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We count extensions up to $K$-isomorphism. An extension of degree $n$ has a maximal unramified extension $E/K$ inside it, with $[E:K]=f\mid n$, and $L/E$ is totally ramified of degree $e=n/f$. The unramified extension of each degree is unique, so there are only finitely many possibilities for $E$.

Fix $E$ and $e$. The coefficient space of monic degree-$e$ Eisenstein polynomials is compact: its nonconstant nonleading coefficients lie in the compact maximal ideal, and its constant coefficient lies in the compact set ${\mathfrak{m}}_E\setminus {\mathfrak{m}}_E^2$. Each such polynomial $F$ has a root $\alpha$ generating a degree-$e$ extension. For sufficiently close coefficients of a polynomial $G$, the numbers $G(\alpha )$ and $G^{\prime }(\alpha )-F^{\prime }(\alpha )$ are arbitrarily small, while $F^{\prime }(\alpha )\ne 0$ because the characteristic is zero. The strong form of Hensel lemma, applied in $E(\alpha )$, supplies a root $\beta$ of $G$ as close to $\alpha$ as desired. Choose the neighborhood so that the Krasner's lemma inequality holds. Then $E(\alpha )\subseteq E(\beta )$, and both degrees are $e$, so the extensions are equal. Compactness gives a finite cover by such neighborhoods, proving finiteness of degree-$n$ extensions. If extensions are realized as subfields of a fixed algebraic closure rather than counted up to isomorphism, each has only finitely many embeddings, so that count is finite too.

For odd $p$, write $K^{\times }={\pi }^{\mathbb{Z}}{\mathcal{O}}_K^{\times }$. The square map is bijective on $1+{\mathfrak{m}}_K$ by Hensel lemma, and the finite residue field has cyclic multiplicative group of even order. ThusThe quadratic extensions from square classes correspond to its three nontrivial classes. Representatives are $u,\pi ,u\pi$, where $u$ is a unit with nonsquare residue. There are $\boxed{3}$ quadratic extensions: $K(\sqrt{u})$ is unramified, while $K(\sqrt{\pi })$ and $K(\sqrt{u\pi })$ are totally ramified. For completeness, two nonsquare parameters give the same extension only if their ratio is a square: writing $\sqrt{b}=x+y\sqrt{a}$ and squaring forces $x=0$, hence $b/a=y^2$.

Logarithms on sufficiently deep units. Suppose $\mathrm{char}K=0$, let $\ell$ be the residue characteristic, and put $e=v(\ell )$. For every integer $r>e/(\ell -1)$ the p-adic logarithm and p-adic exponential function, evaluated in $K$, converge on the required domains:The bounds $v(m!)=e{v}_{\ell }(m!)\le e(m-1)/(\ell -1)$ and $v(m)\le e(m-1)/(\ell -1)$ show that, for $x,y\in {\pi }^r{\mathcal{O}}_K$, every term beyond the linear one has valuation strictly greater than $v(x)$ or $v(y)$. The series therefore converge and preserve these lattices; in particular $\exp (y)\in 1+{\pi }^r{\mathcal{O}}_K$. Their formal composition and addition identities are valid by convergence, giving inverse continuous group homomorphismsThus the logarithm isomorphism on deep principal units isThe right-hand group is torsion-free.

Roots of unity in ${\mathbb{Q}}_p$. For odd $p$, take $\ell =p,e=1,r=1$. The principal unit group $1+p{\mathbb{Z}}_p$ is torsion-free by the logarithm isomorphism on deep principal units. The Teichmuller representative splitting leaves precisely the $p-1$ roots in ${\mu }_{p-1}$.

For $p=2$, take $r=2$. Each odd unit is uniquely $\pm u$ with $u\in 1+4{\mathbb{Z}}_2$, since its residue modulo four is either one or minus one. This latter group is torsion-free. ConsequentlyThis computes the roots of unity in the p-adic numbers.

The units of ${\mathbb{Q}}_2(i)$. Set $\pi =1-i$. Its polynomial $T^2-2T+2$ is Eisenstein, soThe Eisenstein polynomial facts justifying this are stated in the next solution. Write $U_s=1+{\pi }^s{\mathcal{O}}_K$. Reduction gives ${\mathcal{O}}_K^{\times }=U_1$, and the successive quotients of principal-unit groupshave order two for $s\ge 1$. Hence $[U_1:U_3]=4$. The four roots in ${\mu }_4=\{1,i,-1,-i\}$ have distinct images modulo $U_3$, sinceThey exhaust the quotient and intersect $U_3$ trivially. Since $r=3>2/(2-1)$, the logarithm isomorphism on deep principal units gives $U_3\cong ({\mathcal{O}}_K,+)$. Therefore the unit decomposition of the 2-adic Gaussian field isIn particular these four elements are all its roots of unity.

Quadratic extensions. A nonzero element of $K$ is uniquely a power of $\pi$ times a unit, so $K^{\times }\cong {\pi }^{\mathbb{Z}}\times {\mathcal{O}}_K^{\times }$. Squaring on the decomposition above acts as squaring on ${\mu }_4$ and multiplication by two on the additive ring ${\mathcal{O}}_K={\mathbb{Z}}_2\oplus {\mathbb{Z}}_{2}i$. Thus the square-class group of the 2-adic Gaussian field isBy quadratic extensions from square classes, in characteristic different from two the nontrivial square classes classify quadratic extensions $K(\sqrt{a})$: two such extensions are $K$-isomorphic exactly when $a/b$ is a square. There are therefore $\boxed{2^4-1=15}$ quadratic extensions up to $K$-isomorphism.