Galois Theory HT25, Determinant and discriminant

Determinant:

\[\delta := \prod_{1 \le i < j \le n} (\alpha_j - \alpha_i)\]

Discriminant:

\[\Delta := \delta^2 = \prod_{1 \le i < j \le n} (\alpha_j - \alpha_i)^2\]

For all $g \in G$

\[g \cdot \delta = \text{sgn}(g)\cdot \delta\]

Consider the Vandermonde matrix

\[V := \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\\\ x_1 & x_2 & x_3 & \cdots & x_n \\\\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1} \end{pmatrix}\]

Recall that swapping any two columns of a matrix changes the sign of the determinant, i.e.

\[\tau (\det V) = \det(\tau(V)) = -\tau\]

for any transposition $\tau \in S _ n$. Since the transposition generate $S _ n$, it follows

\[g \cdot \det V = \text{sgn}(g) \det V\]

for all $g \in S _ n$.

Recall also that

\[\det V = \prod_{1 \le i < j \le n} (x_j - x_i)\]

Hence by setting each $x _ i = \alpha _ i$, it follows that $\det V = \delta$, and hence

\[g \cdot \delta = \text{sgn}(g) \delta\]

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Proof Vandermonde matrix has determinant $\prod _ {1 \le i < j \le n} (x _ j - x _ i)$:

For a fixed pair $1 \le i < j \le n$, consider the substitution map

\[\Psi_{i, j} : \mathbb Z[x_1, \ldots, x_n] \to \mathbb Z[x_1, \ldots, x_n]\]

which sends $x _ j$ to $x _ i$ and sends $s _ k$ to $x _ k$ for all $k \ne j$.

Then $\ker \Psi _ {i, j} = \langle x _ i - x _ j \rangle$ and therefore $\Psi _ {i, j}(\det V) = 0$.

Since the $n \choose 2$ linear polynomials $\{x _ j - x _ i \mid 1 \le i < j \le n\}$ are coprime in $\mathbb Z[x _ 1, \ldots, x _ n]$ and the ring is a UFD, it follows

\[\prod_{1 \le i < j \le n} (x_j - x_i) \mid \det V\]

(since $\det V$ must be included in the kernel).

Both $\prod _ {1 \le i < j \le n} (x _ j - x _ i)$ and $\det V$ are monic polynomials of degree $n \choose 2$ and therefore

\[\prod_{1 \le i < j \le n} (x_j - x_i) = \det V\]

\[K^{G \cap A_n} = F(\delta)\]

where $\delta$ is the determinant

\[\delta := \prod_{1 \le i < j \le n} (\alpha_j - \alpha_i)\]

$G \le A _ n$ if and only if $\Delta$ is a square in $F$, where $\Delta$ is the discriminant (the square of the product of the roots).

Note that by a previous result we have that for all $g \in G$,

\[g \cdot \delta = \text{sgn}(g) \delta\]

and hence

\[\delta \in K^{G \cap A_n}\]

since all permutations in $G \cap A _ n$ are even and so have positive sign. Hence we have the inclusion of fields

\[F \subseteq F(\delta) \subseteq K^{G \cap A_n}\]

Case 1, $G \le A _ n$: If $G \le A _ n$, it follows that $G \cap A _ n = G$ and so these are all equalities:

\[F = F(\delta) = K^{G \cap A_n}\]

which implies $\delta \in F$ and so $\Delta = \delta^2$ is a square in $F$.

Case 2, $G \not\le A _ n$: If $G \not\le A _ n$, then there exists some odd permutation $\tau \in G$ and $\tau \cdot \delta = -\delta$. By the assumption that $f$ has no repeated roots, $\delta \ne 0$, and since $\text{char }F \ne 2$, $-\delta \ne \delta$. Hence $\delta \notin F$ since there is at least one permutation $\tau$ that doesn’t fix it. Then $\Delta$ cannot be a square in $F$, since if it was, $\delta = \sqrt \Delta$ would also belong to $F$. This establishes (2).

Now consider the result that says:

The indexes of the intermediate fields correspond to the indexes of the quotient groups:

If $F \subseteq L \subseteq K$ is an intermediate field and $H = \text{Gal}(K/L)$, then

\[[L : F] = \vert G/H \vert\]

and

\[[K : L] = \vert H \vert\]

Applied to $L = K^{G \cap A _ n}$, it follows that

\[[K^{G\cap A_n} : F] = [G : (A_n \cap G)]\]

By standard group theory results,

\[[G : (A_n \cap G)] = [GA_n : A_n]\]

which must equal $2$ since $G$ contains a transposition and $A _ n$ contains all even permutations, so $GA _ n = S _ n$ and $[S _ n : A _ n] = 2$.

Therefore $K^{G \cap A _ n} = F(\delta)$ (since it follows that $[K^{G\cap A _ n} : F(\delta)]$ must be $1$ by the tower law as $F \ne F(\delta)$) and this field has degree two over $F$. This establishes (1).