Consider the field $L = \mathbb{Q}_3(\alpha)$ where the minimal polynomial of $\alpha$ over $\mathbb{Q}_3$ is $f = x^4 - 3x^2 + 18$. Over $L$, the polynomial $f$ splits into $$ f = (x-\alpha)(x+\alpha)(x-\alpha')(x+\alpha') $$ where $\alpha' = \frac{(2\alpha^2-3)u}{\alpha}$ and $u = \sqrt{\frac{2}{-7}}$ (cf. this post, credits go to user KCd).
Question What is $\alpha/\alpha'$ modulo $\pi_L$ (a uniformizer in $L$)?
I know that result is only unique up to sign but I hope that does not cause any further difficulties.
I computed via Newton polygons that the valuations of $\alpha$ and $\alpha'$ are both $1/2$ (here, I assume $v(3) = 1$), so the reduction of $\alpha/\alpha'$ is non-zero.
I tried to apply a technique similar to an answer from this post:
I considered the polynomial $$ \frac{f(\alpha x)}{\alpha^4} = x^4 - \frac{3}{\alpha^2} x^2 + \frac{18}{\alpha^4} $$ whose roots are $\pm 1$ and $\pm \frac{\alpha'}{\alpha}$.
This means that the polynomial $$ \frac{x^4 - \frac{3}{\alpha^2} x^2 + \frac{18}{\alpha^4}}{x^2-1} = x^2 + \left(1-\frac{3}{\alpha^2}\right) $$ has the roots $\pm \frac{\alpha'}{\alpha}$.
Now I would like to compute the reduction of this polynomial. However, I am not sure that the reduction of $3/\alpha^2$ must be in this case. This seems to be a similar problem but it seems to be as easy an in the last mentioned post since we only have the second power of $\alpha$ (instead of the fourth power of $\alpha$) here.
Let $\beta=\frac{\alpha^2}{3}$. It follows that $\beta^2-\beta+2=0$, hence $\beta=\frac{1 \pm \sqrt{-7}}{2}$.
It follows that $(\frac{\alpha}{\alpha’})^2=\frac{-3+\sqrt{-7}}{4}$, if we choose $\beta=1+\sqrt{-7}$.
Now, note that in $\mathbb{F}_9$, $-7$ is actually a fourth power, as $(-7)^{8/4}=49=1$.
Thus the reduction of $\frac{\alpha}{\alpha’}$ in the residue field is $\frac{(-7)^{1/4}}{2}=-(-7)^{1/4}$.
If we want a more implicit description, we can check that by changing signs and considering the inverse ratio, we find that $\alpha/\alpha’$ can be any generator of $\mathbb{F}_9^{\times}$.