Computing the reduction of a quotient over the $5$-adic numbers

Question

Let $\alpha_1,\alpha_2,\alpha_3$ be the roots of $f = X^3 - 135 X - 270 \in \mathbb{Q}_5[X]$ over its splitting field $L$.

Because $$g = f/(X-\alpha_1) = X^2 + \alpha_1 X + \alpha_1^2 - 135$$ and $$g = (X-\alpha_2) (X - \alpha_3) = X^2 - (\alpha_2 + \alpha_3) X + \alpha_2 \alpha_3,$$ we obtain the relations $$\alpha_2 \alpha_3 = \alpha_1^2 -135 \quad \text{and} \quad \alpha_2 + \alpha_3 = - \alpha_1.$$

Now let $\lambda = \frac{\alpha_3 - \alpha_1}{\alpha_2 - \alpha_1} \in L$.

Question What is the reduction $\bar{\lambda}$ of $\lambda$?

Why do I need this?

I have an elliptic curve

$$ Y^2 = X (X-1)(X-\bar{\lambda}) $$ over $\mathbb{F}_5$ (the residue field of $L$) and need to count points over $\mathbb{F}_5$. Therefore, I need to know what exactly $\bar{\lambda}$ is.

What did I do to approach my problem?

By computing the Newton polygons of $F$ and $h := g(X+\alpha_1) = (X - (\alpha_2 - \alpha_1))(X - (\alpha_3 - \alpha_1))$, I was able to compute the valuations of their roots $\alpha_1,\alpha_2,\alpha_3$ (roots of $f$) and $n := \alpha_2 - \alpha_1, z := \alpha_3 - \alpha_1$ (roots of $h$) which are all $1/3$ (assuming $v(5) = 1$). This implies that the valuation of $\lambda$ is $0$, so $\bar{\lambda} \neq 0$.

Another approach was using $\alpha_2 = \alpha_1 - \alpha_3$ (follows from the observation above) which gives $$ \lambda = \frac{\alpha_3-\alpha_1}{-\alpha_3} = -1 + \frac{\alpha_1}{\alpha_3}. $$ Here, we also have a fraction where both enumerator and denominator have valuations $1/3$, respectively, but maybe this is better to work with.

Also, similarly as for $g$, we obtain the relations

$$ z + n = -3 \alpha_1 \quad \text{and} \quad zn = 3\alpha_1^2 - 135. $$

But I do not know how to make the transition to $\lambda = z/n$.

Also, I tried to compute the roots of $h$ in terms of $\alpha_1$ with the formula to compute roots of quadratic polynomials. But there, I cannot get rid of denominators which all seems to have valuations greater than $0$.

Thank you!

Answer 1

The main difficulty with your current approach is that the non-leading coefficients of $f(x)/(x-\alpha_1)$ all have positive valuation, and so you don't get a whole lot of information by reducing it modulo the maximal ideal.

Let $\alpha = \alpha_1$. Since $f(x)$ is eisenstein, $\alpha$ is a uniformizer and the ring of integers of $K$ is $\mathbb Z_5 [\alpha]$. As you observed, it will be enough to determine the ratios $\alpha_2/\alpha_1$ and $\alpha_3/\alpha_1$. These are all roots of the following polynomial in $\mathcal O_K[x]$: $$g(x) = \frac 1 {\alpha^3} f(\alpha x) = x^3 - \frac {135}{\alpha^2} x - \frac{270}{\alpha^3}$$

Let $w$ be the valuation on $K$ normalized so that $w(\alpha) = 1$, and so $w(5) = 3$. Then clearly $w(135/\alpha^2) = 1$, while $w(270/\alpha^3) = 0$. Let's try to express the latter in terms of $\alpha$. We know that $\alpha^3 = 270(\frac 1 2 \alpha + 1)$, and so $$\frac{270}{\alpha^3} = \frac{270}{270(\frac 12 \alpha + 1)} = \frac 1 {1 - (-\frac 1 2 \alpha)} = 1 - \frac 1 2 \alpha + \frac 1 4 \alpha^2 + ...$$

Thus, modulo $\alpha$ (a uniformizer), we see that $$g(x) = x^3 - 1 = (x-1)(x^2 + x + 1)\mod (\alpha)$$ and hence its roots are the 3rd roots of unity. Since it's separable, the factorization lifts by Hensel's lemma. The second polynomial is irreducible in $\mathbb F_5$, hence also in $K$ (the extension is totally ramified) so we will want to jump up to a splitting field $L$.

Now everything works out: $1 = \alpha_1/\alpha_1$ is obviously a root of $g$, the other two roots come from $\alpha_2/\alpha_1$ and $\alpha_3/\alpha_1$ and reduce to primitive third roots of unity modulo the uniformizer of $L$. This means $$\bar\lambda = \frac{\zeta - 1}{\zeta^2 - 1}$$ modulo the uniformizer of $L$. This is only well-defined up to the automorphism exchanging $\zeta$ and $\zeta^2$, but hopefully that won't cause you too many issues.

It is not too surprising that things turned out this way. You may have seen that every tamely ramified extension can be obtained by adjoining an $n$th root of unity and an $n$th root of something in the ground field, and if you look carefully through a proof of that (the only I have in mind is by Krasner's lemma) you'll see that it's essentially implicit in the above.