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An instructor asked me to factor $x^{2020} + x^{2019} + \cdots + x + 1$ on $\mathbb Q[x]$, which he considers to be tricky.

This polynomial is trivial to factor on $\mathbb C[x]$ and $\mathbb R[x]$. Just notice $x^{2021} - 1 = (x-1)(x^{2020} + x^{2019} + \cdots + x + 1)$ and we know

$$ \begin{equation} \begin{aligned} x^{2020} + x^{2019} + \cdots + x + 1 =& \prod_{k=1}^{2020} \left[x - \cos(\frac{2k\pi}{2021}) - i\sin(\frac{2k\pi}{2021})\right] \\ =& \prod_{k=1}^{1010} \left[x - \cos(\frac{2k\pi}{2021}) - i\sin(\frac{2k\pi}{2021})\right] \left[x - \cos(\frac{2k\pi}{2021}) + i\sin(\frac{2k\pi}{2021})\right] \\ =& \prod_{k=1}^{1010} \left[x^2 - 2\cos(\frac{2k\pi}{2021})x + 1\right] \end{aligned} \end{equation} $$

It's not easy to "combine" the factorization over $\mathbb R[x]$ to form a factorization over $\mathbb Q[x]$ though.

Should I go the other ways and try to prove it's irreducible? Maybe not. Since $2021 = 43 \times 47$, it seems reducible.

Jyrki Lahtonen
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nalzok
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    You're right to suspect a connection to the factorization of $2021$. Indeed, you should look into cyclotomic polynomials, which are the standard tool used to complete this factorization. – Greg Martin May 29 '20 at 04:35
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    Finding two irreducible factor is easy: \begin{align} 1 + x + \cdots + x^{2020} =& (1 + x + \cdots + x^{42}) + \ &(x^{43} + x^{44} + \cdots + x^{1\times 43 + 42})+ \cdots + \ &(x^{46\times 43} + x^{46\times 43} + \cdots + x^{46\times 43 + 42}) \ = & (1 + x + \cdots + x^{42})(1 + x^{43} + (x^{43})^2 + \cdots + (x^{43})^{46}). \end{align} and same procedure for $1+ \cdots + x^{46}$ is possible. Remaining polynomial is also irreducible but showing it is not that trivial; I also suggest you to study about cyclotimic polynomials. – dust05 May 29 '20 at 06:10
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    What Greg said. $x^{43}-1$ and $x^{47}-1$ are both factors of $x^{2021}-1$. And a text book application of Eisenstein's criterion shows that $(x^{43}-1)/(x-1)$ and $(x^{47}-1)/(x-1)$ are irreducible. The hard part is to prove that the remaining factor of degree $42\cdot46$ is irreducible. This was prove by Gauss, and has been explained on our site also. – Jyrki Lahtonen May 29 '20 at 06:19
  • @JyrkiLahtonen According to this, $x^{2021} - 1 = \phi_1(x)\phi_{43}(x)\phi_{47}(x)\phi_{2021}(x) = \phi_1(x)\phi_{43}(x)\phi_{47}(x)[\phi_{47}(x^{43})/\phi_{47}(x)] = \phi_1(x)\phi_{43}(x)\phi_{47}(x^{43})$, which leads to dust05's conclusion. AFAIK both $\phi_{43}(x)$ and $\phi_{47}(x)$ can be proved to be irreducible over $\mathbb{Q}$ by Eisenstein's criterion. What do you mean by "to prove that the remaining factor of degree $42 \cdot 46$ is irreducible"? I'm having some problems proving $\phi_{47}(x^{43})$ is irreducible, though. – nalzok May 29 '20 at 06:38
  • @nalzok It isn't. The irreducible factors are $\Phi_{43}(x)$, $\Phi_{47}(x)$ and $\Phi_{2021}(x)$. The last one has degree $42\cdot46$. Proving that it is irreducible is not trivial, but is covered in the linked question. – Jyrki Lahtonen May 29 '20 at 06:46

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