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A prime is completely split in an extension of degree $n$ exactly when there are $n$ primes above it. Indeed, the fundamental identity for prime decomposition is ${\sum }_{\mathfrak{q}\mid \mathfrak{p}}{e}_{\mathfrak{q}}{f}_{\mathfrak{q}}=n$, and every summand is a positive integer. Thus $n$ primes means that every ramification index and residue degree equals one, and conversely.

By part (a), the number of primes of $L$ above $\mathfrak{p}$ is the number of right $D$-orbits on $H\backslash G$. This set has $[G:H]=[L:K]$ elements. It has that many orbits exactly when every orbit is a singleton, equivalently whenThis condition says $gD{g}^{-1}\subseteq H$ for every $g$, orthe subgroup core.

Because $M$ is the Galois closure of $L/K$, it is generated by all the fields $g(L)$. The fixed field of $gH{g}^{-1}$ is $g(L)$, so the fixed field of the core is their compositum, namely $M$. The Galois correspondence therefore gives ${\mathrm{Core}}_G(H)=1$. HenceIn the Galois extension $M/K$, the number of primes above $\mathfrak{p}$ is $[G:D]$, so $D=1$ is equivalent to complete splitting in $M$. This proves that complete splitting is preserved by Galois closure:No prior unramifiedness assumption is needed: complete splitting already forces all ramification indices to be one, and the decomposition-group argument includes the ramified case.