Odd degree of extension field in Couveignes Square root method I was reading the Couveignes method to find the square root for the Number Field Sieve (reference here page 4 first line). It says that for this method the degree of the extension $K/\mathbf Q$ must be odd so that $\operatorname{Norm}(-x)=-\operatorname{Norm}(x)$ for any non-zero element $x$ of $K$. Thus only one of the square roots has positive norm.
I am new to abstract algebra and I was trying to implement this method.
Can anyone please explain to me how an odd degree implies $\operatorname{Norm}(-x)=-\operatorname{Norm}(x)$?
 A: Let $\sigma_i, i=1,...,n$   the $\Bbb{Q}$-isomorphisms of $K$
where $n=[K;\Bbb{Q}]$ , let $x\in K$ , a primitive element of $K$ so by definition
$N(x)=\prod_{i=1}^{i=n}\sigma_i(x)$ then
$N(-x)=\prod_{i=1}^{i=n}\sigma_i(-x)=(-1)^n\prod_{i=1}^{i=n}\sigma_i(x)=(-1)^nN(x)$
so the equality $N(-x)=-N(x)$  , then  give $n$ odd.
Edit: In Number Field Sieve, we take $K=\Bbb{Q}(\alpha), \alpha$ is a root of irreducible polynomial $f(x)$. We also have $\Bbb{Q}(\alpha)$ is isomorphic to $\Bbb{Q}[x]/p_\alpha$, where $p_\alpha$ is the minimal polynomial of $\alpha$. Hence, with this isomorphism established we have degree of $p_\alpha$ is equal to degree to $K$. In the above discussion the primitive element can be easily taken as $\alpha$. 
A: The answer depends on your definition of the norm map. There is already an answer using embeddings and roots of the defining polynomial. But let me give you also a more algebraic approach.
Another way to define the norm comes from treating $K$ as a $\mathbf Q$-vector space of dimension $n$, where $n$ is the degree of the extension $K|\mathbf Q$. Now for an element $x \in K$ we consider the map
$$ \mu_x \colon K \longrightarrow K,\ z \longmapsto x\cdot z, $$
that is, the map which is just multiplication by our fixed element $x$.
Then the fact that $K|\mathbf Q$ is a field extension implies that the map $\mu_x$ is $\mathbf Q$-linear. Moreover, we know that
$$ \operatorname{Norm}(x) = \det(\mu_x).$$
In particular we can use all the properties of the determinant to derive properties of the norm map.
For example, we immediately obtain that (since $\mu_{xy} = \mu_x \circ \mu_y$)
$$ \operatorname{Norm}(xy) = \operatorname{Norm}(x)\operatorname{Norm}(y). $$
Moreover $\operatorname{Norm}(-1) = (-1)^n$, since the map $x \mapsto -x$ has determinant $(-1)^n$. Combining this with the multiplicative property of the norm and the fact that $n$ is odd we obtain
$$ \operatorname{Norm}(-x) = - \operatorname{Norm}(x).$$
