February 17th
Today I learned Ostrowski's theorem for function fields, from here . The analog here is that all of our nontrivial places $|\bullet|:F(x)\to\RR$ which are trivial on $F$ (!) are equivalent to $|\bullet|_\infty$ or $|\bullet|_\pi$ for a monic irreducible $\pi\in F[x].$ Here, $|\bullet|_\infty=c^{-\deg\bullet}$ and $|\bullet|_\pi=c^{\nu_\pi(\bullet)}$ for some positive constant $c \lt 1.$ We remark quickly that being trivial on $F$ is somewhat reasonable because, when $F$ is finite, these are our "torsion units'' in $F[x].$ So we expect this statement to reflect the case with $\QQ$ more closely.
Anyways, fix $|\bullet|$ a valuation on $F(x).$ Note that $|\bullet|$ is multiplicative, so we can focus on its behavior on $F[x]$ because $F(x)$ is just quotients of $F[x].$ The nice thing about function fields is that $|\bullet|$ is surely nonarchimedean: the embedding $\ZZ\to F[x]$ ends up in $F,$ and $F$ gets sent to $1,$ so\[\max_{n\in\ZZ}|n|=1 \lt \infty\]is bounded. So being archimedean will not differentiate between $|\bullet|_\infty$ and $|\bullet|_\pi$ as in $\QQ.$ Instead, we do casework on $|x|$ because we know $|x|_\infty \gt 1$ while $|x|_\pi\le1.$
On one hand, we might have $|x| \gt 1.$ Then we hope $|\bullet|$ is equivalent to $|\bullet|_\infty.$ For a polynomial $\alpha(x)=\sum_{k=0}^{\deg\alpha}a_kx^k,$ we note that being nonarchimedean implies\[|\alpha|=\left|\sum_{k=0}^{\deg a}a_kx^k\right|\le\max_{0\le k\le\deg a}\left(|a_k|\cdot|x|^k\right).\]Because $a_k\in F$ has $|a_k|=1,$ this is actually the maximum of $|x|^\bullet.$ Well, $|x| \gt 1,$ so the maximum occurs (uniquely!) at $|x|^{\deg\alpha}.$ The importance of a unique maximum is that, because all triangles are isosceles, $|a|$ must be equal to $|x|^{\deg\alpha},$ our unique minimum. That is,\[|\alpha|=(1/|x|)^{-\deg\alpha}\]for a positive constant $1/|x| \lt 1.$ Thus, $|\bullet|$ is equivalent to $|\bullet|_\infty,$ as desired.
On the other hand, we might have $|x|\le1.$ Then we hope $|\bullet|$ is equivalent to $|\bullet|_\pi$ for some monic irreducible $\pi\in F[x].$ In particular, we need to extract the monic irreducible $\pi,$ which we do by thinking about this like $F[x]_\pi.$ Here, $\pi$ is our uniformizer, generating our topology. So, after noting that any $F[x]$ has\[\left|\sum_{k=0}^da_kx^k\right|\le\max_{0\le k\le d}\left(|a_k|\cdot|x|^k\right)\le1,\]and that $|\bullet|$ cannot be trivial on all $F[x]$ (else $|\bullet|$ would be trivial), we're allowed to extract $\pi$ as an element of $F[x]$ satisfying $|\pi| \lt 1$ with least degree. Because multiplying by an element of $F$ won't affect $|\pi|,$ we take $\pi$ monic.
We verify that $\pi$ behaves like we want. Note $\pi$ is non-constant because $|\bullet|$ is trivial on $F$—this is where this condition shows up in this case. Further, $\pi$ is irreducible, for if $\pi=\alpha\beta$ with $\deg\alpha,\deg\beta \lt \deg\pi,$ then $|\alpha|=|\beta|=1$ by the degree of minimality of $\pi.$ But $|\pi| \lt 1,$ so this is impossible.
To finish, fix any $\alpha\in F[x],$ and set $\alpha=\pi^\nu\beta$ where $\nu:=\nu_\pi(\alpha).$ We would like to show that $|\alpha|=|\pi|^\nu,$ where $c:=|\pi| \lt 1$ will be our positive constant. Well, $|\alpha|=|\pi|^\nu\cdot|\beta|,$ so it suffices to show $|\beta|=1.$ We need to use the condition $\pi\nmid\beta,$ so we use the division algorithm to write\[\beta=q\pi+r.\]Here, $\pi\nmid\beta$ requires $r\ne0,$ so $r$ is a nonzero polynomial with degree smaller than $\deg\pi.$ But by minimality, this means $|r|=1,$ so the inequality\[|\beta|\le\max\{|q|\cdot|\pi|,|r|\}\]has $|r|=1$ as a unique maximum—$|q|\cdot|\pi|\le|\pi| \lt 1.$ So because all triangles are isosceles, $|\beta|=1.$ This completes the proof.