Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 1E (Groups, Rings and Modules)

The Eisenstein criterion says that a primitive polynomialis irreducible if there is a prime $p$ such thatBy the Gauss lemma for polynomials, irreducibility over $\mathbb{Q}$ and over $\mathbb{Z}$ agree for primitive polynomials.

LetIf $n=ab$ with $a,b>1$, thenso $P_n$ is reducible.

Conversely, let $n=p$ be prime. Translation by one is an automorphism of $\mathbb{Z}[X]$, andIts leading coefficient is one, every lower coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. The Eisenstein criterion proves that $P_p(X+1)$ is irreducible, hence so is $P_p(X)$. Therefore the geometric-sum irreducibility criterion is

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.