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Two distinct singular points would impose at least two zeros each on the restriction of the cubic to their joining line. A nonzero homogeneous polynomial of degree three cannot have total zero multiplicity four, so that restriction vanishes identically. The line would be a component, contradicting the irreducible polynomial hypothesis. This argument works in every characteristic.

An Artin–Schreier polynomial over a field of characteristic $p$ has the formIf it has one root $\alpha$, all its roots are $\alpha +c$ for $c\in {\mathbb{F}}_p$. It either has a root in the base field and splits completely, or is an irreducible polynomial of degree $p$.

A field has characteristic zero when no positive integer multiple of its identity vanishes. Equivalently, its prime field is isomorphic to $\mathbb{Q}$. Every irreducible polynomial over such a field is a separable polynomial.

For an integral normal variety that is Noetherian, removing a closed subset of codimension at least two does not create new regular functions. Affine-locally, a normal Noetherian domain is the intersection of its localizations at height-one primes inside its fraction field. The complement still contains every height-one point, so a regular function on it lies in each such localization and therefore in the original ring. On projective space this also follows directly by excluding irreducible polynomial factor of the denominators on each polynomial affine chart.

For a discrete valuation ring with uniformizer $\pi$, an Eisenstein polynomial is a monic polynomial whose nonleading coefficients are divisible by $\pi$ and whose constant coefficient is not divisible by ${\pi }^2$. It is irreducible, and adjoining one of its roots gives a totally ramified extension.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension to every finite field extension. Any K-isomorphism $K({\alpha }_i)\to K({\alpha }_j)$ between conjugate roots pulls the absolute value back to an extension on the source field, which must be the given one. Thus $|h({\alpha }_i)|=|h({\alpha }_j)|$ for every $h\in K[X]$. This argument includes inseparable irreducible polynomials: no transitive Galois action or completeness assumption is needed.

Every cyclotomic polynomial ${\Phi }_n$ is an irreducible polynomial over $\mathbb{Q}$. If a monic irreducible factor $f\in \mathbb{Z}[X]$ contains a primitive root $\zeta$, then it also contains ${\zeta }^p$ for every prime number $p\nmid n$: otherwise, writing ${\Phi }_n=fg$, reduction modulo $p$ and the Frobenius endomorphism give $\bar{f}\mid \bar{g}(X{)}^p$. Some irreducible factor would then divide both $\bar{f}$ and $\bar{g}$, giving a repeated factor of $X^n-1$, contrary to separability of roots of unity. Iterating over the prime factors of every integer coprime to $n$ shows that $f$ contains all primitive $n$th roots, so $f={\Phi }_n$.

Let $N_q(n)$ be the number of monic irreducible polynomials of polynomial degree $n$ over ${\mathbb{F}}_q$. Factoring $X^{q^n}-X$ into all such polynomials whose degrees divide $n$ givesThe Möbius inversion formula therefore gives