Notes - Galois Theory HT25, Determinant and discriminant
Flashcards
Suppose:
- $f \in F[t]$ is a polynomial of degree $n$
- $K$ is a splitting field for $f$
- $\{\alpha _ 1, \ldots, \alpha _ n\}$ are the roots of $f$ in $K$
@Define the determinant and the discriminant.
Determinant:
\[\delta := \prod_{1 \le i < j \le n} (\alpha_j - \alpha_i)\]Discriminant:
\[\Delta := \delta^2 = \prod_{1 \le i < j \le n} (\alpha_j - \alpha_i)^2\]Suppose:
- $f$ is a polynomial
- $K$ is a splitting field for $f$
- $K/F$ is a Galois extension
- $G = \text{Gal}(K/F)$ is identified with a subgroup of $S _ n$
- $\delta$ is the determinant of $f$
@State a result which links the determinant and the sign of the permutations in $G$.
For all $g \in G$
\[g \cdot \delta = \text{sgn}(g)\cdot \delta\]Suppose:
- $f \in F[t]$ is a polynomial of degree $n$
- $K$ is a splitting field for $f$
- $\{\alpha _ 1, \ldots, \alpha _ n\}$ are the roots of $f$ in $K$
- $G = \text{Gal}(K/F)$, identified with a subgroup of $S _ n$
- $\text{char}(F) \ne 2$
- $f$ has no repeated roots
@State a result which links the fixed field of $G \cap A _ n$ with a field extension of $F$.
\[K^{G \cap A_n} = F(\delta)\]
where $\delta$ is the determinant
\[\delta := \prod_{1 \le i < j \le n} (\alpha_j - \alpha_i)\]Suppose:
- $f \in F[t]$ is a polynomial of degree $n$
- $K$ is a splitting field for $f$
- $\{\alpha _ 1, \ldots, \alpha _ n\}$ are the roots of $f$ in $K$
- $G = \text{Gal}(K/F)$, identified with a subgroup of $S _ n$
- $\text{char}(F) \ne 2$
- $f$ has no repeated roots
@State a result that tells you whether $G \le A _ n$.
$G \le A _ n$ if and only if $\Delta$ is a square in $F$.