Non-equivalent norms on $\mathbf{Q}[\sqrt2]$

Question

I found the following post on MO, Can somebody help me filling the details?

Just to be sure, here is the restated problem

We work on $\mathbf{Q}[\sqrt2]=\left\{x\in\mathbf{R}\mid\: \exists (a,b)\in\mathbf{Q}^2,\text{ st. } x=a+b\sqrt2\right\}$, and we wish to prove that the two norms $N_1,N_2$ defined by $N_1(a+b\sqrt2)=|a|+|b|$ and $N_2(a+b\sqrt2)=|a+b\sqrt2|$ are not equivalent.

I have the following intuition :

I believe that using the sequence $u_n={(1-\sqrt 2)}^n$ might work as it seems to me that $u_n\to_{n\to\infty \\N1} +\infty $ and $u_n\to_{n\to +\infty \\N2} 0$

So that these norms are not equivalent.

But I'm unable to prove these claims, can someone help me with the calculations required ?

Answer 1

Your intuition is correct.

$N_2(u_n)$ goes to $0$ since $|1-\sqrt 2| < 1$

Claim: if $|x|<1$, then $|x^n|\to_{n\to\infty} 0$.
Proof: Obvious.

Or, let $|x|=1-\epsilon$, where $0\lt\epsilon\le1$. Then $$|x^n|=(1-\epsilon)^n\lt(\frac1{1+\epsilon})^n=\frac1{(1+\epsilon)^n}\le\frac1{1+n\epsilon}$$ where $1+n\epsilon\to_{n\to\infty}+\infty.$ $\checkmark$

$N_1(u_n)$ goes to $\infty$

Let $u_n=a_n-b_n\sqrt2$, where $a_n, b_n\in\mathrm Q$.

Claim: $a_n\ge n$ and $b_n\ge 1$.
Proof: $a_1=b_1=1\ge1$.
Suppose $a_n\ge n$ and $b_n\ge n$ is true. Then $u_{n+1}=u_n(1-\sqrt2)=(a_n+2b_n)-(a_n+b_n)\sqrt2$. Hence $$a_{n+1}=a_n+2b_n\ge n+2\ge n+1.$$ $$b_{n+1}=a_n+b_n\ge1.$$ Induction is complete. $\quad\checkmark$

Hence $N_1(u_n)=|a_n| + |-b_n|\ge n+1\to_{n\to\infty}+\infty.$

Another proof for $N_1(u_n)$ goes to $\infty$

For all $x=a+b\sqrt2$, $N_1(x)=|a|+|b|\ge\frac{|a|+|b\sqrt2|}{\sqrt2}\ge\frac{|x|}{\sqrt2}$.

Let $a+b\sqrt2\,\mapsto\, \overline{a+b\sqrt2}=a-b\sqrt2$ be the conjugation in $\mathbf{Q}[\sqrt2]$. Verify $N_1(x)=N_1(\overline x)$.

$$N_1(u_n)=N_1\left(\overline{(1-\sqrt2)^n}\right)=N_1\left(\overline{1-\sqrt2}^n\right)=N_1((1+\sqrt2)^n)\ge\frac{(1+\sqrt2)^n}{\sqrt2}\to_{n\to\infty}+\infty$$ where the second equality holds because conjugation commutes with multiplication.

Exercise :

1. Show that $N_1(x^n)\to_{n\to\infty}0$ or $+\infty$ unless $x=\pm1$, in which case $N_1(x^n)=1$ for all $n$.

2. Show that $N_2(x^n)\to_{n\to\infty}0$ or $+\infty$ unless $x=\pm1$, in which case $N_2(x^n)=1$ for all $n$.