Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 3

First extract the polynomial content from $f$:The primitive quartic is Eisenstein at the prime $2$: every nonleading coefficient is divisible by $2$, while the constant term is not divisible by $4$. Eisenstein's criterion makes it irreducible over $\mathbb{Q}$, and Gauss's lemma makes it irreducible in $\mathbb{Z}[X]$. The constant factor $2$ is itself irreducible in $\mathbb{Z}[X]$.

The polynomial $g$ is already irreducible over $\mathbb{Q}$: it is Eisenstein at $3$. Its nonleading coefficients, including the missing zero coefficients, are divisible by $3$, and $-21$ is not divisible by $9$. Thus $g$ is its own sole nonconstant irreducible factor over $\mathbb{Q}$.

Over the field ${\mathbb{Z}}_3$, grouping terms givesThe first two factors are linear and therefore irreducible. The quadratic takes the values $1,2,2$ at $0,1,2$ respectively, so it has no root in ${\mathbb{Z}}_3$. A degree-two polynomial over a field is reducible exactly when it has a root, proving its irreducibility.

All three factorizations are unique up to order and multiplication by units. The ring $\mathbb{Z}$ is a unique factorization domain, hence so is $\mathbb{Z}[X]$ by Gauss's lemma. A polynomial ring over a field is a Euclidean domain, so both $\mathbb{Q}[X]$ and ${\mathbb{Z}}_3[X]$ have unique factorization. The units are $\pm 1$ in the first ring, nonzero rational constants in the second, and $1,2$ in the third.