my question refers to a argumentation step of the proof of VII, Prop. 5.5 in Silverman's "The Arithmetic of Elliptic Curves" (page 199):
Setting:
$E/K$ is a elliptic over local field $K$. In the proof there are used following abbreviations:
HERE we have the proposition:
...AND the proof:
My question is (red taged lines) how he concludes that $\lambda$ is integral (therefore $\lambda \in R$)?
Ideas: We can rewrite the equation $256(1-\lambda (1- \lambda))^3 - j \lambda ^2 (1- \lambda)^2 =0$ to
$0 = 256 \lambda^6 +$ polynomial of degree $5 \ge$ with coefficients from $\mathbb{Z} \cup j\mathbb{Z}$. If $char(k) >0$ I can invert $256$ without leaving $\mathbb{Z}$ because $char(k) \neq 2$. That's already clear.
But how to cope with this if $char(k)=0$?
And why I need the information that $j \in R$ in the proof?
If $j\in R$ and $\frac{1}{256}\in R$ (which is true if and only if $\text{char} k\neq 2)$ you have that $\lambda$ solves a polynomial with coefficient in $R$ so is in $R$ because $R$ is normal as you pointed out.
There is no $\text{char} k=0$ case because the residue field is finite and you need $j\in R$ to use normality