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If I have a field extension of $\mathbb{Q}$ given by $\mathbb{Q}(\alpha)$ and the only thing I know about the primitive element $\alpha$ is it's minimal polynomial $p(x) = a_0 + a_1x + ... + x^n$ such that $p(\alpha) = 0$ how can I find the norm of an element $\beta \in \mathbb{Q}(\alpha)$?

I've got partial notes that seem to claim that I can define a linear map $T_\beta$ such that when I take the determinant of the associated matrix I'll end up with the norm of $\beta$ but I can't find the rest of my notes and I was hoping someone knows how to construct the matrix for $T_\beta$

EDIT: $\beta$ is given as $\beta = b_0 + b_1\alpha + ...$ with $\{\alpha^i\}$ forming a power basis for $\mathbb{Q}(\alpha)$

AvatarOfChronos
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2 Answers2

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Let $\beta = q(\alpha)$ where $q(X) = \sum_{i=0}^{n-1} b_i X^i$ and all $b_i\in\mathbb{Q}$. Represent $p(X) = \prod_{i=1}^n(X-\alpha_i)$ where the $\alpha_i$ are $\alpha$ and its conjugates. Then $\beta_i=q(\alpha_i)$ are $\beta$ and its conjugates. By definition of the norm, $$N(\beta) = \prod_{i=1}^n \beta_i = \prod_{i=1}^n q(\alpha_i) = \operatorname{Res}(p,q)$$ where $\operatorname{Res}$ is the polynomial resultant. It can be expressed solely in terms of the rational coefficients $a_0,\ldots,a_{n-1}$ of $p$ and $b_0,\ldots,b_{n-1}$ of $q$. It has indeed a representation as a determinant of a Sylvester matrix. It can be computed recursively in a way similar to the euclidean GCD algorithm. Writing out the resultant as a polynomial expression for general $q$ and $n\geq 3$ yields unwieldy expressions however.

ccorn
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  • I'm trying to understand your answer so I've tried a concrete example. Let $\alpha$ solve $1 + x + x^2 + x^3 + x^4 == 0$ and then let $\beta \in \mathbb{Q}(\alpha)$ be given by $\beta = 1+23\alpha$. So from what I understand I have $q(X) = 1+23X$ and $p(X) = 1+X+X^2+X^3+X^4$ and I should look at Res($p,q$) which I get to be 1 (using mathematica to calculate the resultant of the polynomials defined above) whereas mathematica says "AlgebraicNumberNorm[1 + 23 $\alpha$]] = 268181". I think I've missed something in your answer. – AvatarOfChronos May 05 '14 at 20:24
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    Pari/GP's polresultant(1+x+x^2+x^3+x^4,1+23*x) yields 268181. – ccorn May 05 '14 at 20:27
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    Wolfram Alpha agrees. – ccorn May 05 '14 at 20:29
  • I've figured out what I was doing wrong in mathematica. I tried to define the Polynomials and then call "Resultant[Q[x],P[x],x]" and for some reason it doesn't like that and returns 1. I've got it working now. – AvatarOfChronos May 05 '14 at 20:31
  • Thanks for your help by the way. Is there a good reference for information about the resultant? I've been reading wikipedia/mathworld but would love to see something more in depth. – AvatarOfChronos May 05 '14 at 20:34
  • Lang's Algebra: it has a chapter on polynomials. Succinct and very succulent. =) – Pedro May 05 '14 at 20:44
  • @AvatarOfChronos Try chapter 1.3 of Polynomials by Victor V. Prasolov. I'd actually refer you the whole book though! – Balarka Sen May 05 '14 at 20:45
  • many thanks for the references. – AvatarOfChronos May 05 '14 at 20:48
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Think of the linear map from $\mathbb Q(\alpha)$ into itself given by $$T_\beta(\gamma) = \beta \gamma $$ the norm of $\beta$ is just the determinant of the matrix associated to this linear map, for instance if you write $$ \beta \alpha^j = a_{0j} + a_{1j}\alpha + \dots + a_{(n-1)j} \alpha^{n-1}\quad \quad j = 0,1,\dots,n-1 $$ then the matrix associated to $T_\beta$ relative to the base $\{1,\alpha,\dots,\alpha^{n-1}\}$, is $$ M_\beta = \begin{pmatrix} a_{00} &\dots&a_{0(n-1)}\\\dots\\a_{(n-1)0} &\dots&a_{(n-1)(n-1)}\end{pmatrix}$$ you can see that the characteristic polynomial of this matrix is $$ \det (M_\beta - XI_n) = (-1)^n\prod_{i=1}^n (X-\beta^{(i})$$ where $\beta^{(i}$ is the $i$-th conjugate of $\beta$. Putting $X=0$, we get that norm of $\beta$ in the extension $\mathbb Q(\alpha)/\mathbb Q$ is $$ N_{\mathbb Q(\alpha)/\mathbb Q}\beta = \prod_{i=1}^n \beta^{(i} = \det M_\beta $$

Esteban Crespi
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