Exercise 13.3.14

Proof.

(a)

We proved in Exercise 12.1.9(b) that the symmetry group of $$\underset{\sigma \in H}{\sum }\sigma \cdot x_1^{n-1}{x}_2^{n-2}\dots x_{n-2}^2{x}_{n-1}$$

is $H$.

Thus, $A_n$ is the symmetry group of

$$D=\underset{\sigma \in A_n}{\sum }\sigma \cdot x_1^{n-1}{x}_2^{n-2}\dots x_{n-2}^2{x}_{n-1}.$$

(b)

We proved in Ex. 2.4.1 the formula $$\sqrt{\mathrm{\Delta }}=\underset{1\le i<j\le n}{\prod }(x_i-x_j)=\left|\begin{array}{cccc}\hfill x_1^{n-1}\hfill & \hfill x_2^{n-1}\hfill & \hfill \cdots \hfill & \hfill x_n^{n-1}\hfill \\ \hfill x_1^{n-2}\hfill & \hfill x_2^{n-2}\hfill & \hfill \cdots \hfill & \hfill x_n^{n-2}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \text{⋱}\hfill & \hfill ⋮\hfill \\ \hfill x_1\hfill & \hfill x_2\hfill & \hfill \cdots \hfill & \hfill x_n\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill \cdots \hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right|$$

By definition of the determinant, if $A={(a_{i,j})}_{1\le i,j\le n}$

$$\det <!--\; FUNCTION\; APPLICATION\; -->(A)=\underset{\sigma \in S_n}{\sum }\mathrm{sgn}(\sigma )\underset{i=1}{\overset{n}{\prod }}{a}_{\sigma (i),i},$$

and

$$\det <!--\; FUNCTION\; APPLICATION\; -->(A)=\det <!--\; FUNCTION\; APPLICATION\; -->({}^{t}A)=\underset{\sigma \in S_n}{\sum }\mathrm{sgn}(\sigma )\underset{i=1}{\overset{n}{\prod }}{a}_{i,\sigma (i)}.$$

Applying this formula to $a_{i,j}=x_j^{n-i}$, with $\phi =x_1^{n-1}{x}_2^{n-2}\cdots x_{n-1}$ and $\tau =(12)$, we obtain

$$\begin{array}{llll}\hfill \sqrt{\mathrm{\Delta }}& =\underset{\sigma \in S_n}{\sum }\mathrm{sgn}(\sigma )x_{\sigma (1)}^{n-1}{x}_{\sigma (2)}^{n-2}\cdots x_{\sigma (n-1)}& \hfill & \\ \hfill & =\underset{\sigma \in S_n}{\sum }\mathrm{sgn}(\sigma )\sigma \cdot x_1^{n-1}{x}_2^{n-2}\cdots x_{n-1}& \hfill & \\ \hfill & =\underset{\sigma \in A_n}{\sum }\sigma \cdot \phi -\underset{\sigma \in S_n\setminus A_n}{\sum }\sigma \cdot \phi & \hfill & \\ \hfill & =\underset{\sigma \in A_n}{\sum }\sigma \cdot \phi -\underset{\sigma \in A_n}{\sum }(\mathit{\tau \sigma })\cdot \phi & \hfill & \\ \hfill & =D-\tau \cdot D=D-D^{\prime }.& \hfill & \end{array}$$

(c)

Since $S_n=A_n\coprod <!--\; FUNCTION\; APPLICATION\; -->\tau A_n$, $$D+D^{\prime }=\underset{\sigma \in S_n}{\sum }\sigma \cdot \phi .$$

For each $\chi \in S_n$,

$$\chi \cdot (D+D^{\prime })=\underset{\sigma \in S_n}{\sum }(\mathit{\chi \sigma })\cdot \phi =\underset{{\sigma }^{\prime }\in S_n}{\sum }{\sigma }^{\prime }\cdot \phi =D+D^{\prime }({\sigma }^{\prime }=\mathit{\chi \sigma }).$$

Therefore the symmetry group of $D+D^{\prime }$ is $S_n$.

Since $(D(f)+D^{\prime }(f)=(D+D^{\prime })({\alpha }_1,\dots ,{\alpha }_n)$, where $D+D^{\prime }$ is a symmetric polynomial, $D(f)+D^{\prime }(f)\in F$.

Moreover

$$D{D}^{\prime }=\underset{\sigma \in A_n}{\sum }\sigma \cdot \phi \underset{{\sigma }^{\prime }\in A_n}{\sum }(\tau {\sigma }^{\prime })\cdot \phi .$$

Similarly, if $\chi \in A_n$, then $\chi \cdot D=D$. Since

$$D^{\prime }=\underset{{\sigma }^{\prime }\in A_n}{\sum }(\tau {\sigma }^{\prime })\cdot \phi =\underset{{\sigma }^{″}\in A_n}{\sum }({\sigma }^{″}\tau )\cdot \phi ,$$

$\chi \cdot D^{\prime }=D^{\prime }$.

Therefore $\chi \cdot (D{D}^{\prime })=D{D}^{\prime }$ for all $\chi \in A_n$, and

$$\tau \cdot (D{D}^{\prime })=\underset{\sigma \in A_n}{\sum }(\mathit{\tau \sigma })\cdot \phi \underset{{\sigma }^{\prime }\in A_n}{\sum }{\sigma }^{\prime }\cdot \phi =D^{\prime }D=D{D}^{\prime },$$

so that $\chi \cdot (D{D}^{\prime })=D{D}^{\prime }$ for all $\chi \in S_n$, and $D{D}^{\prime }$ is a symmetric polynomial. Thus $D(f)D^{\prime }(f)=(D{D}^{\prime })({\alpha }_1,\dots ,{\alpha }_n)\in F$.

This proves

$$D_f(y)=(y-D(f))(y-D^{\prime }(f))=y^2-(D(f)+D^{\prime }(f))y+D(f)D^{\prime }(f)\in F[y].$$

Finally, the evaluation mapping $(x_1,\dots ,x_n)\mapsto ({\alpha }_1,\dots ,{\alpha }_n)$ gives

$$\sqrt{\mathrm{\Delta }(f)}=\underset{1\le i<j\le n}{\prod }({\alpha }_i-{\alpha }_j),$$

thus

$$\mathrm{\Delta }(f)=\underset{1\le i<j\le n}{\prod }{({\alpha }_i-{\alpha }_j)}^2.$$

(d)

The discriminant of $D_f(y)$ is ${(D(f)-D^{\prime }(f))}^2=\sqrt{\mathrm{\Delta }(f)}\ne 0$, since we assumed that $f$ is separable. Therefore $D_f(y)$ is separable, so that the roots $D(f),D^{\prime }(f)$ are simple roots.

Since $D_f(y)$ is the resolvent associated to $\phi$ with symmetry group $A_n$, Proposition 13.3.2 shows that $G_f\subset A_n$ if and only if $D_f(y)$ has a root in $F$, if and only if $D_f(y)$ splits over $F$.