# Is the result true when the valuation is trivial and $\dim(X)=n$?

Here I proved the following result:

Proposition: Let $(K,|\cdot|)$ be a valued field with non-trivial valuation and let $X$ be a vector space over $(K,|\cdot|)$. Two norms $p_1,p_2$ on $X$ are equivalent (i.e., they induce the same topology) iff there are constants $c_1$ and $c_2$ such that $c_1p_1\leq p_2\leq c_2p_1$

And here Eric Wofsey showed, with a counterexample, that the proposition is false when the valuation is trivial (i.e. $|x|=1$, for each $x\in K^*$) and the space $X$ is infinite-dimensional.

Question: Is the proposition true when the valuation is trivial and $X$ is finite-dimensional?

## 1 Answer

Yes, it is. In fact, we can say more. If $K$ has the trivial valuation, $X$ is a finite-dimensional $K$-vector space, and $p$ is any norm on $X$, then there exist positive constants $c_1$ and $c_2$ such that $c_1\leq p(x)\leq c_2$ for all nonzero $x\in X$.

To get $c_2$, note that if $\{e_1,\dots,e_d\}$ is a basis for $X$, then writing $x=\sum a_i e_i$, we get $p(x)\leq\sum |a_i|p(e_i)\leq \sum p(e_i)$. So we can take $c_2=\sum p(e_i)$.

To get $c_1$, let $X_n\subseteq X$ be the span of all the vectors $x\in X$ such that $p(x)<1/n$. We wish to show $X_k=0$ for some $k$ (since we can then take $c_1=1/k$). Let $m=\min\{\dim(X_n):n\in\mathbb{N}\}$ and let $k\in\mathbb{N}$ such that $\dim(X_k)=m.$ Since the $X_n$ are a descending sequence of subspaces of $X$, $m\leq\dim(X_n)\leq\dim(X_k)$ for all $n\geq k$. Then $X_k=X_n$ for $n\geq k$, so $X_k=\bigcap\{X_n:n\geq k\}.$ Now suppose $x\in X_k$. Then for all $n\geq k$, $x$ can be written as a linear combination of vectors of norm $<1/n$. But as in the previous paragraph, this implies that $p(x)<d/n$, where $d=\dim X$. Since $n$ is arbitrary, this means $p(x)=0$ so $x=0$. Thus $X_k=0$, as desired.