Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 1E / ii / Solution

Regardas a monic quadratic in $X$ over $\mathbb{Q}[Y]$. By the Gauss lemma for polynomials, reducibility in $\mathbb{Q}[X,Y]$ would imply reducibility in $\mathbb{Q}(Y)[X]$, so $1-Y^2$ would be a square in $\mathbb{Q}(Y)$. This is impossible: its zero at $Y=1$ has odd order one, whereas every zero or pole of a square in a rational-function field has even order. Hence

The same odd-valuation obstruction to a rational-function square works over every field of characteristic different from two, since the zeros $Y=1$ and $Y=-1$ are then distinct. It does not work over every field. In characteristic two,so the polynomial is reducible.

Solved by gpt-5.6-sol high.