If $K/F$ is a finite algebraic extension, it can be defined by a polynomial $f(x)\in F[x]$. The polynomial discriminant, $\mathrm{disc}(f)$, is well-defined up a factor of a non-zero square. The **discriminant root field** of the extension is $F(\sqrt{\mathrm{disc}(f)})$, which is well-defined.

If $n=[K:F]$, then the Galois group $G$ for $K/F$ is a subgroup of $S_n$, well-defined up to conjugation. The discriminant root field can alternatively be described as the fixed field of $G\cap A_n$.

Note if $F=\mathbb{Q}_p$, there are only a small number of possibilities for quadratic extensions of $\mathbb{Q}_p$. If $p$ is odd, we let $u$ denote any unit in $\mathbb{Z}_p$ which is not a square modulo $p$. Then, the quadratic extensions of $\mathbb{Q}_p$ are $\mathbb{Q}_p(\sqrt{u})$ (which is the unramified quadratic extension), $\mathbb{Q}_p(\sqrt{p})$, and $\mathbb{Q}_p(\sqrt{pu})$.

For $p=2$, there are seven quadratic extensions of $\mathbb{Q}_2$. Here we let $u$ denote any value in $\mathbb{Z}_2$ which is congruent to $5$ modulo $8$ (for example, $u=5$) so again $\mathbb{Q}_2(\sqrt{u})$ is the quadratic unramified extension; the other six quadratic extensions are $\mathbb{Q}_2(\sqrt{-1})$, $\mathbb{Q}_2(\sqrt{-u})$, $\mathbb{Q}_2(\sqrt{2})$, $\mathbb{Q}_2(\sqrt{2u})$, $\mathbb{Q}_2(\sqrt{-2})$, and $\mathbb{Q}_2(\sqrt{-2u})$.

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- Review status: reviewed
- Last edited by John Jones on 2023-04-07 13:18:15

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**History:**(expand/hide all)

- 2023-04-07 13:18:15 by John Jones (Reviewed)
- 2020-11-27 12:30:35 by John Jones (Reviewed)
- 2020-11-26 12:06:28 by John Jones (Reviewed)
- 2018-05-23 14:46:52 by John Cremona (Reviewed)

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