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I am trying to solve lots of exercises from Neukirch's textbook "Algebraic Number Theory". Exercise 1.8.3 (Chapter I, section 8, exercise 3) asks me to prove:

Let $L/K$, $L'/K$ be two separable extensions. $P$ is an prime ideal of $K$ that is totally split in $L$ and $L'$ then $P$ is also totally split in the compositum $LL'$.

I know a solution using property of decomposition group but decomposition group only shows up at the following section, Section 9. Please provide me some hints for a solution without using decomposition group if possible. Thanks.

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Here are some suggestions.

One way to think of it is this, if $E/F$ is an extension and $\mathfrak{P}$ a prime factor of $\mathfrak{p}$ is of degree one ($f=1$) iff every $x\in \mathcal{O}_E$ is congruent mod $\mathfrak{P}$ to an element $y\in \mathcal{O}_F$. (You can rewrite this as an equality of residue fields.)

Thus you want to verify this condition for every prime divisor in the composite field, knowing that it holds in the individual fields. For this you may use the fact that the integers of the composite are of the form $a_1b_1+\cdots +a_nb_n$ where the $a$ is in one field and the $b$'s in the other.

Look at prop 2.11.

Rene Schipperus
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