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In title, factor $x^{17}+1$ into a product of irreducibles over $\mathbb{R}$. I know it factors as $$(x+1)(x^{16}-x^{15}+\dots+1)$$

but I have no real justification for why the second factor is irreducible besides "mathematica says it's true and I don't want to try to factor it". I know it has no real zeros, since -1 is the only real zero of the original function and $x^{17}+1$ is coprime with its derivative, but that doesn't rule out that its the product of other irreducible polynomials of degrees higher than 1. I could also show the ideal generated by $x^{16}-x^{15}+\dots+1$ is maximal but that sounds horrific and I don't want to do it. Eisenstein's criterion also fails since the constant term is 1.

Joseph DeGaetani
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    Do you know about cyclotomic polynomials? The second factor is $\Phi_{17}(-x)$. It can be proved that $\Phi_n(x) $ is irreducible. – Paramanand Singh Aug 17 '21 at 17:10
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    The second factor is not irreducible over $\mathbb R$, as the only irreducible polynomials over $\mathbb R$ are the linear ones and the quadratics with negative discriminant. The base field is key here. You wrote that you want a factorization over $\mathbb R$ but it seems like you want it over $\mathbb Q$. – paul blart math cop Aug 17 '21 at 18:39

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Suppose you seek to prove that $x^4-x^3+x^2-x+1$ is irreducible. You can't apply the Eisenstein criterion to this polynomial as such, but if you render $x=y-1$ then you have $y^4-5y^3+10y^2-5y+5$ which is $5$-Eisenstein.

Try the same substitution with $x^{16}-x^{15}+x^{14}-...+1$ and see what happens.

Oscar Lanzi
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    Beautiful, thank you so much. For the sake of the question, using that and expanding gives $17-136y+680y^2-2380y^3+6188y^4-12376 y^5+19448y^6-24310 y^7+24310y^8-19448y^9+12376y^{10}-6188y^{11}+2380 y^{12}-680 y^{13}+136y^{14}-17y^{15}+y^{16}$, which is 17-Eisenstein. Is there a particular reason you chose $x=y-1$, and is there any quicker way to arrive at this result? This is a problem from a previously given timed exam and expanding that would eat up basically the entirety of the time. – Joseph DeGaetani Aug 17 '21 at 17:00
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    Say you have $x^5+1$, following on with my example. With $x=y-1$ you now have $(y-1)^5+1$. Expand the binomial power and observe the constant term cancels out, thus $y=x+1$ is a factor. The remaining terms have coefficients inherited from the binomial-power expansion, which (apart from the leading term) are all multiples of $5$ in my case because the original exponent $5$ was prime. The same happens with other prime initial exponents including $17$. – Oscar Lanzi Aug 17 '21 at 17:10
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    @JosephDeGaetani: if $p$ is a prime number then all the binomial coefficients $\binom{p} {r} $ for $r=1,2,\dots,p-1$ are divisible by $p$. The transformation $x=y-1$ makes use of binomial coefficients via binomial theorem. – Paramanand Singh Aug 17 '21 at 17:16
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Consider $\omega_k=\cos(\frac{2k\pi}{17})+i\sin(\frac{2k\pi}{17})$, with $k\in\{0,...,16\}$. Now since $f(x)=x^{17}+1$ splits in $\mathbb{C}$ and $f(-\omega_k)=0, \forall k\in\{0,...,16\}$, we can write $f$ as, $$f(x)=\prod_{k=0}^{16}(x+\omega_k)=(x+1)\prod_{k=1}^{16}(x+\omega_k).$$ Using the fact that $f(-\omega_k)=0\implies f(-\overline{\omega_k})=0$ we have that, $$(x+1)\prod_{k=1}^{16}(x+\omega_k)=(x+1)\prod_{k=1}^8(x+\omega_k)(x+\overline{\omega_k})=(x+1)\prod_{k=1}^8(x^2+2\operatorname{Re}(\omega_k)+\lvert\omega_k\rvert^2)=$$

$$=(x+1)\prod_{k=1}^8(x^2+2\cos(\frac{2k\pi}{17})+1).$$ At this point notice that the polynomials $f_k(x)=(x^2+2\cos(\frac{2k\pi}{17})+1)\in\mathbb{R}[x]$ are degree 2 and have negative discriminant, for any $k\in\{1,...,16\}$, and therefore they are irreducible.

gbaz
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  • I realized a little later that since the zeros were the 17th roots of unity, I could write them in the form $(x-a)(x-\overline{a})$, but didn't actually want to try to write those with real coefficients, nor did I know if it would've been acceptable to write then in condensed form with the justification that the above polynomial expands to real coefficients! – Joseph DeGaetani Jan 30 '23 at 20:41