Proposition III.3.1 in Silverman's The Arithmetic of Elliptic Curves

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?

• 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? Jun 21, 2021 at 18:31
• @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! Jun 21, 2021 at 19:25

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.