# Subring of quaternion algebra

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.)

• "I don't immediately see" -- as long as you saw it you can congratulate yourself. (The "immediately" notion belongs to trivia). – Wlod AA Apr 11 '20 at 19:07
• @WlodAA Please read and understand something like this. – Robert Furber Apr 11 '20 at 19:48
• @RobertFurber, what does your link have to do with @‍WlodAA's comment, which seemed to be purely to the effect that it's OK if the amount of work required to understand a claim is not proportional to the length of the claim? – LSpice Apr 11 '20 at 21:58
• @RobertFurber, I don't see how you have contributed to "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. – Wlod AA Apr 11 '20 at 22:28

That's because $$w$$ is a valuation (the main point being the nonarchimedean triangle inequality), which can be proved by reducing to the commutative case; see for instance Lemma 13.3.2 in https://math.dartmouth.edu/~jvoight/quat.html.
• $$w$$ is a valuation on every commutative subfield of $$E$$.
• $$w$$ is multiplicative by definition.
• Let $$a,b\in E$$ with $$b\neq 0$$. Then $$w(a+b) = w((ab^{-1}+1)b) = w(ab^{-1}+1)+w(b)$$ by multiplicativity. Then $$w(ab^{-1}+1) \ge \min(w(ab^{-1}),0)$$ since $$w$$ is a valuation on $$F(ab^{-1})$$, and finally $$\min(w(ab^{-1}),0)+w(b) = \min(w(a),w(b))$$ again by multiplicativity. Put together we get $$w(a+b)\ge\min(w(a),w(b))$$ as desired.