$p$ splits completely in $\mathbb{Q}(\zeta_l)$ implies it splits in $\mathbb{Q}(\sqrt{-l})$ Let $l>3$ be a prime such that $l\equiv 3 (\textrm{mod } 4)$. Let $p$ be an odd prime such that $p\equiv 1 (\textrm{mod } l)$. Then we can prove directly that $p$ splits in $\mathbb{Q}(\sqrt{-l})$. 
My question is, how to deduce that $p$ splits in $\mathbb{Q}(\sqrt{-l})$ from the fact that $p$ splits completely in $\mathbb{Q}(\zeta_l)$?
 A: It's a general fact that if $K \subset L$ is an  inclusion of algebraic number fields and $p$ splits completely in $L$, then it splits completely in $K$.
To see this, consider the prime factorization of $(p)$ in $\mathcal O_K$; it is a product of a bunch of primes $\mathfrak p$, each with some $e$ and $f$.  To say that $(p)$ splits completely is to say that each $e$ and $f$ equals $1$.
Now consider the factorization of each of the primes $\mathfrak p$ in $\mathcal O_L$.  Each of them factors into the product of a bunch of $\mathfrak q$'s, again each $\mathfrak q$ has an $e$ and $f$.
Now combine these various factorizations to get the factorization of $(p)$ in $\mathcal O_L$.  If $\mathfrak q$ is a factor in $\mathcal O_L$ of one of the factors  $\mathfrak p$ of $(p)$ in $\mathcal O_K$, then the $e$ and $f$ for $\mathfrak q$ thought of as a factor of $(p)$ in $\mathcal O_L$ are the product of the $e$ and $f$ for $\mathfrak q$ thought of as a factor of $\mathfrak p$ and the $e$ and $f$ for $\mathfrak p$ thought as as a factor of $(p)$. 
You are assuming that $(p)$ splits completely in $L$, so these products must equal $1$.  Thus each of the separate $e$'s and $f$'s (for $\mathfrak q$ over $\mathfrak p$ and for $\mathfrak p$ over $(p)$) must equal $1$.  In particular, $(p)$ splits completely in $K$. 
