Given a non-archimedean normed field $(F,|\cdot|)$, show we can extend the norm to $F[x]$ by setting $|a_0+a_1x+\ldots+a_nx^n|\mapsto \max(|a_0|,|a_1|,\ldots,|a_n|).$
I am having trouble showing the multiplicativity of this extended norm. We have to show that
\begin{equation*}
\max_k \left\vert \sum_{i+j=k} a_ib_j \right\vert=\max_i |a_i|\cdot \max_j |b_j|.
\end{equation*}
Since $|-|$ is non-archimedean, we have
$$
\left\vert \sum_{i+j=k} a_ib_j \right\vert\leq \max_{\substack{i,j \\ i+j=k}} |a_i | |b_j| \leq \max_{i,j}|a_i| |b_j|=\max_i |a_i|\cdot \max_j |b_j|.
$$
Taking $\max_k$ on both sides gives one equality of what we need to show. However, I need to inspect when equalities occur.
Any hints?
Best Answer
Let $$ f=a_0+a_1x+a_2x^2+\dots+a_nx^n $$ and $$ g=b_0+b_1x+b_2x^2+\dots+b_mx^m. $$ Let $r$ be the smallest index such that $|a_r|=|f|$ and $s$ be the smallest index such that $|b_s|=|g|$.
Now, consider $|fg|$. In particular, let's look at $|(fg)_{r+s}|$. As you note, $$ (fg)_{r+s}=\sum_{i=0}^{r+s}a_ib_{r+s-i}. $$ We observe that $|a_rb_s|=|f||g|$. Moreover, for $i\not=r$, $|a_ib_{r+s-i}|<|f||g|$ by the assumptions on $r$ and $s$ (either $i<r$ or $r+s-i<s$). One of the properties of non-Archimedean norms states when the norm of a sum is equal to the maximum norm of the summands, this is exactly the current case.