The number of abolute value on $\mathbb{Q}(\sqrt{2})$ Let $|.|$ be the usual absolute value on $\mathbb{Q}$.
The number of absolute value on $Q(\sqrt{2})$ extending |.| is 2 
since $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ in $\mathbb{R}[x]$. 
Let $\hat{Q(\sqrt{2})}_i , i=1,2$, be the corresponding completion of $Q(\sqrt{2})$ with extended absolute value $|.|_i$. 
$(x-\sqrt{2})$ and $(x+\sqrt{2})$ are of degree 1 and then we have $\hat{Q(\sqrt{2})}_i=\mathbb{R}$,$i=1,2$. 
Then $|.|_i=|.|$, $i=1,2,$ on $\mathbb{R}$. This implies the restricted absolute value of $|.|_i=|.|,i=1,2$ on $Q(\sqrt{2})$ are the same?! Why I obtain a contradiction? Which step I did wrong?
 A: Mariano's comments say it all, but maybe it is helpful to see things more explicitly. Instead of $\mathbb{Q}(\sqrt 2)$, one should really write $K = \mathbb{Q}[x]/(x^2-2)$, and look at its elements as
$$a+bx$$
with $a,b \in \mathbb{Q}$ and the relation $x^2=2$ (which means $(a+bx)(c+dx) = (ac+2bd) + (ad+bc)x$.)
Now there are two injections $i_1, i_2: K \hookrightarrow \mathbb{R}$:
$$i_1(a+bx) = a+b\sqrt2$$
$$i_2(a+bx) = a-b\sqrt2.$$
$i_1$ induces the absolute value $$|a+bx|_1 = |a+b\sqrt2|,$$
 whereas $i_2$ induces the absolute value $$|a+bx|_2 = |a-b\sqrt2|.$$
Both completions $\hat{(K, |\cdot|_1)}$ and $\hat{(K, |\cdot|_2)}$ are indeed isomorphic to $\mathbb{R}$. More precisely, $i_1$ and $i_2$ extend to two different isomorphisms
$$f_1: \hat{(K, |\cdot|_1)} \simeq (\mathbb{R}, |\cdot|)$$
$$f_2: \hat{(K, |\cdot|_2)} \simeq (\mathbb{R}, |\cdot|);$$
the difference between them still basically being that
$$f_1(x) = \sqrt2$$
$$f_2(x) = -\sqrt2.$$
So in a way, it is true that each $f_i$ identifies its respective absolute value $|\cdot|_i$  with the usual absolute value on $\mathbb{R}$; but $f_1$ sends $x$ to $\sqrt2$, and $f_2$ sends it to $-\sqrt 2$.
In yet other words: It is true that the restriction of the absolute value on $\mathbb{R}$ to the subset $i_1(K) = i_2(K) \subset \mathbb{R}$ is the same; but $|\cdot|_1$, which is the pullback  of that restriction via $i_1$ to the abstract field $K$, is different from $|\cdot|_2$, the pullback of that restriction to $K$ via $i_2$.
