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Let me explain Serre's definition in the case of a finite extension $K$ of $Q_p$ ($p$ prime). In his notation, $K_s$ is an algebraic closure of $K$, $K_t$ is the maximal tamely ramified extension of $K$ in $K_s$, $K_{nr}$ is the maximal unramified extension of $K$ in $K_t$, and $I_t=Gal(K_t|K_{nr})$. The residue field $k_s$ of $K_s$ is an algebraic closure of the finite field $F_p$.

So inside $k_s$ you have all the finite extensions $F_{p^n}$ of $F_p$, with norm maps $F_{p^m}^\times\rightarrow F_{p^n}^\times$ whenever $n|m$. The first thing to observe is that there is a natural identification $\theta$ of $I_t$ with the inverse limit of this system of norm maps (prop. 2)

As such, for each power $q=p^n$ of $p$, there is a natural projection $\theta_{q-1}:I_t\rightarrow F_q^\times$, which you might call the fundamental character of level $n$. When you compose it with some power $\varphi^i$ of the automorphism $\varphi:x\mapsto x^p$ of $F_q^\times$, you get all the other fundamental characters of level $n$. There are thus $n$ fundmental characters of level $n$, namely $\varphi^i\circ\theta_{q-1}$ for $i\in[0,n[$.

Here is how to think of $\theta_{q-1}$ concretely. Take a uniformiser $\pi$ of $K_{nr}$ and adjoin its $(q-1)$-th root $\root q-1\of\pi$ to get the extension $L_{q-1}$ of $K_{nr}$ in $K_t$; this extension does not depend on the choice of $\pi$. On the one hand, the group $Gal(L_{q-1}|K_{nr})$ is a quotient of $I_t$. On the other hand, $Gal(L_{q-1}|K_{nr})$ is naturally isomorphic to the group $\mu_{q-1}$ of $(q-1)$-th roots of $1$ (under $\zeta\mapsto(\root q-1\of\pi\mapsto\zeta.\root q-1\of\pi$)), which is naturally isomorphic (under reduction modulo $\pi$) to the group $F_q^\times$. The character $\theta_{q-1}$ is just the composite $$ I_t\to Gal(L_{q-1}|K_{nr})\to \mu_{q-1} \to F_q^\times. $$

I'll leave it to the experts to elucidate their role in Serre's modularity conjecture. I believe they serve to define the optimal weight of an odd irreducible representation $Gal(\bar Q|Q)\rightarrow GL_2(\bar F_p)$.