Past exam of the number theory course of the University of Oxford / 2009 / A / 1

The alternating polynomial given by the Vandermonde product is alternating: interchanging two variables changes its sign. One proof writes it as a Vandermonde determinant, up to a fixed sign, and interchanging two variables interchanges the corresponding determinant rows. Since transpositions generate $S_n$,Thus every even permutation fixes $\delta$, while a transposition negates the nonzero formal polynomial, so it is not invariant under all of $S_n$. This uses $n\ge 2$ and characteristic different from two, as in the subsequent rational setting. For $n=1$ both groups are trivial and the last distinction does not arise.

The discriminant is a symmetric polynomial in the four roots, so the Fundamental theorem of symmetric polynomials expresses it polynomially in the elementary symmetric polynomial. For this depressed quartic they are $e_1=e_2=0$, $e_3=-r$, $e_4=s$. Scaling each root by $t$ multiplies the discriminant by $t^{12}$, while $r,s$ scale by $t^3,t^4$. Thus only weighted-degree-twelve monomials can appear, giving ${\Delta }_f=A{r}^4+B{s}^3$.

For $r=0$, use $f^{\prime }=4X^3$ and the monic discriminant identity ${\Delta }_f=(-1{)}^{n(n-1)/2}{\prod }_i{f}^{\prime }({\alpha }_i)$. At $n=4$ this gives $4^4({\prod }_i{\alpha }_i{)}^3=256s^3$, so $B=256$. For $s=0$, the roots are zero together with those of $g=X^3+r$. The cubic's discriminant is $-\prod g^{\prime }({\beta }_i)=-27r^2$. The extra squared differences from zero multiply this by $(\prod {\beta }_i{)}^2=r^2$, giving $-27r^4$. HenceThese computations are polynomial identities, so they extend to parameter values with repeated roots as well.

For a polynomial with $n$ distinct roots, the criterion isWrite $\delta ={\prod }_{i<j}({\alpha }_i-{\alpha }_j)\ne 0$. Part (a) gives $\sigma \delta =\mathrm{sgn}(\sigma )\delta$. If every Galois permutation is even, $\delta$ is fixed by the entire Galois group and therefore belongs to its fixed field $\mathbb{Q}$. Thus ${\Delta }_f={\delta }^2$ is a rational square. Conversely, if ${\Delta }_f=q^2$ with $q\in {\mathbb{Q}}^{\times }$, then ${\delta }^2=q^2$ in a field implies $\delta =\pm q$. Every automorphism fixes this nonzero element, so its permutation sign must be one.

The distinct-root hypothesis is necessary in the statement's intended degree-$n$ permutation action. The source says “arbitrary degree” without stating that hypothesis. If repeated roots were allowed, zero discriminant would carry no sign information: $(X-1{)}^2(X^2-2)$ has discriminant zero, while its nontrivial automorphism swaps $\sqrt{2},-\sqrt{2}$ and fixes the two rational-root positions, an odd permutation. The criterion above applies to the squarefree polynomial, or to the squarefree part with its degree and root action adjusted accordingly.

Set $Y=X+1$. The polynomial becomes $Y^4+Y+1$, and translating all roots leaves their pairwise differences unchanged. Part (b) givesIt is nonzero and lies strictly between $15^2=225$ and $16^2=256$. An integer which is a rational square is an integer square, so this is not a rational square. The roots are distinct because the discriminant is nonzero. By (c), $\boxed{G_f⊈A_4}$.