If $M_1:K$ and $M_2:K$ are normal extensions then $K(M_1,M_2):K$ is normal.

Question

By normal extension of $K$ I mean the splitting field for a $S\subseteq K[x]$, and by $K(M_1,M_2)$ I mean the least field that contains $K, M_1,$ and $M_2$. (Suppose $L:M_1:K$ and $L:M_2:K$ so that this makes sense.)

I $\textit{guessing}$ that if $S_1\subseteq K[x]$ and $S_2\subseteq K[x]$ are sets such that $M_1$ and $M_2$ are their splitting fields over $K$ respectively, then $K(M_1,M_2)$ is probably a splitting field for $S_1\cup S_2$. The elements of $S_1\cup S_2$ definitely split over $K(M_1,M_2)$, but I'm having trouble checking that if $K(M_1,M_2):H:K$ and $H\subsetneq K(M_1,M_2)$ then there's some element of $S_1\cup S_2$ that doesn't split over $H$.

Answer 1

Definitions being used: (I state them because apparently my book using non-standard terminology.)

$L:K$ means $L$ extends $K$. (Contains $K$ as a subfield.)

$L:K$ is called normal if there exists some $S\subseteq K[x]$ such that $L$ is the least field extension of $K$ where every element of $S$ splits. (i.e. $L$ is a splitting field for $S$.)

If $L:K$, we say a subset $S\subseteq K[x]$ splits over $L$ if every element of $S$ splits over $L$.

We will write $M_1M_2$ for $K(M_1,M_2)$.

Let $S_1\subseteq K[x]$ and $S_2\subseteq K[x]$ be sets such that $M_1$ and $M_2$ are their splitting fields over $K$ respectively. Then, $S_1\cup S_2$ clearly splits over $M_1M_2$ but let's check minimality. Suppose $H$ is a field such that $H:K$ and $S_1\cup S_2$ splits over $H$. In particular $S_1$ splits over $H$ and so, by minimality of $M_1$, we have $H:M_1$. Analogously, we get $H:M_2$ and combining these we get $H:M_1M_2$.

Answer 2

It seems to be you mean the compositum $\;M_1\vee M_2=M_1M_2\;$. Let

$$f:M_1M_2\to\overline K$$

be any $\;K\,-$ embedding of $\;M_1M_2\;$ to an algebraically closed field $\;\overline K\;$ containing $\;K\;$ .

Let us look at the restrictions

$$f_1:=f\uparrow_{M_1}:M_1\to\overline K\;,\;\;f_2:=f\uparrow_{M_2}:M_2\to\overline K$$

Since $\;M_i/K\;$ are normal extensions, both $\;f_1,\,f_2\;$ are actually automorphisms:

$$f_i\in\text{Aut}\,(M_i/K)$$.

But since any element in the compositum $\;M_1M_2\;$ is a finite sum of products of elements in $\;M_1,\,M_2\;$, we have that

$$f(m_1m_2...)=f(m_1)f(m_2)\ldots=f_1(m_1)f_2(m_2)\ldots\in M_1M_2$$

since $\;f_i(m_i)\in M_i\;,\;\;m_i\in M_i\;\implies f\in\text{Aut}\,(M_1M_2/K)\;$

and $\;M_1M_2/K\;$ is normal