I recently came across the following claim in a paper concerning polyonomials in $\mathcal{R}_p = \mathbb{Z}_p[X]/\langle X^d + 1\rangle$ (where $\mathbb{Z}_p = \mathbb{Z}/\mathbb{Z}p$) and their automorphism $\sigma_i: X \mapsto X^i$.
$$\sigma_i \langle X - \alpha \rangle = \langle X^i - \alpha \rangle = \langle X - \alpha^{i^{-1}} \rangle$$
where the latter (according to the authors) follows as the generators have the same roots in (in some extension of the base field $\mathbb{Z}_p$), where $X - \alpha$ is irreducible modulo $p$ for prime $p$.
I can see that they would have a same root in an appropriate extension, but I don't understand how that leads to the claim that they are equal in the base field itself.
It's been a while since I took algebra but what I remember is that two principal ideals of a ring $\langle a \rangle$ and $\langle b \rangle$ are equal if there is a unit $u$ in the ring such that $a = bu$. Is it simple enough to find such a unit here? How do we get the claim (ie., generators have same roots therefore principal ideals are equal) from this?
The precise description of the ring $\mathcal{R}_p$ added to the question explains what is going on. I lead off with a very simple toy example to get the ball rolling.
I consider the case of $p=13$, $d=16$ (the parameter $d$ is always a power of two). As further prescribed in section 2.2. of the linked article we can then choose $\ell=2$ as $$ p-1=12\equiv 2\ell\pmod{4\ell}, $$ or, equivalently, $4=2^2=2\ell$ is the highest power of two that divides $p-1$.
Modulo $p=13$ we have $$(\pm 5)^2=25\equiv-1\pmod{13},$$ so the fourth roots of unity modulo $p$ are $\zeta=5$ and $\zeta^3=-\zeta=8$. Consequently the polynomial $X^{16}+1$ factors as $$ X^{16}+1=(X^8-5)(X^8-8)\pmod{13}, $$ which you can also easily verify by hand.
As a part of the toy example let's consider the (ring) automorphism $\sigma_3$ of $\mathcal{R}_{13}$ defined by $X\mapsto X^3$. If I understood correcly, the question is then about the equality of the principal ideals (line 12 of section 2.2. in the source) $$\sigma_3(\langle X^8-5\rangle)=\langle X^{24}-5\rangle$$ and $$\langle X^8-5^{1/3}\rangle.$$ Here $8^3=5$ (compare this with the usual calculation with fourth roots of unity: $i^3=-i$), so we can use $5^{1/3}=8$ ($\dagger$). Furthermore, because the arithmetic of polynomials of $X$ is done modulo $X^d+1=X^{16}+1$, we have $X^{16}=-1$ in the quotient ring $\mathcal{R}_{13}$. Therefore $$ X^{24}-5=X^{16}\cdot X^8-5=-X^8-5=-(X^8-8). $$ In other words, the two generators $X^8-8$ and $X^{24}-5$ are gotten from each other by multiplication with $-1$.
Adding another example with more maximal ideals. Need a higher power of two dividing $p-1$, so let's use $p=41$ and $d=32$, when $$ p-1\equiv 2\cdot4\pmod{4\cdot4}. $$ This time $\ell=4$, so we get four maximal ideals generated by the factors $$x^{32}+1=(x^8-3)(x^8-14)(x^8+3)(x^8+14).$$ Here $3^4=81\equiv-1\pmod{41}$ and $3^8\equiv1\pmod{41}$. Hence $3$ is of order eight, and so are $3^3=27\equiv-14$ and their negatives.
Let's again study the effect of $\sigma_3$ on the maximal ideals (an automorphism will always map a maximal ideal to another, and the question is just which). Because $3\cdot3\equiv1\pmod 8$ we have $\zeta^9\equiv\zeta$ for every primitive eighth root $\zeta\in\Bbb{Z}_{41}$. This means that we can use $$\zeta^{1/3}=\zeta^3=\frac{\zeta^4}{\zeta}=-\frac1{\zeta}.$$ This time $X^{32}=-1$, and therefore $X^{24}=X^{-8}\cdot X^{32}=-X^{-8}$. Consequently $$ \begin{aligned} \sigma_3(X^8-\zeta)&=X^{24}-\zeta\\ &=-X^{-8}-\zeta\\ &=-\zeta X^{-8}(\frac1\zeta+X^8). \end{aligned} $$ As $-\zeta X^{-8}$ is a unit of the ring $\mathcal{R}_{41}$, this means that $$ \langle\sigma_3(X^8-\zeta)\rangle=\langle X^8-(-1/\zeta)\rangle =\langle X^8-\zeta^{1/3}\rangle $$ exactly as the authors claimed!
I'm aware that I haven't yet answered the title question about zeros of polynomials in an extension field. Nevertheless, I hope that that the examples lift some of the fog. A particular fact that may become as a surprise to the unwary is that $X$ and its powers are units in this ring.
($\dagger$) The element $5$ has no less than three cube roots in $\Bbb{Z}_{13}$. I use $8$ because it is the only one that is another fourth root of unity (the other two both have order twelve). I think that is the intention of the authors also.