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For a number field $K$, why does the square root of the discriminant $d_K$ lie in the field $K$, i.e., $\sqrt{d_K} \in K$?

I know that $d_K$ is an element of $\mathbb{Q}$, so surely it lies in $K$. In the case of a quadratic extension $\mathbb{Q}(\sqrt{d})$, I know that the discriminant is either $d$ or $4d$ depending on whether $d$ is congruent to $1$ or $2,3 \pmod{4}$. For $d\equiv 1 \pmod{4}$, obviously $\sqrt{d_K}$ lies in $\mathbb{Q}(\sqrt{d})$. I don't see how it follows for the $d\equiv 2,3 \pmod{4}$, or any other non-quadratic number field in general. Any help will be appreciated!

Yang Awotwi
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1 Answers1

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This is false, in general. Counter-example: Consider $K=\mathbb{Q}(\sqrt[3]2)$. The discriminant is $-2^2\cdot3^3$. The square root does not lie in $K$, which can be shown since $\sqrt{-3} \notin K$. However, this is true if the field $K$ is `normal' i.e if $K=\mathbb{Q}[a]$ and all conjugates of $a$ lie in $K$, then the $\sqrt{d_K}=|\sigma_i(a_j)| \in K$. (Since $\sigma_i(a_j) \in K, \forall i,j$, where $\sigma_i$'s are the embeddings of $K$ and $a_j$'s are the conjugates of $a$ over $\mathbb{Q}$)

Vishnu Kadiri
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