Prove $x^n+x^{n-1}+2$ is irreducible in $\mathbb{Q}[x]$ for $n\geq 2$

Question

I want to prove $f(x)=x^n+x^{n-1}+2$ is irreducible in $\mathbb{Q}[x]$ for $n\geq 2$. I think I'm very close to doing this, and I have used the following theorem:

Generalized Eisenstein criterion: Let $f(x)\in\mathbb{Z}[x]$ be primitive and let $p$ be a prime number. Write $f(x)=f_nx^n+ ... + > f_mx^m + ... + f_1x + f_0$ with $m\leq n$. Suppose that $p$ does not divide $f_nf_m$, and $p|f_i$ for all $i<m$, and $p^2$ does not divide $f_0=f(0)$. Then $f(x)$ has an irreducible factor in $\mathbb{Q}[x]$ of degree $\geq m$.

This is what I got so far:

Let $f(x)=x^n+x^{n-1}+2$. We see that $f(x)\in\mathbb{Z}[x]$ and that it is primitive ($\gcd(1,1,2)=1$, and $1$ a unit in $\mathbb{Z}$). Let $m=n-1$ and let $p=2$ a prime number. Then $f_nf_m=1$ and $2$ does not divide $1$. $2$ does divide all coefficients $f_i$ with $i<m$, since only $f_0\neq 0$, and $2|0$, and $2|2$ and $f_0=2$. We also see that $p^2=2^2=4$ does not divide $f_0$. By theorem 2 we have that $f(x)$ has an irreducible factor $g(x)$ in $\mathbb{Q}[x]$ of degree $\geq > m$. So $\deg(g(x))$ is either $n-1$ or $n$. Suppose $f(x)$ is reducible in $\mathbb{Q}[x]$ and can be written as $f(x)=g(x)h(x)$, with $h(x)\in\mathbb{Q}[x]$.

If $\deg(g(x))=n$, then $h(x)$ is a constant term and in $\mathbb{Q}$. Every non-zero element of $\mathbb{Q}$ is a unit, and $h(x)$ cannot be zero, because then $f(x)\equiv 0$ which is not the case. So if $\deg(g(x))=n$, we have that $f(x)$ irreducible.

If $\deg(g(x))=n-1$, then $h(x)$ is a linear term, not constant, and thus there is an element $a\in\mathbb{Q}$ such that $h(a)\equiv 0$.

Assuming this is all correct, how do I show that either $h(x)$ with degree $1$ is a unit (which I think cannot happen in $\mathbb{Q}[x]$), or $h(x)$ cannot exist for some reason, (maybe because $f(x)$ cannot have roots)?

And a question about my proof so far: $h(x)$ is linear, so it has a root (zero point). In what number field does this root lie? As you can see in my prove I thought it would be $\mathbb{Q}$ since we are talking about $h(x)\in\mathbb{Q}[x]$

This is also a special case of Is $x^j+x^k+2$ irreducible whenever $j+k$ is odd? — Sil, Jun 12 '21 at 10:44

score 2 · Accepted Answer · answered Mar 14 '19 at 13:45

2

It is indeed true that $f$ has no roots in ${\mathbb Q}$. If $\frac{p}{q}$ (with $p,q$ coprime integers and $q>0$) were a root of $f$, then $p^n+p^{n-1}q+2q^n=0$, so that $q$ divides $p^n$. By Gauss' lemma, $q$ divides $1$, so $q=1$. Then $p^n+p^{n-1}=(-2)$, so $p$ divides $2$, and is therefore one of $1,2,-1,-2$. It is easy to see that none of those is a solution.

answered Mar 14 '19 at 13:45

Ewan Delanoy

61,600

What do you mean by "so that $q$ divides $p^n$? You got the equation before that by filling in the fraction and multiplying everything by $q^n$ right? – The Coding Wombat Mar 14 '19 at 13:59
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I mean that, from $p^n+p^{n-1}q+2q^n=0$ you deduce $p^n=q(-p^{n-1}-2q^{n-1})$, so $p^n$ is $q$ times something, so $q$ divides $p^n$ – Ewan Delanoy Mar 14 '19 at 16:53
How does Gauss's lemma give us that $q \mid 1$? – Mar 04 '20 at 12:55
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@WesleyStrik We know that $q$ dvides $p^n \times 1$, but $q$ is coprime to $p^n$, so $q$ divides $1$ – Ewan Delanoy Mar 04 '20 at 13:35
That makes more sense, just a prime divisibility argument. – Mar 04 '20 at 13:38

score 1 · Answer 2 · answered Mar 14 '19 at 13:28

1

Hint: For $n>2$ apply Eisenstein to $f(x+2)$ and study $n=1,2$.

answered Mar 14 '19 at 13:28

Tsemo Aristide

87,475

1

Does this continue on the partial proof I've given? – The Coding Wombat Mar 14 '19 at 13:36
And I forgot to add a criterion: $n\geq 2$. – The Coding Wombat Mar 14 '19 at 13:37

Prove $x^n+x^{n-1}+2$ is irreducible in $\mathbb{Q}[x]$ for $n\geq 2$

2 Answers2