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Let $K$ be a finite extension of $\mathbb Q$. Then I have to show that there exists a finite extension $E$ over $K$ such that $E/K$ is not normal.

It is clear that $E/\mathbb Q$ is not normal as well.

I tried taking $K=\mathbb Q(\alpha)$, $|K:\mathbb Q|=n$ and consider $L=\mathbb Q(\beta)$, where $\beta$ is root of the irreducible polynomial $X^m-2$. $m$ is chosen to be relatively prime with $n$, so that the compositum has degree $mn$, i.e., $|KL:\mathbb Q|=mn$, where $KL=\mathbb Q(\alpha,\beta)$. Then by natural irrationality $KL/L$ is Galois of degree $n$. I tried to show $KL$ play the role for $E$, but I couldn't pull it back to the other chain.
Any help will be highly appreciated.Thank You.

user371231
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1 Answers1

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Let $n := [K : \Bbb Q]$. Let $p$ be a prime with $p - 1 > n$.

Let $\alpha = 2^{1/p} \in \Bbb R$. Note that $[\Bbb Q(\alpha) : \Bbb Q] = p > n$ and thus, $\alpha \notin K$. Let $E := K(\alpha)$. Note that in particular, we have $$p \mid [E : \Bbb Q]. \tag{1}$$

Claim. $E/K$ is not normal.

Proof. For the sake of contradiction, assume that $E/K$ is normal.
Consider the polynomial $f = X^p - 2 \in K[X]$. Note that $\alpha$ satisfies $f$.
Let $g \in K$ be the minimal (irreducible) polynomial satisfied by $\alpha$ over $K$. Then, we can write $$f = g h$$ for some $h \in K[X]$.

Let $d := \deg(g) \le p$. Then, we have $[E : K] = d$.
Since $\alpha \notin K$, it follows that $g$ is not linear. By assumption of normality, we see that $g$ has another root $\beta \in E$.

Define $\gamma := \beta/\alpha \in E$. Then, we have $\gamma \neq 1$ and $\gamma^p = 1$. Thus, $\gamma$ is a primitive $p$-th root of $1$ and hence, $$(p - 1) \mid [E : \Bbb Q]. \tag{2}$$

(This is because $[\Bbb Q(\gamma) : \Bbb Q] = p - 1$.)

From $(1)$ and $(2)$, we conclude that $$p(p - 1) \mid [E : \Bbb Q]. \tag{3}$$ On the other hand, we have $$[E : \Bbb Q] = [E : K][K : \Bbb Q] = dn < p(p - 1),$$ contradicting $(3)$. $\Box$

Aryaman Maithani
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  • Just too many doubts in the above proof. Are you assuming $K/\mathbb Q$ a Galois extension ? – user371231 Jun 27 '21 at 10:33
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    We can observe that $[E:K] \leq p$ and hence $[E:\mathbb {Q}] \leq pn$. But $[E:\mathbb{Q}] =[E:\mathbb{Q}(\alpha)] [\mathbb{Q} (\alpha):\mathbb{Q}] $ so $[E:\mathbb{Q}] $ is divisible by both $p, n$ and hence by $pn$ and so is greater than or equal $pn$. It follows that $[E:\mathbb{Q}] =pn$ and $[E:K] =p$ and $f(x) $ is irreducible over $K[x] $. +1 btw. – Paramanand Singh Jun 27 '21 at 15:43
  • @user371231: What exactly are your doubts? No, I'm not assuming $K/\Bbb Q$ to be Galois. – Aryaman Maithani Jun 27 '21 at 15:52
  • Why $E$ is not a splitting field of $f(x)$? – user371231 Jun 27 '21 at 16:14
  • @user371231: Well, whatever I've done shows that $E/K$ is not normal. So, in particular, it is not a splitting field of $f(x)$. – Aryaman Maithani Jun 27 '21 at 16:18
  • As much as I got from your answer you basically proved that $f(X)$ is irreducible over $K$. How it is enough for non normality of $E/K$? – user371231 Jun 27 '21 at 16:23
  • @user371231: No, I didn't prove that $f(x)$ is irreducible over $K$. I showed that if assume $E/K$ to be normal, we get a contradiction. – Aryaman Maithani Jun 27 '21 at 16:24
  • The splitting field of $f$ over $\mathbb Q$ has degree $p(p-1)$.Please correct me if i am wrong. Since $f$ is irreducible over $K$ also is the splitting field of degree $p(p-1)$ over $K$ as well ? – user371231 Jun 27 '21 at 16:29
  • @user371231: 1. Yes, the degree of that splitting field (over $\Bbb Q$) you mention is correct. 2. I have not shown that $f$ is irreducible over $K$. 3. I think it would be easiest for you to ask me doubts regarding my steps. I have not used splitting fields anywhere, for instance. Note that adjoining the root of a polynomial does not always give a splitting field. – Aryaman Maithani Jun 27 '21 at 16:32
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    Now I got your answer, choosing $p>n+1$ really helped. Thank You. – user371231 Jun 27 '21 at 16:40
  • Also by the comment of @ParamanandSingh, since $f$ is irreducible over $K$, we have $|E:\mathbb Q|=np<p(p-1)$. Thus $E/Q$ is not normal, and hence $E/K$ is not normal. – user371231 Jun 27 '21 at 16:46
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    Very nice! ${}{}$ – Jyrki Lahtonen Jul 01 '21 at 16:38