I'm reading an article about elliptic curves, but since I'm not very experienced on this subject, I ended up getting stuck. The problem starts as:
"Let $K/\mathbb{Q}$ be a number field and $E/K$ an elliptic curve defined over $K$. Let $v\in M_K$ be a finite place of good reduction for $E$, and fix a minimal Weierstrass equation for $E$ at $v$, $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ Then..."
After this there are some equations concerning the local canonical height which the article wants to prove, but the problem I have is not really about this equations. Its all about the following statement made during the proof:
$``$ The integrality of the Weierstrass equation implies EASILY that \begin{equation}\tag{1} v(x^{-1})<0 \iff v(x/y)<0 \quad" \end{equation}
After playing with the Weierstrass equation and it's integrality ($v(a_i)\ge 0$ for all the $i$'s), indeed I was able to conclude
$$v(x)<0 \iff v(y)<0 \quad\text{and in this case }\quad 2v(y)=3v(x)$$ Thus $$v(x)<0 \Rightarrow v(y/x)<0$$ Hence $$v(x/y)\le0 \Rightarrow v(x^{-1})\le0$$
But this was the closest I could get to the statement $(1)$. I have already tried (without success) to work on the Weierstrass equation in a lots of different ways to explicit the $(x/y)$ and somehow manage to relate $v(x/y)$ and $v(x^{-1})$ as expected in $(1)$.
At this point, due to the "easily" on the text and since I'm not very experienced, I'm starting to think it is some kind of standard trick or I'm missing something very obvious. I would appreciate so much any kind of help.
I'm sorry for my English, it is not my native language.
Thanks a lot!!
If $y$ is a unit, then (1) says $v(x^{-1})<0$ iff $v(x) < 0$ which is a contradiction unless $v(x)=0$. So we can look for counterexample by finding a point where $v(y)=0$ but $v(x) \ne 0$.
Consider the curve $y^2=x^3-4$ over $\mathbb{Q}$. Its discriminant is $-2^8 3^3$, hence it is a global minimal Weierstrass equation (no prime in the discriminant occurs with multiplicity $\ge 12$ - see Silverman Arithmetic of Elliptic Curves, Chapter VII, section 1). Take the point $x=5$, $y=11$, and valuation $v$ at the prime 5, which is a prime of good reduction for this curve. Then $v(y)=0$, $v(x)=1$, and we have our counterexample. In fact, $v(x^{-1}) = v(1/5) = -1$ but $v(x/y) = v(5/11) = 1$.