I am following the book *Introduction to Quadratic Forms over Fields* by T. Y. Lam. In section VI.2, the author proves that, over an arbitrary local field $F$, there is a unique quaternion division algebra, namely $D=\left(\frac{\pi,u}{F}\right)$ where $\pi$ is a uniformizer and $u$ is such that $F(\sqrt{u})$ is the unique unramified quadratic extension of $F$. (This has been proved before for nondyadic local fields; the point here is to extend the statement to the dyadic case.) The proof starts as follows:

Step 1:Define a homomorphism $w':E\setminus\{0\}\to\mathbb{Z}$ by $w'(x)=v(\mathrm{N}(x))$, where $\mathrm{N}$ denotes the (anisotropic) norm form of $E$. Let $d$ be the unique positive integer such that $w'(E\setminus\{0\})=d\mathbb{Z}$. Since $w'(\pi)=v(\mathrm{N}(\pi))=v(\pi^2)=2$, $d$ must be either $2$ or $1$. We may "normalize" $w'$ by setting $w(x)=\frac{w'(x)}{d}$ for $x\in E\setminus\{0\}$, and (by convention) $w(0)=\infty$. Let $B=\{x\in E:w(x)\geq 0\}$,which is a subring of $E$(...).

I don't immediately see that the part on bold is true (namely, the fact that $E$ is closed under taking sums). Is there any obvious reason why this is so? (Non-obvious reasons are also welcome.)

Quality, Quantity, Relation, Manner"? Was your comment to the OP post helpful to user50139 (or to anybody at all? -- that was a rhetoric question since the answer is clear, the answer is "NO"). #### Sorry, everybody, for getting distracted by RF. $\endgroup$ – Wlod AA Apr 11 '20 at 22:28