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Let $j$ be an integer.

Does there exist an elliptic curve $E_j$ over $\mathbf{Q}$ with $j$-invariant equal to $j$ whose minimal discriminant we can write down in a practical way?

For example, can we write down an elliptic curve $E_j$ as above for which we know the reduction over $\mathbf{Z}$ and we know a number field $K$ such that $E_j$ obtains semi-stable reduction over $K$.

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For a give $j\in\mathbb Z$, there is not a unique $E$ over $\mathbb Q$ with $j$ as $j$-invariant. So you formulate differently your question. – user18119 Oct 15 '12 at 11:41
I forgot that every twist of an elliptic curve has the same j-invariant. I accordingly changed the formulation of the question. – Harry Oct 15 '12 at 16:46
up vote 1 down vote accepted

If by "practical way" you mean a fast algorithm, I think that's possible. In $char \neq 2,3$ $$j=1728 \frac{c_4^3}{c_4^3-c_6^2}$$ (see ) Therefore given a $j$-invariant we can solve for $c_4$ and $c_6$, and if they can be chosen to be rational, we already have a curve $E$ over $\mathbb Q$ to start with. I assume for convenience that $j \neq 0,1728$.

Given $E$, we can use the process Stein and Watkins applied to construct their table, see

The relevant part is

For each $(c_4 , c_6 )$ pair (again with $c_4 \le 1.44 \cdot 10^{12}$ ) which satisfies the $|\Delta| \le 10^{12}$ condition, we then determine whether this curve is minimal —not only in the traditional sense for its minimal discriminant, but also whether it is has the minimal discriminant in its family of quadratic twists. For $p \ge 5$, this is rather easy to determine; unless $p^6 | \Delta$ and $p | c_4$, the curve is minimal for quadratic twists (the only difference between this and the standard notion of minimality is that the exponent here is 6 instead of 12). If both the above conditions hold, then we throw the curve out, as $E_\tilde{p}$ where $\tilde{p} = \left( \frac{-1}{p} \right)p$, is a curve with lesser discriminant.

Here $E_\tilde{p}$ refers to the quadratic twist by $\tilde{p}$. It appears to me you could twist by $\tilde{p}$ and repeat this process until you arrive at a twist of minimal conductor (apart from $2,3$), then you can find its minimal discriminant.

For $2$ and $3$ there are possibly more complications, but you can just do trial and error among the finitely many possible twists.

p.s. When $E$ has nontrivial $p$-torsion over a number field $K$, reduction at all places of $K$ not lying over $p$ must be semistable. (e.g. see the paper "Abelian varieties having purely additive reduction") You can construct many fields this way where the curve becomes semistable.

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