Over $\mathbb{Q}$ you cannot test every number for a root. Instead there are tools that use integer arithmetic. Together they settle most polynomials you will meet. They start from the rational root theorem and build on the degree-2-or-3 test.
- Rational root test. If $\frac{r}{s}$ in lowest terms is a root of $a_{n}x^{n} + \cdots + a_{0}$ with integer coefficients, then $r \mid a_{0}$ and $s \mid a_{n}$. For degree $2$ or $3$, no rational root means irreducible over $\mathbb{Q}$.
- Gauss's lemma. If a polynomial with integer coefficients factors over $\mathbb{Q}$, it factors into polynomials with integer coefficients of the same degrees. So we may work in $\mathbb{Z}[x]$.
- Eisenstein's criterion. If a prime $p$ divides every coefficient except the leading one, does not divide the leading coefficient, and $p^{2}$ does not divide the constant term, then the polynomial is irreducible over $\mathbb{Q}$.
- Reduction mod $p$. If $p$ does not divide the leading coefficient and the reduction mod $p$ is irreducible over $\mathbb{Z}_{p}$ (of the same degree), then the polynomial is irreducible over $\mathbb{Q}$.
Why it works
Why Eisenstein works. Suppose $f = gh$ in $\mathbb{Z}[x]$ (Gauss's lemma) with both factors of positive degree, and reduce mod $p$. All coefficients of $f$ but the leading one vanish, so $\bar{f} = \bar{a}_{n}x^{n}$. Since $\mathbb{Z}_{p}[x]$ has unique factorization, $\bar{g}$ and $\bar{h}$ must also be monomials $\bar{b}x^{k}$ and $\bar{c}x^{n-k}$ with $k, n - k \ge 1$ (their degrees cannot drop, because $p$ does not divide the leading coefficient of $f$). So $p$ divides the constant terms of both $g$ and $h$, and $p^{2}$ divides their product, the constant term of $f$: a contradiction. $\square$ Why reduction works: a factorization $f = gh$ over $\mathbb{Z}$ reduces to a factorization $\bar{f} = \bar{g}\bar{h}$ of the same shape.
Worked example: Eisenstein
Show that $x^{4} + 10x + 5$ is irreducible over $\mathbb{Q}$.
- Take $p = 5$: $5$ divides $10$ and $5$ (and the zero coefficients), $5 \nmid 1$, and $25 \nmid 5$.
Answer: Irreducible by Eisenstein at $5$.
Worked example: A cubic
Is $x^{3} - 3x - 1$ irreducible over $\mathbb{Q}$?
- Rational roots must be $\pm 1$.
- $f(1) = 1 - 3 - 1 = -3$ and $f(-1) = -1 + 3 - 1 = 1$.
- No rational root, and a cubic with no root is irreducible.
Answer: Irreducible. (Eisenstein does not apply to it at any prime.)
Worked example: A shift makes Eisenstein work
Show that $f(x) = x^{4} + x^{3} + x^{2} + x + 1$ is irreducible over $\mathbb{Q}$.
- $f(x) = \frac{x^{5} - 1}{x - 1}$, so $f(x + 1) = \frac{(x + 1)^{5} - 1}{x} = x^{4} + 5x^{3} + 10x^{2} + 10x + 5$.
- Eisenstein at $5$ applies to $f(x + 1)$.
- A factorization of $f(x)$ would give one of $f(x + 1)$, so $f$ is irreducible.
Answer: Irreducible.
Worked example: Reduction mod 2
Show that $x^{3} + 3x + 7$ is irreducible over $\mathbb{Q}$.
- Mod $2$ it becomes $x^{3} + x + 1$, which has no root in $\mathbb{Z}_{2}$ ($f(0) = 1$, $f(1) = 1$), so it is irreducible over $\mathbb{Z}_{2}$.
- The leading coefficient $1$ is not divisible by $2$, so $x^{3} + 3x + 7$ is irreducible over $\mathbb{Q}$.
Answer: Irreducible.
Common mistake
Treating a failed test as a proof of reducibility. If Eisenstein does not apply at any prime, or the reduction mod $3$ factors, that proves nothing: $x^{3} - 3x - 1$ above is irreducible although Eisenstein fails. Also never reduce mod a prime that divides the leading coefficient: $2x^{2} + x$ over $\mathbb{Q}$ is reducible ($x(2x + 1)$), but mod $2$ it becomes $x$, of lower degree.
Key idea
Over $\mathbb{Q}$: rational root test (enough in degree 2 or 3), Eisenstein's criterion (possibly after a shift $x \mapsto x + 1$), and reduction mod a prime not dividing the leading coefficient. A test that fails proves nothing.