Solution
= Solution
\b[True.] If $M/K$ is normal, the <finite normal extension as a splitting field> criterion writes $M$ as the splitting field over $K$ of some $f\in K[T]$. Since $K\subseteq L\subseteq M$, the same field $M$ is the splitting field of $f$, now regarded as a polynomial over $L$. Hence $M/L$ is normal.