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Is $x^{20} +x^{15}+ x^{10}+x^5+1$ irreducible in $\mathbb{Q}[x]$?

I think if $y=x^5$, then $P(y)=y^{4} +y^{3}+ y^{2}+y+1 =\displaystyle \frac{y^5-1}{y-1}$, $5$ is prime, then $P(y)$ is irreducible in $\mathbb{Q}[y]$ or $\mathbb{Q}[x^5]$ , how do I show that it is irreducible in $\mathbb{Q}[x]$?

andre
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    Are you familiar with cyclotomic polynomials or Eisenstein's criterion? – anon Oct 24 '13 at 03:59
  • see this : http://math.stackexchange.com/questions/215042/irreducibility-of-xp-1-ldots-x1 – jim Oct 24 '13 at 04:10
  • I´m familiar with Eisenstein's criterion – andre Oct 24 '13 at 04:20
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    Let $f(x) = x^{20}+x^{15}+x^{10}+x^5+1 \in \mathbb{Z}[x]$, we have $$\begin{array}{rrcl} & (x^5-1)f(x) &=& x^{25} - 1\ \implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1\ \implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1 \pmod 5\ \end{array}$$ Simplifying this, what can you say about the coefficients of $f(x+1)$? – achille hui Oct 24 '13 at 05:03
  • just one doubt if f(x+1) is irreducible , then f(x) is too because homomorphism , i can say that f($x^5+1$) is irreducible , then f($x^5$) is irreducible too? – andre Oct 24 '13 at 13:04
  • For the $1^{st}$ part of your statement, $f(x)$ reducible $\implies f(x) = g(x)h(x)$ for some non-trivial $g(x), h(x) \implies f(x+1) = g(x+1)h(x+1)$ reducible. The contra-positive is $f(x+1)$ irreducible $\implies f(x)$ irreducible. So the $1^{st}$ part of statement is true. For the $2^{nd}$ part, $f(x^5)$ reducible $\implies f(x^5) = g(x)h(x)$ for non-trivial $g(x),h(x)$. However, there is no obvious transform to express $f(x^5+1)$ in terms of $g(x), h(x)$. This means there is no reason to believe irreducibility of $f(x^5+1)$ will lead to irreducibility of $f(x^5)$. – achille hui Oct 24 '13 at 19:36

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Let $f(x) = x^{20}+x^{15}+x^{10}+x^5+1 \in \mathbb{Z}[x]$, we have $$\begin{array}{rrcl} & (x^5-1)f(x) &=& x^{25} - 1\\ \implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1\\ \implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1 \pmod 5\\ \iff & ((x^5+1)-1)f(x+1) &=& (x^5+1)^5 - 1 \pmod 5\\ \iff & x^5 f(x+1) &=& (x^{25}+1) - 1 \pmod 5\\ \iff & f(x+1) &=& x^{20} \pmod 5 \end{array}$$ This implies aside from the leading term $x^{20}$, the coefficients of $x^{k}$ in $f(x+1)$ where $0 \le k < 20$ are all divisible by $5$. Notice the constant term in $f(x+1)$ is $f(1) = 5$ which is not divisible by $5^2$. By Eisenstein's criterion, $f(x+1)$ is irreducible over $\mathbb{Q}[x]$. As a corollary, so does $f(x)$.

user26857
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achille hui
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