2
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ Looks good so far. Now you can use the fact that $[K(\alpha) : L]$ is the degree of the minimal polynomial of $\alpha$ over $L$. $\endgroup$ Nov 18 '19 at 20:35
0
$\begingroup$

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. $\blacksquare$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.