Let $K/\mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $\mathbb Q(\sqrt{d_K})$ where $d_K = \mathrm{disc}(K)$.

Question

Let $K/\mathbb Q$ be a number field and a Galois extension. Show $K$ contains the field $\mathbb Q(\sqrt{d_K})$ where $d_K = \mathrm{disc}(K)$.

I'm guessing this will somehow use the fact that the $d_K$ in an integer but I'm not sure how this being an Galois extension is used.

Answer 1

Let $ [K:\mathbb{Q}] = r $ and $ \text{Gal}(K/\mathbb{Q} )= \{ \sigma_1, \cdots, \sigma_r \} $. If $ \alpha_1, \cdots, \alpha_r $ is a $ \mathbb{Z} $-basis for the ring of integers $ O_K $ of $ K $, then one definition of the discriminant is $ d_K = [\det (\sigma_i(\alpha_j))_{i,j}]^2 $. As $ K/ \mathbb{Q} $ is Galois, the entries $ \sigma_i(\alpha_j) $ of the matrix are in $ K $. So just expanding this determinant out shows you that $ K $ contains a square root of $ d_K $.