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Show that if $p$ is a prime number, then $\mathbb{Q} ( \sqrt{p})/\mathbb{Q} $ is a field extension of degree $2$. Find the Galois group of $\mathbb{Q} ( \sqrt{p})/\mathbb{Q}$ and its fixed field.

I think that I must use that the minimal polynomial is $x^2 - p \in \mathbb{Q}$, but then I don't know how to argue and how to find the Galois group $G(\mathbb{Q} ( \sqrt{p})/\mathbb{Q} )$.

Also, wouldn't the fixed field be $\mathbb{Q}$, by definition of the Galois Group? I'm confused with this too.

I hope you could help me with this.

Thanks in advance!

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Hint: $\mathbb{Q}(\sqrt{p})$ is the splitting field of $x^2-p$ and so $\mathbb{Q} ( \sqrt{p})/\mathbb{Q}$ is a Galois extension. Therefore, its Galois group has order $2$. What are the groups of order $2$?

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