# Elliptic curves over Spec Z

I want to show that there are only finitely many elliptic curves over Spec $\mathbf Z$ without appealing to Siegel's theorem or Shafarevich' theorem.

Firstly, I think (but I am not sure) that such an elliptic curve has a Weierstrass equation $y^2=x^3+Ax+B$ with $A$ and $B$ in $\mathbf Z$. Is this true? (I think there are some problems at the primes $2$ and $3$.)

Then, by the fact that this is an elliptic curve over $\mathbf Z$, we have that the discriminant $-16(4A^3+27B^2)$ is an element of $\mathbf Z^\times = \{\pm 1\}$. Is this true?

But the absolute value of the discriminant is at least $16$, so it is never $1$. QED

Is this a correct proof?

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How do you get the Weierstrass equation? – Martin Brandenburg Jan 29 '13 at 23:07
I wanted to be sure myself, but let $E$ be an elliptic curve over $\mathbf Q$ with good reduction over $\mathbf Z$. Then its minimal regular model coincides with its minimal Weierstrass model. – Harry Jan 29 '13 at 23:53

Yes your equation is not the good one at $p=2, 3$. An elliptic curve over $\mathbb Z$ has an equation $$y^2+(a_1x+a_3)y=x^3+a_2x^2+a_4x+a_6, \quad a_i\in \mathbb Z$$ with discriminant equal to $\pm 1$. The discriminant of this equation is $$\Delta=2^{-8}\mathrm{disc}(4(x^3+a_2x^2+a_4x+a_6)+(a_1x+a_3)^2).$$ You can find an explicit formula of the discriminant $\Delta$ in terms of the $a_i$ in Silverman's book. I am told that Tate proved there is no elliptic curve of $\mathbb Z$ by direct computations.