An Irreducible polynomial in $\mathbb{Q}[x]$

Question

I want to prove that $x^4+x^3+x^2+x+1$ is irreducible in $\mathbb{Q}[x]$. I noticed that this polynomial can be rewritten as $x(x+1)(x^2+1)+1$. As seen, it has no proper divisor because of the constant term $1$. However, I need a rigorous proof. Can anybody help me?

Thanks in advance.

score 2 · Answer 1 · answered Oct 26 '19 at 01:48

2

This is the cyclotomic polynomial for $\mathbb{Q}(\zeta_5)$.

Let me prove a more general case:

Proposition: If $p$ is a prime number, then the polynomial $f(x) = x^{p - 1} + x^{p - 2} + \cdots + x + 1$ is irreducible over $\mathbb{Q}$.

Proof: It suffices to show that $g(x) = f(x + 1)$ is irreducible.

Since $f(x) = \frac{x^p - 1}{x - 1}$, we have $$g(x) = \frac{(x + 1)^p - 1}{x} = x^{p- 1} + \binom{p}{p - 1}x^{p-2} + \cdots + \binom{p}{2}x + \binom{p}{1}.$$ Notice that

all the coefficients of $g$ are divisible by $p$, except the leading coefficient $1$;
the constant term is $p$, which is not divisible by $p^2$.

It then follows from Eisenstein's criterion that $g$ is irreducible over $\mathbb{Q}$.

answered Oct 26 '19 at 01:48

WhatsUp

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Nice answer, but I do believe this is a dup – Rushabh Mehta Oct 26 '19 at 01:50
Why this statement is true: "It suffices to show that g(x)=f(x+1) is irreducible" ? – Hussein Eid Oct 26 '19 at 01:50
@DonThousand Ah... all right ^_^ – WhatsUp Oct 26 '19 at 01:50
@HusseinEid Check the duplicate answer please! – Rushabh Mehta Oct 26 '19 at 01:51
@DonThousand Reading through the duplicate answer, I think it might worth leaving the answer here, since the accepted answer there uses much deeper mathematics, while this proof of mine is more elementary and accessible to e.g. students of lower levels. – WhatsUp Oct 26 '19 at 02:00
@HusseinEid Because if $f(x) = a(x)b(x)$ is reducible then it follows that $g(x) = f(x + 1) = a(x + 1)b(x + 1)$ is also reducible. – WhatsUp Oct 26 '19 at 02:02

An Irreducible polynomial in $\mathbb{Q}[x]$

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