The field of Laurent series over $\mathbb{C}$ is quasi-finite

How can I prove that the field of Laurent series over $\mathbb{C}$ is quasi-finite, which means that it has a unique extension of degree $n$ for all $n \geqslant 1$ ?

The article says that the extension of degree $n$ is $\mathbb{C}((T^{1/n}))$. I think that I understand why it is an extension of degree $n$ (the polynomial $X^n – T$ is irreducible by Eisenstein’s criterion ?), but how can I prove that it is unique please ?

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The assertion is a consequence of the ramification theory of valuations: the field of Laurent series $\mathbb{C}((t))$ carries the discrete valuation $v(a_kt^k+a_{k+1}t^{k+1}+\ldots ):=k$, $a_k\neq 0$. It is complete with respect to this valuation; consequently $v$ has a unique extension $w$ to every finite extension field $K$ of $\mathbb{C}((t))$.

Assume $[K:\mathbb{C}((t))]=n$, then by the fundamental equality of valuation theory one has $(\Gamma :\mathbb{Z})[\overline{K}:\mathbb{C}]=n$, where $\Gamma$ is the value group of $w$ and $\overline{K}$ is the residue field of $w$. Since $\mathbb{C}$ is algebraically closed, one gets $\overline{K}=\mathbb{C}$ and thus $(\Gamma :\mathbb{Z})=n$.

Now assume for the moment that $K/\mathbb{C}((t))$ is Galois. Ramification theory then yields that the Galois group $G$ is isomorphic to $\Gamma /\mathbb{Z}$, which is cyclic of order $n$. In particular every intermediate field of $K/\mathbb{C}((t))$ is Galois with cyclic Galois group.

Since every finite extension $L/\mathbb{C}((t))$ is contained in a finite Galois extension $K/\mathbb{C}((t))$, this shows that every finite extension $L/\mathbb{C}((t))$ is Galois with cyclic Galois group.

Since in a finite cyclic group for every divisor $d$ of the group order there exists exactly one subgroup of index $d$, for every degree $d$ there exists exactly one extension $K/\mathbb{C}((t))$ of degree $d$.