The common topology determines which elements have absolute value below one, because
$$
|x|_i<1\quad\Longleftrightarrow\quad x^n\longrightarrow0
$$
in that topology. It therefore also determines all comparisons: $|x|_i<|y|_i$ exactly when $|x/y|_i<1$.

Choose $a\in K^\times$ with $|a|_1>1$; then also $|a|_2>1$. For every $x\in K^\times$ and positive integers $m,n$, the preceding observation gives
$$
|x|_1^n<|a|_1^m
\quad\Longleftrightarrow\quad
|x|_2^n<|a|_2^m.
$$
Thus the two real numbers
$$
\frac{\log|x|_1}{\log|a|_1}
\quad\hbox{and}\quad
\frac{\log|x|_2}{\log|a|_2}
$$
have the same rational upper cuts and are equal. Setting
$$
c=\frac{\log|a|_1}{\log|a|_2}>0
$$
gives $|x|_1=|x|_2^c$ for every $x$. Hence the two absolute values are <equivalent absolute values>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-136/2/a/solution