Equivalence of norms on valued fields I got confused about notions of equivalence of norms.
Let $K$ be a field with an absolute value $|\cdot|$. Some theorem showed two norms $|\cdot|_1,|\cdot|_2$ are equivalent if and only if $|\cdot|_1=|\cdot|_2^c$ for some $c>0$.
On the other hand, two norms $\Vert\cdot\Vert_1,\Vert\cdot\Vert_2$ on a vector space are equivalence if and only if $D\Vert\cdot\Vert_2\leqslant\Vert\cdot\Vert_1\leqslant C\Vert\cdot\Vert_2$ for some $C,D>0$.
It seems that I could view the absolute value as a norm on $K$ as a vector space over itself. But then the two notions of equivalence seems to be contradicting to each other. One is exponential, another is linear.
 A: Absolute values are multiplicative on the inputs.  Vector space norms are not, or rather they have a totally different relation to multiplication, namely the scalar multiplication: $||cv|| = |c|\,||v||$. That isn't the same thing as $|xy| = |x||y|$ for absolute values because you're dealing with different objects: two elements of a field being multiplied or a scalar and a vector being multiplied.
In particular, you need to pick an absolute value on the scalar field before you can even speak about a vector space norm, since part of the conditions on a norm is its relation to scalar multiplication.
One situation where we can regard something as both an absolute value and a vector space norm is when we extend an absolute value from one field to another. Suppose $K$ is a field with an absolute value $|\cdot|$ on it and $L$ is a larger field to which $|\cdot|$ can be extended as an absolute value, and let $|\cdot|'$ be such an absolute value on $L$. Then we can regard $|\cdot|'$ as both
(i) an absolute value on $L$ since $|xy|' = |x|'|y|'$ for all $x$ and $y$ in $L$,
(ii) a $K$-vector space norm on $L$ since $|cx|' = |c|'|x|' = |c||x|'$, as we assume $|\cdot|'$ on the subset $K$ equals $|\cdot|$.
Sometimes $|\cdot|$ may extend from $K$ to an absolute value on $L$ in more than one way, but when $|\cdot|$ makes $K$ complete and $L$ is finite-dimensional over $K$ then there is at most one extension of $|\cdot|$ to an absolute value on $L$. Indeed, suppose $|\cdot|'$ and $|\cdot|''$ are extensions of $|\cdot|$ to an absolute value on $L$.  Viewing them as $K$-vector space norms on $L$, they must be equivalent vector space norms because of the equivalence of all norms on a finite-dimensional vector space over a field $K$ complete with respect to an absolute value when we use that complete absolute value on $K$ in the definition of the vector space norm (see Theorem 3.2 here). Therefore there are positive constants $C$ and $D$ such that
$C|x|'' \leq |x|' \leq D|x|''$ for all $x$ in $L$.  Replacing $x$ with $x^n$, using multiplicativity, and taking $n$th roots,
$\sqrt[n]{C}|x|'' \leq |x|' \leq \sqrt[n]{D}|x|''$. Now let $n \to \infty$ and we get $|x|'' \leq |x|' \leq |x|''$, so $|\cdot|'$ and $|\cdot|''$ are equal on $L$.
Example: the standard absolute value on $\mathbf C$ is the unique absolute value on $\mathbf C$ that extends the standard absolute value on $\mathbf R$ since $\mathbf R$ is complete with respect to its standard absolute value and $\mathbf C$ is finite-dimensional over $\mathbf R$.
A: I would just like to emphasize a few points differently than in KCd's answer.
You are right that if $K$ is a field, and $\lvert \cdot \rvert_1$ and $\lvert \cdot \rvert_2$ are two absolute values on $K$, then they induce the same topology on $K$ if and only if there is some $c > 0$ such that $\lvert \cdot \rvert_1 = \lvert \cdot \rvert_2^c$.
But the condition you state for equivalence of norms is imprecise. What is true is: Let $V$ be a vector space over a field $K$ with a given absolute value $\lvert \cdot \rvert_K$. Let $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ be two norms on $V$ with respect to that same absolute value, i.e. $\Vert av \Vert_i = \lvert a \rvert_K \cdot \Vert v \Vert_i$ for all $a \in K, v \in V, i\in \{1,2\}$. Then these two norms induce the same topology on $V$ if and only if there are $C, D > 0$ such that $D\Vert\cdot\Vert_2\leqslant\Vert\cdot\Vert_1\leqslant C\Vert\cdot\Vert_2$.
Now the case of $\dim_KV=1$ is actually instructive to see the difference: Here, it is easily seen that the norms on $V$ with respect to a fixed absolute value $\lvert \cdot \rvert_K$ on $K$ are exactly the scalar multiples of $\lvert \cdot \rvert_K$, or to be more precise, for any chosen basis vector $v \in V$, we have that $\Vert av \Vert = \lvert a \rvert_K \cdot B$ for all $a \in K$, where $B := \Vert v \Vert$; and conversely for any $B > 0$ this defines a norm on $V$ with respect to the given absolute value on $K$. Now if even more specifically we say $V=K$, we can choose $1_K$ as basis vector, and see that the norms on $K$ as $K$-vector space with respect to $\lvert \cdot \rvert$ are all of the form $\Vert a \Vert = B\cdot \lvert a \rvert_K$. They happen to be all equivalent to each other too. But only one of them is an absolute value, namely $\Vert \cdot \Vert = \lvert \cdot \rvert_K$ itself.
One could of course ask: Let $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ be two norms on a $K$-vector space $V \neq 0$, with respect to possibly different absolute values $\lvert \cdot \rvert_i$, $i =1,2$, on $K$. When do these norms induce the same topology? Well, the absolute values actually have to be equivalent for that to happen. (If they were not, via this post choose $a \in K$ with $\lvert a \rvert_1 < 1$ but $\lvert a \vert_2 \ge 1$; then for any $v \neq 0$ the sequence $a^n v$ goes to $0$ w.r.t. $\Vert \cdot \Vert_1$ but not w.r.t. $\Vert \cdot \Vert_2$.)
So let's say wlog $\lvert \cdot \rvert_1 = \lvert \cdot \rvert_2^c$ with $c \ge 1$; then $\Vert \cdot \Vert_2^{1/c}$ is still a norm, and it's still equivalent to $\Vert \cdot \Vert_1$, but now it's a norm w.r.t. $\lvert \cdot \rvert_1$. So by the criterion for norms with same absolute value, there are $C, D > 0$ such that
$$D \cdot \Vert\cdot\Vert_2^{1/c} \leqslant\Vert\cdot\Vert_1\leqslant C \cdot \Vert\cdot\Vert_2^{1/c}$$
and I guess the existence of $c, C, D$ making that true is as general a criterion as you can get; and, the case of the absolute value now fits in neatly.
