Let $K = \mathbb{Q}_5$ and $f = X^3 - 135X - 270 \in K[X]$.
Let $\alpha_1,\alpha_2,\alpha_3$ be the roots of $f$ over its splitting field $L$.
Consider the $M = \mathbb{Q}_5(\alpha_1,\sqrt{C})$ of $K$ where $C = (\alpha_2 - \alpha_1)^3$.
Question: What is $\lambda := \frac{270}{C} \mod \pi_M$? (where $\pi_M$ is a uniformizer in $M$)?
What I tried:
I tried to apply the technique from one of the answer of this previous of mine. However, this technique seems to require an explicit uniformizer in that post.
Since the ramification index of $M/K$ is $6$, I must find some element with valuation $1/6$ (note: I let my valuation satisfy $v(5) = 1$). And here, I am lost already.
Furthermore, I know that the residue field of $M/K$ is $\mathbb{F}_5$.
More valuations which may be helpful: It is $v(\alpha_1) = v(\alpha_2) = v(\alpha_2 - \alpha_1) = 1/3$ and $v(\sqrt{C}) = 1/2$. Hence $v(C) = 1 = v(270)$, so $\lambda \mod \pi_M$ is non-zero in $\mathbb{F}_5$.
Could you please help me with this problem? I am grateful if you could help me with the uniformizer or help me to find another approach.
Thank you!
Sorry it’s taken me so long, but I was too tired last night. I’ll use $(\alpha,\beta)$ instead of $(\alpha_1,\alpha_2)$.
First, as you probably saw, the extension $K(\alpha,\beta)=L\supset K$ has $e=3$, $f=2$. (In case you didn’t see this, we have $f(X)=(X-\alpha)\bigl(X^2+\alpha X+(\alpha^2-135)\bigr)$. If you divide the roots of the quadratic factor by $\alpha$, you get a polynomial $\equiv X^2+X+1\pmod{(\alpha)}$. Thus you need to adjoin the cube roots of unity to $\Bbb F_5$ to get the right residue-class field.)
As you recognize, $v_5(\beta-\alpha)=\frac13$, so that its cube has valuation $1$; square root of that requires a valuation of $1/2$, so your uniformizer (element of valuation $\frac16$) is clear.
Finally, to get a good handle on your $\lambda$, I decided to work computationally over $\Bbb Q_5=K$. I defined $K(\alpha)$ as you did, but I worked with $L$ defined as $K(\alpha,\omega)$ where $\omega^2+\omega+1=0$, and found that we could take $\beta$ to be $$ \beta=\dots043;\times5 + \dots014;\times5\alpha + \dots324;\alpha^2 + (\dots141;\times5 + \dots331;\alpha + \dots203;\alpha^2)\omega\,, $$ where I hope you’re comfortable with $5$-ary expansion: “$\dots203;$” means $3\times5^0+0\times5^1+2\times5^2$, each notation like this to be read modulo $5^3$. Notice that, modulo $\alpha^2$, the above is $\equiv\alpha\omega$.
And now, my computation package rendered $\lambda=\frac{270}{(\beta-\alpha)^3}$ as $$ \lambda=\dots211; + \dots323;\alpha + \dots314;\times5\alpha^2 + (\dots012; + \dots442;\alpha +\dots222;\alpha^2)\omega\,, $$ in other words $\equiv1-\omega\pmod{(\alpha)}$, when we remember that $\alpha$ is still the uniformizer in $L$.
Last of all, notice that your $\lambda$ already is in $L$, so we don’t need to go to $M$ to describe it.
Addendum:
This is a significant simplification of my first answer. There, I claimed that $K\bigl(\alpha,(\beta-\alpha)^{3/2}\bigr)$ is the same as $K\bigl(\alpha,(\beta-\alpha)^{1/2}\bigr)$, an essentially insupportable claim (and likely untrue). This did not affect the applicability of my answer to your question, and it’s now gone.