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I am supposed to determine the number of functions in Gal $(\sqrt[4]{3},i)$ without solving for what the exact functions are. That is a separate problems.

Below is the general outline of my proof. I attempt to follow an example in the text and am looking for your thoughts.

Proof

First show that $[\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]=4$

Step 1: Prove that $m_\sqrt[4]{3}=x^4-3 \in \mathbb{Q}[x]$. Hence $[\mathbb{Q}(\sqrt[4]{3}):Q]=4$.

Next show that $[\mathbb{Q}(i):\mathbb{Q}]=2$.

Step 2: Prove that $m_i=x^2+1\in \mathbb{Q}[x]$. Hence $[\mathbb{Q}(i):Q]=2$.

Thus $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3}][\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})]\cdot4$.

Also, $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(i)]\cdot2$.

This implies $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}] \geq 8$.

Next, $m(x)$ is the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb{Q}(i)$.

Then deg $m(x) \leq$ deg $m_\sqrt[4]{3}(x)$.

This implies $[\mathbb{Q}(\sqrt[4]{3}:i):\mathbb{Q}] \cdot$ deg $m(x) \leq 4$.

This implies $[\mathbb{Q}(\sqrt[4]{3}:i):\mathbb{Q}] \leq 4 \cdot 2 = 8$.

Therefore the number of elements in Gal $(\sqrt[4]{3},i) = 8$

NUG
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  • As a sidenote, the appropriate term is the number of automorphisms of the field extension $\mathbb{Q}(\sqrt[4]{3},i)/\mathbb{Q}$ – Jef Apr 08 '17 at 15:02

1 Answers1

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Your ideas are good but there's a flaw in your reasoning: you cannot conclude that $[\mathbb{Q}(\sqrt[4]{3},i) : \mathbb{Q} ] \geq 8$ since $4$ and $2$ are not coprime. If you want to give a quick proof, start as you did: $$[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})][\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})]\cdot4$$ And note that $[\mathbb{Q}(\sqrt[4]{3})(i):\mathbb{Q}(\sqrt[4]{3})]$ is either $1$ or $2$, depending on whether the polynomial $X^2+1$ is reducible or not over $\mathbb{Q}(\sqrt[4]{3})[X]$. Since $i \not\in \mathbb{Q}(\sqrt[4]{3})$ because this last field is contained in the reals, the polynomial $X^2+1$ is indeed irreducible and is thus the minimal polynomial of $i$ over $\mathbb{Q}(\sqrt[4]{3})$, so $[\mathbb{Q}(\sqrt[4]{3})(i):\mathbb{Q}(\sqrt[4]{3})] = 2$ and the result follows

Jef
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  • I understand what you mean but I'm confused on which sections I would need to remove from the above proof. – NUG Apr 08 '17 at 16:25