$ to a power of $\zeta$ and such automorphisms commute since $(\zeta^n)^m=(\zeta^m)^n$). So, if $K\subseteq\mathbb{Q}(\zeta)$ is Galois over $\mathbb{Q}$, its Galois group must also be abelian (since it is a quotient of the Galois group of $\mathbb{Q}(\zeta)$). So if the Galois group of $\alpha$ is not abelian, then it cannot be expressed in terms of roots of unity. For instance, $\alpha=\sqrt[3]{2}$ cannot be expressed in terms of roots of unity, since its Galois group is $S_3$.

(In fact, this is the only obstruction: by the Kronecker-Weber theorem, if $\alpha$ has abelian Galois group, then it is in $\mathbb{Q}(\zeta)$ for some root of unity $\zeta$.)

Eric Wofsey
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    Interesting. Does this imply that ancient Greek geometers cannot double the cube even if you give them a device that can divide any angle by an arbitrary positive integer? – kaya3 Dec 12 '22 at 16:02
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    @kaya3: It doesn't obviously imply that. Note that being able to divide angles by arbitrary positive integers is (a priori) stronger than having access to arbitrary roots of unity, since there are constructible angles that are not rational multiples of $\pi$. – Eric Wofsey Dec 12 '22 at 16:10