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.

• Looks good so far. Now you can use the fact that $[K(\alpha) : L]$ is the degree of the minimal polynomial of $\alpha$ over $L$. Nov 18 '19 at 20:35

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. $$\blacksquare$$