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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.)

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  • $\begingroup$ "I don't immediately see" -- as long as you saw it you can congratulate yourself. (The "immediately" notion belongs to trivia). $\endgroup$ – Wlod AA Apr 11 '20 at 19:07
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    $\begingroup$ @WlodAA Please read and understand something like this. $\endgroup$ – Robert Furber Apr 11 '20 at 19:48
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    $\begingroup$ @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? $\endgroup$ – LSpice Apr 11 '20 at 21:58
  • $\begingroup$ @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. $\endgroup$ – Wlod AA Apr 11 '20 at 22:28
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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.

For the sake of completeness I will give the argument here:

  • $w$ 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.
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