Proving the minimal polynomial of a ground field is reducible in an extension Specifically, I'm trying to solve this problem:

Let $\mathbb{K}\subseteq\mathbb{L}\subseteq\mathbb{M}$ be fields such that $\mathbb{M}=\mathbb{K}(\alpha)$ for some $\alpha$ that is algebraic over $\mathbb{K}$. Let $k(x)$ be the minimal polynomial of $\alpha$ over $\mathbb{K}$. Prove that if $\mathbb{K}\neq\mathbb{L}$, then $k(x)$ is not irreducible in $\mathbb{L}[x]$.

Since $\alpha$ is algebraic over $\mathbb{K}$, I believe it follows that $[\mathbb{K}(\alpha):\mathbb{K}]<\infty$. Then by transitivity of degree in field extensions,
$$[\mathbb{K}(\alpha):\mathbb{K}]=[\mathbb{K}(\alpha):\mathbb{L}]\underbrace{[\mathbb{L}:\mathbb{K}]}_{\geq2}.$$
I think the desired result follows from this. I'm looking for assistance verifying my claim and formalizing the proof.
 A: We'll prove the contrapositive statement.
Suppose that $k(x)$ is irreducible in $\mathbb{L}[x]$. Then, since $\alpha$ is algebraic over $\mathbb{K}$, it follows that $[\mathbb{M}:\mathbb{K}]<\infty$. Then, by transitivity of degree of field extensions,
$$[\mathbb{M}:\mathbb{K}]=[\mathbb{M}:\mathbb{L}][\mathbb{L}:\mathbb{K}].$$
Let $l(x)$ denote the minimal polynomial of $\alpha$ over $\mathbb{L}$. Then, $[\mathbb{L}(\alpha):\mathbb{L}]=\deg(l(x))$ and $[\mathbb{M}:\mathbb{K}]=\deg(k(x))$. Note that $\mathbb{K}(\alpha)\subseteq\mathbb{L}(\alpha)$ because $\mathbb{K}\subseteq\mathbb{L}$. Furthermore, since $\mathbb{L}(\alpha)$ is the smallest ring containing $\mathbb{L}$ and $\alpha$, $\mathbb{L}(\alpha)=\langle L\cup\{\alpha\}\rangle\subseteq\mathbb{K}(\alpha)$. Thus, $\mathbb{L}(\alpha)=\mathbb{M}$, and it follows then that
$$\deg(k(x))=\deg(l(x))[\mathbb{L}:\mathbb{K}].$$
Hence, $l(x)\mid k(x)$. But since $l(\alpha)=0$, it must be true that $l(x)=k(x)$, otherwise the minimality of $k(x)$ would be contradicted. Therefore, $[\mathbb{L}:\mathbb{K}]=1$ and the desired result follows from definition of degree of field extension.
$\blacksquare$
