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Let $K$ be a finite extension of $\mathbb{Q}$. For each $\alpha \in K$, multiplication by $\alpha$ is a linear map from $K$ to $K$. Fix an integral basis $\omega_1,\ldots,\omega_n$ for $K$ over $\mathbb{Q}$. Let $M_{\alpha}$ be matrix for the "multiplication by $\alpha$" map with respect to the given basis.

If $\alpha = \sum_{i=1}^{n} a_i \omega_i$, set $$\| \alpha \|_{1} = \max_{1 \leq i \leq n} \| a_i \|$$ and set $$ \| \alpha \|_{2} = \max_{1 \leq i,j \leq n} \| (M_{\alpha})_{ij} \| $$ where $(M_{\alpha})_{ij}$ is the entry in the $i$-th row and $j$-th column of $M_{\alpha}$.

Both $\| \cdot \|_1$ and $\| \cdot \|_2$ are norms on $K$, which is finite-dimensional, thus the norms are equivalent in the sense that there are constants $c,C > 0$ such that $$ c \|\alpha \|_1 \leq \| \alpha \|_2 \leq C \|\alpha \|_1 $$ for all $\alpha \in K$.

Update Since $K$ is over $\mathbb{Q}$ (as opposed to $\mathbb{R}$ or $\mathbb{C}$), it is not necessarily true that all norms on $K$ must be equivalent. So that gives a preliminary question: Are the two norms equivalent?

Update 2: I posted an answer that resolves the question in the first update. In summary, the norms are equivalent because they are equivalent when extended to the vector space given by the formal span $\text{span}_{\mathbb{R}}(\omega_1,\ldots,\omega_n)$. The questions below still stand.

Question How do the constants depend on $K$? Do they depend only on the degree $n$ of $K$, if so how? For a given degree $n$, can we always find a $K$ so that $c$ and $C$ are as small and large, respectively, as we desire?

Side Question Is there any good reference for properties of matrix representations of algebraic numbers? Every book I've seen hardly goes beyond saying that the determinant of $M_{\alpha}$ is the norm of $\alpha$.

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  • $\begingroup$ Let's look at a simple example: $K={\bf Q}(\sqrt2)$, $\alpha=\sqrt2$, let $a,b,c,d$ be (large) integers such that $ad-bc=1$, let $\omega_1=a+b\sqrt2$, let $\omega_2=c+d\sqrt2$. Then $a\omega_2-c\omega_1=\alpha$, so anyway $\|\alpha\|_1$ can be arbitrarily large. You should be able to work out $\|\alpha\|_2$ for these choices. $\endgroup$ – Gerry Myerson Aug 26 at 6:21
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    $\begingroup$ @GerryMyerson The matrix corresponding to $\alpha$ seems to be $\begin{bmatrix} 2bd-ac & 2bd-ac \\ a^2 - 2b^2 & c^2 - 2d^2 \end{bmatrix}$. Recording this now to think about later. $\endgroup$ – MichaelGaudreau Aug 26 at 7:00
  • $\begingroup$ @GerryMyerson The point you are trying to make is not clear to me. Are you suggesting that the two norms are not comparable? Because of the condition $ad-bc=1$, it is not clear by looking at this matrix that, for example, $\|\alpha\|_2$ could be arbitrarily larger or smaller than $\|\alpha\|_1$. $\endgroup$ – MichaelGaudreau Aug 26 at 17:45
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    $\begingroup$ @GerryMyerson Right you are. The norms are not equivalent. They are both norms and $K$ is finite dimensional, but because $K$ is a vector space over $\mathbb{Q}$ rather than $\mathbb{R}$ or $\mathbb{C}$, the norms need not be equivalent. $\endgroup$ – MichaelGaudreau Aug 27 at 1:35
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    $\begingroup$ This is a comment-request that I frequently have to make. It is an unnecessary and unfriendly choice to use both $\alpha$ and $a$ in the same question, even if their function is slightly different. As a certified geezer with less than perfect vision, I find it difficult to distinguish the two letters, especially in subscripts. Don’t bother changing your question on my account; just go and sin no more. $\endgroup$ – Lubin Aug 27 at 3:52
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EDIT: It appears I have misunderstood the question. I'll leave this up, in case what's in it can be used somehow (and I'll delete it, if folks think I ought to).

Take $K={\bf Q}(\sqrt2)$, $\alpha=\sqrt2$, $\omega_1=a+\sqrt2$, $\omega_2=a-1+\sqrt2$. Then (details can be found in the comments) $\|\alpha\|_1=a$, $\|\alpha\|_2\ge a^2-2$. So there is no constant $C>0$ such that $\|\alpha\|_2\le C\|\alpha\|_1$ holds for all $\alpha \in K$.

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  • $\begingroup$ I don't understand your point, why would there not be such a $C$ ? $\endgroup$ – Captain Lama Aug 27 at 2:02
  • $\begingroup$ @Captain, there is no positive constant $C$ such that, for all $a$, $a^2-2\le Ca$. That is, $(a^2-2)/a$ is not bounded above. $\endgroup$ – Gerry Myerson Aug 27 at 2:04
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    $\begingroup$ But the norms change if you change $a$. $\endgroup$ – Captain Lama Aug 27 at 2:22
  • $\begingroup$ @CaptainLama That's true. The norms do depend on $a$, since they depend on the basis. So this doesn't show that there is no such $C$. It only shows that the basis may be chosen so that the two norms corresponding to that basis are at least as "far apart" as one desires. To show there is no such $C$, the basis should be fixed and for each $C$ an element $\alpha$ should be exhibited the violates the inequality. $\endgroup$ – MichaelGaudreau Aug 27 at 4:47
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I have figured out that the norms are equivalent. Here is my argument.

Viewing $K$ as a vector space over $\mathbb{Q}$, it is a subset of $V = \text{span}_{\mathbb{R}}(\omega_1,\ldots,\omega_n)$.

The norm $\| \cdot \|_1$ extends to $V$ with the same formula.

Note $\omega_i \omega_j = \sum_{k=1}^{n} c_{i,j,k} \omega_k$ with $c_{i,j,k}$ integers. We can multiply elements $\alpha = \sum_{i=1}^{n} a_i \omega_i$ and $\beta = \sum_{j=1}^{n} b_j \omega_j$ by $$ \alpha \beta = \sum_{i=1}^{n} \sum_{j=1}^{n} (a_i b_j) (\omega_i \omega_j) $$ Multiplication by $\alpha$ is still a linear map and the matrix $M_{\alpha}$ is formed the usual way (the columns are the coefficients of $\alpha \omega_j$). The norm $\| \cdot \|_2$ extends to $V$ with the same formula.

Now $\| \cdot \|_1$ and $\| \cdot \|_2$ are two norms on $V$, which is a finite-dimensional vector space over $\mathbb{R}$. So they must be equivalent. Thus they are equivalent over the subset $K$.

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