Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 9G / Solution

Gauss lemma for polynomials states that the product of two primitive polynomials over a unique factorization domain is primitive. In particular, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion says that a primitive polynomialis irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. To prove it, suppose $F=GH$ with both factors of positive degree. After removing contents, Gauss's lemma lets us take $G,H\in \mathbb{Z}[X]$ primitive. Reduction modulo $p$ givesso both reduced factors are monomials of positive degree. Their constant terms are therefore divisible by $p$, making $p^2$ divide $a_0=G(0)H(0)$, a contradiction.

An algebraic integer is a complex number satisfying a monic polynomial in $\mathbb{Z}[X]$. Let $\alpha$ be one and let $m_{\alpha }\in \mathbb{Q}[X]$ be its monic minimal polynomial. The usual proof using polynomial content shows that $m_{\alpha }\in \mathbb{Z}[X]$. The setis the kernel of the polynomial-evaluation ring homomorphism, hence an ideal. Every $f\in I_{\alpha }$ is divisible by $m_{\alpha }$ in $\mathbb{Q}[X]$. Because $m_{\alpha }$ is monic, polynomial division of $f$ by $m_{\alpha }$ takes place in $\mathbb{Z}[X]$, and the remainder must vanish by minimality. Thereforewhere $m_{\alpha }$ is monic and irreducible.

Fortranslation givesThis is Eisenstein at $p=3$, so $f(X+1)$ and therefore $f(X)$ are irreducible over $\mathbb{Q}$. HenceIrreducibility and Gauss's lemma also show that $(f)$ is a prime ideal of $\mathbb{Z}[X]$, soIt is not a field. The class of $2$ is nonzero but not a unit: reducing further modulo $2$ gives a nonzero quotientin which the image of $2$ is zero, whereas a unit must remain a unit under every unital quotient map. Thus