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I'm trying to show that the following polynomial is irreducible in $\mathbb{Z}[x]$.

$$ f(x)=x^n + p^{n+1} $$

I already tried:

  1. Eisenstein's criterion.
  2. Substitution $f(x+1)$.
  3. Checking if $f(x)$ is irreducible over $\mathbb{Z}/p\mathbb{Z}$.

But I'm stuck.

Martin Brandenburg
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Márcio
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  • Notice that "irreducible over R" means "irreducible in R[T]" for some variable in the context. This is why I changed it to "irreducible in Z[x]" in your post. In the title it was already correct. – Martin Brandenburg Nov 27 '24 at 06:49

3 Answers3

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A monic polynomial in $\mathbf Z[x]$ is irreducible in $\mathbf Z[x]$ if and only if it is irreducible in $\mathbf Q[x]$, so we just need to prove the polynomial is irreducible in $\mathbf Q[x]$. Here we can make the invertible change of variables $x \mapsto px$ which changes $x^n + p^{n+1}$ to $p^nx^n + p^{n+1} = p^n(x^n + p)$, and this is irreducible in $\mathbf Q[x]$ if and only if $x^n + p$ is irreducible in $\mathbf Q[x]$, and tha can be seen using the Eisenstein criterion.

This is the same idea as in the other answer, just expressed in another way.

Martin Brandenburg
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KCd
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I think maybe it may be solved in this way:

Assume that $x^n+p^{n+1}$ is reducible over $\mathbb{Z}[x]$, then there exists two monic polynomials $g,h \in \mathbb{Z}[x]$ such that $$x^{n}+p^{n+1} = g(x)\cdot h (x)$$

Now let $x = py$ and note that $\deg(g)+\deg(h) = n$, we can cancel out $p^n$ on both sides to get $$ y^n+p = g_1(y)\cdot h_1(y)$$ for some $g_1,h_1 \in \mathbb{Q}[y]$ are monic. Now if $\deg(g_1),\deg(h_1) >1$, this contradicts the irreducibility of $y^n+p$ in $\mathbb{Q}[y]$ by Eisenstein criterion. So exactly one of $g_1,h_1$ has degree $0$, which means it is a constant. WLOG, let $\deg(g_1)=0$. But $\deg(g) = \deg(g_1)=0$, thus $g(x)$ must be identically an integer. Since the leading coefficient of $x^n+p^{n+1}$ is $1$, the only choice is $g=1$. By definition, the given polynomial must be irreducible.

Martin Brandenburg
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Tri
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Some excellent proofs have been given. I provide a field theoretic approach. Let $\alpha := \sqrt[n]{p}$, and $\varepsilon_n := e^{\pi i / n}$. Since $x^n + p$ is irreducible over $\mathbb{Q}[x]$, we know that $ [\mathbb{Q}(\alpha \varepsilon_n) : \mathbb{Q}] = n $. Notice that $$ \mathbb{Q} \left( \alpha^{n + 1} \varepsilon_n \right) = \mathbb{Q} \left( \alpha^n \alpha \varepsilon_n \right) = \mathbb{Q} \left( p \alpha \varepsilon_n \right) = \mathbb{Q} (\alpha \varepsilon_n), $$ which implies that $ \left[ \mathbb{Q} \left( \alpha^{n + 1} \varepsilon_n \right) : \mathbb{Q} \right] = n $. And since $\alpha^{n + 1} \varepsilon_n$ is a root of $x^n + p^{n + 1}$, we may obtain the conclusion.

Long-Ping Li
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