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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 $\operatorname{Norm}(-x)=-\operatorname{Norm}(x)$?

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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 $\alpha$.

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  • $\begingroup$ Thanks a lot. I just wanted to make it clear that are $\sigma_i$ the roots of polynomial which can be seen as an element of K(i.e. minimal polynomial of x)? $\endgroup$ – Mayank Jun 23 '16 at 10:23
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    $\begingroup$ yes $\sigma_i(x)$ are the different roots of $irr(x,\Bbb{Q})$ $\endgroup$ – m.idaya Jun 23 '16 at 10:52
  • $\begingroup$ How is this always true that the number of roots of $irr(x,Q)$ is equal to $[K:Q]$? $\endgroup$ – Mayank Jun 23 '16 at 12:20
  • $\begingroup$ sorry, valid for $x$ a primitive element of $K$, in your case $K/\Bbb{Q}$ finite and separable as number field, thanks for your remarques $\endgroup$ – m.idaya Jun 23 '16 at 12:33
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    $\begingroup$ $N(-x)=\prod_{i=1}^{i=n}\sigma_i(-x)=(-1)^n\prod_{i=1}^{i=n}\sigma_i(x)=(-1)^nN(x)$ is true for all $x\in K$, but $deg(irr(x,\Bbb{Q}))=[K:\Bbb{Q}]$ iff $x$ is a primitive element of $K$. thanks $\endgroup$ – m.idaya Jun 23 '16 at 12:41
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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 $$ \operatorname{Norm}(-x) = - \operatorname{Norm}(x).$$

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