Show that a Field Extension is Unramified Using Transitivity Let $K=\mathbb{Q(\sqrt{5})}, L=\mathbb{Q(\sqrt{7})}, M=\mathbb{Q(\sqrt{35})}$, and $KL=\mathbb{Q(\sqrt{5},\sqrt{7})}$. Show that $KL/M$ is unramified (i.e. every prime ideal of $M$ is unramified in $KL$). 
A hint provided is to only consider the ramification indices of $2$,$5$, and $7$ in extensions $L/\mathbb{Q}, K/\mathbb{Q}, M/\mathbb{Q}$ and to apply transitivity. 
Based on what I know, I need to consider the discriminant of each extension:
disc$(K)=5$, disc$(L)=28=2^2\cdot7$, disc$(M)=140=2^2\cdot5\cdot7$ where I am using the fact that $Q(\sqrt{D})$ has discriminant either 
$D$ if $D \equiv 1$ (mod 4), or $4D$ if $D \equiv 3$ (mod 4). 
I tried to do three different lattices: one for each prime divisor: $(2,5,7)$. However, I am finding that each discriminant is divisible by $5$ in the diagram for $5$. 
Can somebody please explain how to solve this problem? I have been struggling very much with this problem for 4 days now. 
Thanks in advance, any help is greatly appreciated. 
 A: First, note that any primes ramifying in $KL/M$ sit above primes ramifying in $KL/\mathbb{Q}$. One can compute that $$\text{disc}(KL) = \text{disc}(K)^2 \text{disc}(L)^2 = 5^2 \cdot 2^4 \cdot 7^2$$ using e.g. Proposition 1.2.11 in Neukirch, Algebraic Number Theory.  Thus, only primes sitting over $2$, $5$, or $7$ could possibly ramify in $KL/M$.
Let $p \in \{2, 5, 7\}$. For fields $F_1, F_2$ in our scenario, let $e_p(F_1/F_2)$ denote the ramification degree of some primes in $F_1$, $F_2$ sitting over $p$; this is well defined since all extensions in sight are Galois. We know $e_p (M/\mathbb{Q}) = 2$ for all $p \in \{2, 5, 7\}$, since $M/\mathbb{Q}$ is quadratic.
For each $p \in \{2, 5, 7\}$. Note that exactly one of $e_p (L/\mathbb{Q})$ and $e_p (K/\mathbb{Q})$ equals $2$, and the other equals $1$, e.g. we have $e_2 (K/\mathbb{Q}) = 1$ and $e_2 (L/\mathbb{Q}) = 2$. Thus, $$e_2 (KL/\mathbb{Q}) = e_2 (KL/K) e_2 (K/\mathbb{Q}) \le 2\cdot 1 = 2$$ and $$e_2 (KL/\mathbb{Q}) = e_2 (KL/L) e_2 (L/\mathbb{Q}) \ge 1\cdot 2 = 2$$ therefore $e_2 (KL/\mathbb{Q}) = 2$. Similarly, we compute $e_p (KL/\mathbb{Q}) = 2$ for $p \in \{5, 7\}$.
Finally, we have $$2 = e_p (KL/\mathbb{Q}) = e_p (KL/M) e_p (M/\mathbb{Q}) = 2 e_p (KL/M)$$ so $e_p (KL/M) = 1$ for $p \in\{2,5,7\}$. We conclude that $KL/M$ is unramified.
