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Let $K$ be a number field and $L$ be a finite extension of $K$ of degree $n$. I want to define a norm map on $L$. So that, for some $\alpha \in L$, I want to find the matrix $\mu_{\alpha}$, the map of the transformation $x \mapsto \alpha x$ for for any $x\in K$.

For example let $L = \mathbb{Q}(\sqrt{2})$ and $K=\mathbb{Q}$. $L$ has basis $(1,\sqrt{2})$. Let $\alpha = a +b\sqrt{2}$. If we check for the bais elements,we see that $1 \mapsto a + b\sqrt{2}$ and $\sqrt{2} \mapsto 2b+a\sqrt{2}$. Thus, I assume that the transformation matrix for $\mu_{\alpha}$ is given by:

$ M= \left[ {\begin{array}{cc} a & 2b \\ b & a\\ \end{array} } \right] $

However, $1 \mapsto a+b\sqrt{2}$ but we have $[1,0] \left[ {\begin{array}{cc} a & 2b \\ b & a\\ \end{array} } \right] = [a , 2b] = a + 2b\sqrt{2} \neq a+b\sqrt{2} .$

Where is the problem and how can we find the desired transformation matrix?

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1 Answer 1

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You should multiply the matrix with the coordinate vector from the right side. Indeed you would have:

$$\left[ {\begin{array}{cc} a & 2b \\ b & a\\ \end{array} } \right]\left[ {\begin{array}{c} 1 \\ 0 \\ \end{array} } \right]=[a \enspace b]$$

On the other side if you want to keep left multiplication the transformation matrix should be:

$$M = \left[ {\begin{array}{cc} a & b \\ 2b & a\\ \end{array} } \right]$$

In other words, instead of writing the values for the basis elements in the columns you are doing it in a row.


Let's determine the transformation matrix under left multiplication. We know that it's of the form: $M= \left[ {\begin{array}{cc} x & y \\ z & t\\ \end{array} } \right]$, where $x,y,z,t$ are yet to be determined.

Now we know that $[1,0]M = [a,b]$ by the definition of the function. Now just perform the multiplication on the left-hand side and comapre coefficients to get: $x=a$ and $y=b$.

Similarly from $[0,1]M = [2b,a]$ we get that $z=2b$ and $t=a$. Therefore the transformation matrix becomes $M = \left[ {\begin{array}{cc} a & b \\ 2b & a\\ \end{array} } \right]$

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  • $\begingroup$ Can you explain the reason please? $\endgroup$
    – Ninja
    Jul 28, 2018 at 22:28
  • $\begingroup$ @Ninja I added some further explanation. I hope it answers your question $\endgroup$
    – Stefan4024
    Jul 28, 2018 at 22:35
  • $\begingroup$ One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say. $\endgroup$
    – Ninja
    Jul 29, 2018 at 10:53
  • $\begingroup$ @Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,\sqrt{2})$ on the left, which can't be written like this in a basis. $\endgroup$
    – Stefan4024
    Jul 29, 2018 at 11:39
  • $\begingroup$ For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700 $\endgroup$ Jul 30, 2018 at 12:33

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