A multiplicative semi-norm on a ring $A$ is a function $|\,|:A\to \mathbb{R}_{\ge 0}$ that is multiplicative and satisfies the semi-norm conditions:
$|0|=0,|1|=1\\ |fg|=|f||g|,\\ |f+g|\le |f|+|g|.$
I want to see why the set of multiplicative semi-norms on $\mathbb{C}[x]$ that extend the absolute value norm on $\mathbb{C}$ is of the form $f\mapsto |f(x)|$ for some $x\in \mathbb{C}$. It is said here that this follows from Gelfand-Mazur's theorem but I do not see how. Can someone give a proof?
Given a seminorm $p$, let $\mathfrak m_p = \{f \in \Bbb C[x] : p(f) = 0\}$. This ideal is prime, as if $fg \in \mathfrak m_p$, $p(f)p(g) = 0$, so one of $p(f)$ or $p(g)$ is zero. So $\mathfrak m_p = \langle x-a\rangle$ for some $a \in \Bbb C$ or $\mathfrak m_p = \{0\}$.
Suppose $\mathfrak m_p = \{0\}$. Then $p$ is in fact a norm on $\Bbb C[x]$ and extends to a norm on $\Bbb C(x)$ by $p'\left(\frac{a}{b}\right) = p(a)/p(b)$. So $\Bbb C(x)$ has the structure of a normed field. Passing to the completion we now have a complete normed field that contains $\Bbb C(x)$, and is thus infinite dimensional over $\Bbb C$; this contradicts Gelfand-Mazur, and thus $\mathfrak m_p$ must have been maximal.
Write $\mathfrak m_p= \langle x-a_p\rangle$. Then for any $f(x)$, $f(x) = g(x)+c$ where $g \in \mathfrak m_p$. Then the reverse triangle inequality gives $|p(f)-p(c)| \leq p(f-c) = p(g) = 0$, so $p(f) = p(c) = |c| = |f(a_p)|$. Hence $p(f) = |f(a_p)|$ for all $f \in \Bbb C[x]$ as desired.