The direct question is:
Let $F$ be a field, with the trivial norm, that is, $|x|=\begin{cases}1&x\ne0\\0&x=0\end{cases}$. Let $M$ be a free module over $F$ with basis $\left\{x_i\mid i=1,2,\cdots\right\}$. Define two norms $|\cdot|_1$ and $|\cdot|_2$ by $$|\sum_ia_ix_i|_j=\sup_i|a_i|\cdot|x_i|_j=\sup_{a_i\ne0}|x_i|_j,$$ where $|x_i|_1=i$ and $|x_i|_2=i^2$. Then are these two norms metrically inequivalent but topologically equivalent?
I think they are metrically inequivalent because there is no constant $c>0$ such that $i^2<c\cdot i,\,\forall i=1,2\cdots$. And I think they are topologically equivalent because they both generate the discrete topology (the point $0$ is open in both topologies).
Why ask this?
In the book p-adic Differential Equations by Kiran.S.Kedlaya one finds the exercise I.1.(4):
Let $M$ be a module over a non-archimedean field $F$. Prove that any two norms compatible with $F$ are topologically equivalent if and only if they are metrically equivalent.
And I think the above paragraph provides a counter-example to this exercise. Per chance the book is missing the condition the the norm on the filed should not be trivial (in which case I figured out how to prove it).
Any help is appreciated. Thanks in advance.