In J. Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, he proposed this question in exercise 4.51(b).
Take $K/\mathbb{Q}_{3}$ a 3-adic field, and let $L/K$ be a wildly ramified extension of degree 3. From here we know that the ramification index is 3, so $v_{L} = 3v_{K}$.
Suppose we have a curve given by $E: y^{2} = x^{3}+a_{6}$ with $v_{K}(a_{6}) = 1$, and $v_{K}(3)=1$, so $K$ is unramified, we wish to show that $E/L$ has reduction type $III^{*}$, in Kodaira's notation. According to the Tate algorithm given in the book, I am able to proceed until Step 6, where we get the polynomial $$P(T) = T^{3}+a_{6,3}$$ which can be written as $(T+a_{6,3}^{1/3})^{3}$ over the field $\overline{k}$, where $k$ is the residue field of $L$ (since it has characteristic 3). Now, according to Step 8, we should make a translation on $x$ to take the triple root of $P(T)$ to $T=0$.
This is where I am stuck, it appears due to $v_{L}(a_{6}) = v_{L}(3) = 3$, the equation we started with was not minimal, so any translation I attempted will take us to Step 11, where we have to reduce the equation. But if I reduce it using $x = x'\pi_{L}^{2}, y=y'\pi_{L}^{3}$, we would be getting $y^{2} = x^{3}+\pi^{-3}_{L}a_{6,3}$, and the coefficient is no longer in the ring of integers.
Can anybody help me with this one? How do I find a proper translation to solve the problem?
Here's some progress. I think what should happen is the following. I use a simple case in which $a_{6} = \pi_{L}^{3}$ to demonstrate. Note that this is not always the case, as $a_{6,3}^{1/3}$ may not be in the ring. $$P(T) = (T+\pi_{L})^{3}$$ will be the new polynomial, and as in Step 8, we make a translation $x\mapsto x+2\pi_{L}$, and $$y^{2} = x^{3}+6\pi_{L}x^{2}+12\pi_{L}^{2}x+9\pi_{L}^{3}$$. Then apply the reduction $x = x'\pi^{2}, y = y'\pi^{3}$, we get $$y^{2} = x^{3}+\frac{6}{\pi_{L}}x^{2}+\frac{12}{\pi_{L}^{2}}x+\frac{9}{\pi_{L}^{3}}$$ and checking valuations shows that the reduction is limited by $a_{4}$, which has valuation 1. So starting from the beginning of the algorithm, I think we are taken to Type III instead of Type III* as indicated in the book. There certainly is something wrong, I would appreciate if anybody could help.