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Is $\alpha = 2 \cos (2 \pi / 19)$ an algebraic integer?

I know that $\alpha$ is algebraic over $\mathbb{Q}$, since the left-hand side of $$(\cos(2\pi/19)) + i \sin(2 \pi/19))^{19} = 1$$ gives an integral polynomial in $\cos(2\pi/19)$. But I don't think this polynomial is monic (it's an ugly mess), so I don't know how to proceed.

I appreciate any help.

José Carlos Santos
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hdecristo
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  • Wolfram Alpha To be sure, a proof by hand would be more valuable, but still. – lulu Aug 27 '25 at 15:20
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    $2\cos x$ is an algebraic integer whenever $x$ is a rational multiple of $\pi$. This is because $2\cos x = e^{ix} + e^{-ix}$ and because roots of unity are algebraic integers. – darij grinberg Aug 27 '25 at 15:25
  • This seems like it should be a duplicate but surprisingly I can't find one. There are threads about $\cos \frac{2 \pi k}{n}$ being an algebraic number but I can't find one about this more refined question. – Qiaochu Yuan Aug 27 '25 at 15:36
  • @QiaochuYuan Here and here are two related questions. – Bowei Tang Aug 27 '25 at 15:46
  • Writing it as the sum of two algebraic integers as above is easy, but when we define algebraic integers as roots of monic polynomials, the fact that the set of algebraic integers is closed under addition feels like a complete miracle. Now that we are posting links, perhaps it could be useful for the OP to also link to an MSE question explaining this miracle in a nice way, if anybody knows one – Vincent Aug 27 '25 at 16:47
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    @Vincent: you can do this explicitly using Kronecker sums: https://math.stackexchange.com/a/155158/232. – Qiaochu Yuan Aug 27 '25 at 17:06
  • This is really nice. I learned a different proof years ago, but this is far more intuitive – Vincent Aug 28 '25 at 10:22

3 Answers3

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$2 \cos \frac{2 \pi k}{n} = \zeta_n^k + \zeta_n^{-k}$ is a sum of two roots of unity, where $\zeta_n = e^{\frac{2 \pi i}{n}}$.

You can also write down more explicitly a monic polynomial (not necessarily minimal) satisfied by $2 \cos \frac{2 \pi k}{n}$ using Chebyshev polynomials.

Qiaochu Yuan
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You can write explicitly a polynomial with root $\exp \frac{2 \pi i}{19}$

$$\frac{x^{19}-1}{x-1} = \sum_{k=0}^{18} x^k$$

Now divide the above equality by $x^9$ and get

$$(x^9 + x^{-9}) + (x^8 + x^{-8}) + \cdots + (x + x^{-1}) + 1 = 0$$

Some algebra tells us that $x^{k}+x^{-k}$ is a polynomial of degree $k$ in $(x+ x^{-1})$. So in the end an expression of degree $9$ in $(\exp \frac{2 \pi i}{19}+\exp -\frac{2 \pi i}{19})$ equals $0$.

Note: in general, we get a reciprocal equation for $\exp\frac{ 2 \ell \pi i}{n}$

$$\sum_{k=0}^{n-1} x^k = 0$$

If $n$ is odd, then proceed as above to get an equation of degree $\frac{n-1}{2}$ in $x+\frac{1}{x} = 2 \cos \frac{ 2 \ell \pi i}{n}$.

If $n$ is even, ( so $n-1$ odd) the above equation will be divisible by $x+1$. After division, get again a reciprocal equation of even degree, so proceed as above.

This is high school level algebra.

$\bf{added:}$

One could obtain explicitely an equation for $y =2 \cos \frac{2pi}{19}$ as follows:

We know that with $x=\exp \frac{2 \pi i}{19}$ we have

$$\sum_{k=0}^{18} x^k = 0 \\ x^2 - y x + 1 = 0$$

Therefore, the two above equations in $x$ have a common root, so their resultant is $0$. Now one can calculate the resultant as a determinant of a $20 \times 20 $ matrix ($20 = 18+2$) :

\begin{eqnarray} \left | \begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -y & 1 \\ \end{matrix} \right | \end{eqnarray}

which evaluates to $$\left(y^9+y^8-8 y^7-7 y^6+21 y^5+15 y^4-20 y^3-10 y^2+5 y+1\right)^2$$

orangeskid
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Your constant is in exactly the form used in Reuschle (1875 ). There is a simple method due to Gauss, some thirty years before Galois. The method is written in modern terms in Cox, Galois Theory I have the second edition (2012). Chapter 9, section 2 is Gauss and Roots of Unity. For that matter, Chapter 9 is called Cyclotomic Extensions, pages 229-254. He does one or two examples. Reuschle does many examples

On page numbered 23; he uses $\eta$

$$ \eta^9 + \eta^8 -8 \eta^7 -7 \eta^6 + 21 \eta^5 + 15 \eta^4 -20 \eta^3 - 10 \eta^2 +5 \eta +1 $$

Will Jagy
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