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I want to show that $f(x)=x^{4k} - 3x ^{3k} + 4x^{2k}-2x^k +1$ is irreducible in $\mathbb{Q}$ for all $k\in \mathbb{N}$. When $k=1$, it is easy to show; however I have trouble in proving this while $k\ge 2$. I have tried lots of irreducibility tests, but I have not found a way to prove this. Can anyone give me, at least, a hint?

Sil
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Gheehyun Nahm
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  • You can try Einstein's Criterion. – Scientifica Feb 08 '15 at 18:00
  • @Scientifica I have tried it, but for instance $k=2$, I believe that this does not work for any $n \in \mathbb{Z}$, even if you substitute $x+n$ for $x$. – Gheehyun Nahm Feb 08 '15 at 18:01
  • I didn't really understand what you said, but seems I was talking about another criterion.Sorry for the confusion. The one I'm talking about states that if a polynomial $P=\sum\limits_{k=0}^na_kX^k$ with integer coefficients (which is the case for your polynomial) has any rational root $\frac{p}{q}$ where $\gcd (p,q)=1$ then $p|a_0$ and $q|a_n$. Try this for your polynomial. – Scientifica Feb 08 '15 at 18:11
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    I'm quite sure that Einstein hadn't worked with polynomials. I suppose you meant "Eisenstein's Criterion"; but it's not useful in this setting. –  Feb 08 '15 at 18:16
  • @Scientifica I believe I did not understand your comment fully, but I am quite sure that there is no rational root of this polynomial, since this is tantamount with $x^k = 1 + e^{2\pi i/5}$. – Gheehyun Nahm Feb 08 '15 at 18:16
  • @AhmedHussein My bad, it's Eisensteins not Einstein. Furthurmore I was talking about another different thing. Sorry for the confusion I caused. – Scientifica Feb 08 '15 at 18:20
  • @user213955 I'll post my comment as an answer so that you understand better. – Scientifica Feb 08 '15 at 18:21
  • @user213955 I was mistaken as you said as a comment in my deleted answer. Sorry for that. About "Furthermore, Sakakibara-kun is so cute! :) (Which is irrelevant to this question)" then here was my irrelevant answer: " Oh you're an anime fan too! Nice to meet you! We can discuss about anime by e-mails (we can't here^^). My e-mail is in my profil page (click on my username)". – Scientifica Feb 08 '15 at 18:52
  • im not sure but i think you can use Einstein's Criterion – ali Feb 09 '15 at 14:30
  • @ali I believe that I have proven that using Eisenstein's Criterion does not work in this situation. – Gheehyun Nahm Feb 09 '15 at 15:21
  • you are right. i post an answer using a little galios theory. – ali Feb 09 '15 at 16:03

1 Answers1

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Lemma: If $F$ contains a primitive $k$th root of unity then $f(x)=x^k-b$ is irreducible over $F$ if $b$ has not any $n$th root in $F$, $n>1$.

Proof: We know $A=\{\sqrt[k]{b},w\sqrt[k]{b},w^2\sqrt[k]{b},...,w^{k-1}\sqrt[k]{b}\}$ is a subset of $K=F(\sqrt[k]{b})$ so $K/F$ is Galois. Its Galois group is a subgroup of $\mathbb Z_k$ because the roots of minimal polynomial of $\sqrt[k]{b}$ are in $A$, so $\phi:G\to \mathbb Z_k:\phi(\eta)=i$ if $\eta(\sqrt[k]{b})/\sqrt[k]{b}=w^i$ is an injective homomorphism. If $g$ is the minimal polynomial of $\sqrt[k]{b}$, then $g(0)=\prod_{j\in G}{w^j\sqrt[k]{b}}=\sqrt[k]{b^{\deg(g)}}$, so $g(0)\in F \iff \deg(g)=k$, so $g=f$ and $f$ is irreducible.

Let $f=x^{4k}-3x^{3k}+4x^{2k}-2x^k+1$. To prove $f$ is irreducible it is sufficient to show $[K:\mathbb Q]=4k$ where $K=\mathbb Q(\sqrt[k]{1+e^{2\pi i/5}})$. By using the tower lemma we have $$[K:\mathbb Q]=[K:F][F:\mathbb Q]=4[K:F]\ (F=\mathbb Q(e^{2\pi i/5}))$$ so it is sufficient to show $[K:F]=k$ or $x^k-(1+e^{2\pi i/5})$ is irreducible over $F$. But $1+e^{2\pi i/5}$ hasn't any $n$th root in $F$, so by the lemma it is irreducible over $F(w)$, so over $F$.

user26857
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ali
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