Past exam of the algebra 2 course of the University of Oxford / 2014 / 2 / c

Use the reduction criterion: a primitive integer-coefficient polynomial whose leading coefficient remains nonzero modulo a prime number $p$ is irreducible over $\mathbb{Q}$ if its reduction is irreducible over ${\mathbb{F}}_p$. This follows from Gauss lemma for polynomials, since a nontrivial factorization over the integers would reduce to a nontrivial factorization of the same degrees. For a monic polynomial the factors may be taken monic, so neither degree can drop.

Modulo $2$ the polynomial is $\overline{f}=t^4+t+1$. Its values at $0$ and $1$ are both $1$, so it has no linear polynomial factor. A reducible quartic polynomial without a linear polynomial factor must factor into two irreducible quadratic polynomials. The only monic irreducible quadratic polynomial over ${\mathbb{F}}_2$ is $q=t^2+t+1$: among monic quadratic polynomials a nonzero constant is necessary, and $t^2+1$ has a root, while $q$ has none.

But $t^2\equiv t+1(\mathrm{mod}q)$, so $t^3\equiv 1$ and $t^4\equiv t$. ConsequentlyThus $q$ does not divide $\overline{f}$, excluding the remaining possible factorization. We obtain

Eisenstein's criterion states that an integer-coefficient polynomial is irreducible over $\mathbb{Q}$ if some prime number divides every nonleading coefficient, does not divide the leading coefficient, and its square does not divide the constant coefficient. Equivalently apply this to the primitive part when phrased for primitive polynomials.

Take $p=2$. Every nonleading coefficient of $t^5+4t+2$ is divisible by $2$ (including the zero coefficients), its leading coefficient is $1$, and its constant coefficient $2$ is not divisible by $4$. Therefore

A cubic polynomial over a field is reducible if and only if it has a root in that field: a nontrivial factorization of degree three has a linear polynomial factor, and conversely a root supplies one by the factor theorem.

The rational root theorem, or part (b), shows that a rational root of this monic polynomial must be an integer. Such an integer must divide the constant coefficient $4$, so the only candidates are $\pm 1,\pm 2,\pm 4$. Direct evaluation givesNone vanishes. Hence there is no rational root, provingThe same no-root argument alone would not decide a quartic polynomial, which could factor as two quadratic polynomials; its sufficiency here uses degree three.