Irreducibility of $g(y)=y^n+qy+q^2$ in $\mathbb{Q}[y]$ I am trying to prove that $g(y)=y^n+qy+q^2$ is irreducible in $\mathbb{Q}[y]$, (with $q$ a prime number and $n\in \mathbb{N}$).
I know I cannot use Eisenstein's criterion, since $q^2$ divides $q^2$.
I have tried to work with the equality $g(y)=g_1(y)g_2(y)$ in $\mathbb{Z_p[y]}$, but I could not conclude.
A nearly identical question is posed here: Is $x^n+px+p^2$ irreducible in $\mathbb{Z}[x]$?, but I do not fully understand the answer given.
Any hints as to how to approach this problem would be really helpful. Thanks in advance.
 A: I will write $p$ for the prime involved, so
the polynomial $g$ is $g=y^n+py+p^2$.
(Usually, $q$ stays for a power of some prime. Well, i also use it as a prime - but only in case $p$ is already occupied with some other job around.)
$g$ is a monic polynomial with integer coefficients. By Gauss' Lemma it is irreducible in the polynomial ring $\Bbb Q[y]$ if and only if it is irreducible in the polynomial ring $\Bbb Z[y]$.
Let us show the irreducibility over $\Bbb Z[y]$. (We will use the same idea of reduction modulo $p$ in the proof.)

Assume that $g=uv$ is a product of two monic polynomials $u,v\in\Bbb Z[y]$
of degree $\ge 1$. We consider all involved coefficients modulo $p$, thus passing from $\Bbb Z[y]$ to $\Bbb F_p[y]$. (Here $\Bbb F_p=\Bbb Z/p$ is the field with $p$ elements, its elements are integers taken modulo $p$.) We denote the passage by $g\to \bar g$. So we obtain a decomposition in $\Bbb F_p[y]$
$$
y^n=\bar g =\overline{u\cdot v}=\bar u\cdot \bar v\ .
$$
Such a decomposition is only of the shape $y^n=y^s\cdot y^t$, with $s+t=n$, $s,t\ge 1$. We lift it, so $u,v$ have the shapes
$$
\begin{aligned}
u &= y^s + pU\ , &\deg U &< s\ ,\\
v &= y^t + pV\ , &\deg V &< t\ ,\\[3mm]
&\qquad\text{ giving}\\[3mm]
g 
&= uv\\
&=(y^s+pU)(y^t+pV)\\
&=y^{s+t} + p(y^tU + y^sV) + p^2UV\ .
\end{aligned}
$$
Working modulo $y$, we pick the free coefficient, so $p^2$ is the free coefficient of $g$, thus of $p^2UV$, so the free coefficients in $U,V$ are among $\pm 1$. How can we obtain a coefficient $p$ modulo $p^2$ in degree one in $g$? This is only possible when either $s=1$ or $t=1$. (Recall $g=y^{s+t} + p(y^tU + y^sV) + p^2UV$.)
So $g$ has a linear factor, say it is $u$, this may be only for $U=\pm 1$, so $u=y\pm p$, so $\mp p$ is a root of $g$, there is no positive root of $g$, so the root is $-p$, so $(-p)^n +p(-p)+p^2=0$, which never happens.
So $g$ is irreducible over $\Bbb Z[y]$, thus also over $\Bbb Q[y]$.
