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I'm reading Silverman's The Arithmetic of Elliptic Curves, and I have a question concerning his proof of proposition III.3.1a, which states that any elliptic curve $E$ over $K$ is isomorphic to a plane curve given by a Weierstrass equation. Using Riemann-Roch and II.5.8, he notes that there exist $x,y\in K(E)$ such that $\{1,x\}$ is a basis for $\mathcal L(2(O))$ and $\{1,x,y\}$ a basis for $\mathcal L(3(O))$. Since $\mathcal L(6(O))$ has dimension $6$, there exists a linear relation $$ A_1+A_2x+A_3y+A_4x^2+a_5xy+A_6y^2+A_7x^3=0. $$ He notes that the $A_i$ can be chosed in $K$. For this he refers to II.$5.8$, which says that for any $D\in\operatorname{Div}_K(E)$, $\mathcal L(D)$ has a basis consisting of function in $K(E)$. I understand that II.$5.8$ was used to claim that we can find $x,y\in K(E)$ (instead of in $\overline K(E)$), but I don't see why they need to also be $K$-linearly independent (instead of $\overline K$-linearly independent). Could someone explain?

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    $\begingroup$ You have a $K$-vector subspace $V$ spanned by $1,x,y,x^2,xy,y^2,x^3$ inside $\mathcal{L}(6(0))$, and $V \otimes_K \overline K \subset \mathcal{L}(6(0))$ has dimension at most 6, so $V$ has dimension at most 6 over $K$, so there is a linear relation as claimed. Does this address your question? $\endgroup$
    – Watson
    Jun 21, 2021 at 18:31
  • $\begingroup$ @Watson Ah, I hadn't thought of using the tensor product like that, but that's of course how extension of scalars should work - or at least... I still have to proof that $V\otimes_K\overline K$ embeds into $\mathcal L(6(0))$. I'll have to think about that. Thanks for your help! $\endgroup$
    – Sha Vuklia
    Jun 21, 2021 at 19:25

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Let me clarify the situation. You have a $K$-vector subspace $V$ spanned by $S := \{ 1,x,y,x^2,xy,y^2,x^3 \} \subset K(E)$ inside $\mathcal{L}(6(0))$. Denote by $V'$ the $\overline K$-span of $V$, inside $\mathcal{L}(6(0)) \subset \overline K(E)$ as well.

Since $\dim_{\overline K}\big( \mathcal{L}(6(0)) \big) = 6$, $W$ has dimension at most 6 over $\overline K$, so $V$ has dimension at most 6 over $K$, and thus there is a linear relation as claimed.

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  • $\begingroup$ Yes, thanks! My problem was essentially that I wasn't aware of some basic results concerning the extension of scalars of vector fields. For whoever in the future isn't aware of this either, they are stated (and proven) here: blag.nullteilerfrei.de/2013/02/26/… $\endgroup$
    – Sha Vuklia
    Jun 21, 2021 at 19:46

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