Explicitly computing the reduction of some quotient over the $5$-adics where both enumerator and denominator have the same positive valuation

Question

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!

Answer 1

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.