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I am trying to understand the statement that was mentioned here.

If $f(x)$ is a monic polynomial in ${\mathbb Z}[x]$ and all roots have absolute value 1, then all roots are roots of unity.

I was wondering why we need to assume that $f(x)$ is monic. Are there any non-monic polynomials where the conclusion does not hold?

Oscar Lanzi
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ghc1997
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  • @ThePhoenix thank you! – ghc1997 May 17 '23 at 11:48
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    @ThePhoenix That can be an answer, I'd say. – aschepler May 17 '23 at 11:54
  • To give a more theoretical answer, if $f$ is monic it means that its roots are algebraic integers, rather than just general algebraic numbers. It is not true that an algebraic integer with abs. value 1 is a root of unity, but it is true if all the roots of its minimal polynomial have abs. value 1, which is what your assumption implies. See https://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity – SomeCallMeTim May 17 '23 at 12:24
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    Interestingly enough, a stronger claim is true. If all roots of a monic polynomial in $\mathbb Z[x]$ have absolute value less than or equal to $1$, then all roots are either $0$ or roots of unity. It surprises me that this weaker version of the theorem (the one in the link) is the one I most often see, since most proofs apply to the stonger version just fine. – Christian E. Ramirez May 17 '23 at 12:40
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    Why is this question tagged [tag:linear-algebra]? – J. W. Tanner May 17 '23 at 13:18

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ThePhoenix gives the example $5x^2-6x+5=0$, but $2x^2-3x+2=0$ will do with smaller coefficients. The roots are complex conjugates with the discriminant $-7$, and their product is $2/2=1$ so they must have unit modulus. But the actual roots are given by

$\dfrac{3\pm i\sqrt7}{4},$

and the arguments of these roots are $\pm\cos^{-1}(3/4)$. This can't be a rational multiple of $\pi$ since the cosine is a rational number that isn't half an integer, and roots of unity must have arguments that are rational multiples of $\pi$. $\rightarrow\leftarrow$

Oscar Lanzi
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