Eisenstein's Irreducibility Criterion

Eisenstein's criterion is the standard route to proving $\Phi_p(x) = 1+x+\cdots+x^{p-1}$ is irreducible over $\mathbb{Q}$ for prime $p$, which is foundational in constructing cyclotomic field extensions $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ used throughout algebraic number theory and in classical results such as Gauss's constructibility criterion for regular $n$-gons. In Galois theory courses, $x^n-p$ for prime $p$ gives instant, verifiable examples of degree-$n$ irreducible polynomials, used to build field extensions of prescribed degree and to exhibit Galois groups such as cyclic or non-abelian groups explicitly. In computational algebra and computer algebra systems, Eisenstein-type tests (and their generalization to Newton polygons over $p$-adic valuations) are cheap, fast pre-filters run before more expensive factorization algorithms such as those based on Berlekamp/Zassenhaus or LLL-lattice methods. In algebraic geometry and commutative algebra, the same total-degeneration argument generalizes to Eisenstein-type criteria for irreducibility of polynomials over more general local rings, tying into valuation theory and Newton polygons used in studying ramification in local fields. In cryptography, irreducible polynomials over finite fields (constructed or verified via reduction mod $p$ ideas closely related to this circle of techniques) are used to build the finite field arithmetic underlying AES and elliptic-curve cryptosystems.

:::quiz[If $f \in \mathbb{Z}[x]$ is Eisenstein at $p$ but has content $c(f) = 3 \neq \pm 1$, what can we conclude?]

• $f$ is irreducible in $\mathbb{Z}[x]$
• $f$ is irreducible in $\mathbb{Q}[x]$, but reducible in $\mathbb{Z}[x]$ as $3\cdot(f/3)$ ✓
• $f$ is reducible in $\mathbb{Q}[x]$
• The criterion cannot be applied at all

Ask the tutor: How do I decide which linear substitution $x\mapsto x+c$ will reveal an Eisenstein prime when the original polynomial isn't Eisenstein at any prime?

Think of the coefficients of $f$ as a vector, and imagine reducing every coefficient modulo $p$. Condition (2) says that the reduced polynomial $\bar f \in (R/pR)[x]$ has every coefficient below degree $n$ killed; condition (1) says the top coefficient survives. So in the quotient ring, $f$ collapses entirely to a single term:

This total collapse is the heart of the mechanism. If $f = gh$ were a nontrivial factorization in $R[x]$ with $\deg g = r$, $\deg h = s$, $r+s=n$, $r,s\ge 1$, then reducing mod $p$ gives $\bar a_n x^n = \bar g \bar h$ in $(R/pR)[x]$. Since $p$ is prime, $R/pR$ is an integral domain, and unique factorization of $x^n$ there forces both $\bar g$ and $\bar h$ to be (unit multiples of) pure powers of $x$ — meaning all of the lower coefficients of $g$ and of $h$ are individually divisible by $p$. In particular the constant terms of $g$ and $h$ are both divisible by $p$, so their product $a_0$ is divisible by $p^2$. This is precisely what condition (3) forbids, giving a contradiction. Geometrically, picture the coefficient vector of $f$ living in $R^{n+1}$, with a "shell" at $p$-divisibility. Eisenstein's condition says all coefficients except the leading one lie inside the shell $p \mid a_i$, but the constant term sits in the thin ring $p \mid a_0$, $p^2 \nmid a_0$ — one layer only, not two. Any factorization would need to distribute this single layer of $p$-divisibility between two constant terms, which is arithmetically impossible unless one of the factors contributes zero layers — forcing that factor to have unit constant term and hence, chasing back up, forcing the whole "other" factor to be a non-constant polynomial divisible by $p$ throughout, which contradicts $p \nmid a_n$.

Example 1: An asymmetric Eisenstein polynomial over $\mathbb{Z}$. Let

Test $p=3$. The coefficients are $a_5=4,\ a_4=-6,\ a_3=15,\ a_2=-21,\ a_1=33,\ a_0=-6$.

• $p=3$ does not divide $a_5 = 4$: condition (i) holds.
• $3 \mid -6,\ 15,\ -21,\ 33$: indeed $-6=3(-2)$, $15=3(5)$, $-21=3(-7)$, $33=3(11)$; condition (ii) holds for $a_4,a_3,a_2,a_1$.
• $a_0 = -6 = 3(-2)$, so $3\mid a_0$; and $9 \nmid -6$ since $-6/9$ is not an integer: condition (iii) holds.

All three Eisenstein conditions hold at $p=3$. The content of $f$: $\gcd(4,6,15,21,33,6)=1$, so $f$ is primitive. By the theorem, $f$ is irreducible in $\mathbb{Z}[x]$, and hence (by Gauss's Lemma) irreducible in $\mathbb{Q}[x]$. Note the asymmetry deliberately built in: the leading coefficient $4$ is not itself $1$, and it is not divisible by $3$ even though $4$ is even — this shows the criterion does not require the leading coefficient to be a unit in $\mathbb{Z}$, only that it avoid the specific prime $p$. Also $\gcd(4, -6)=2\ne 1$ momentarily suggests $2$ might be a useful test prime, but $2$ fails condition (i) since $2\mid a_5=4$; the criterion must be checked prime-by-prime, and only $p=3$ works here.

Example 2: The cyclotomic polynomial $\Phi_5(x)$, via a shift trick. Let $q=5$ be prime and consider

Eisenstein's criterion does not apply to $\Phi_5$ directly: all its coefficients equal $1$, and no prime divides the non-leading coefficients. The standard trick is to substitute $x \mapsto x+1$ and apply the criterion to the shifted polynomial; irreducibility of $g(x):=\Phi_5(x+1)$ in $\mathbb{Q}[x]$ is equivalent to irreducibility of $\Phi_5(x)$, since the substitution $x\mapsto x+1$ is an automorphism of $\mathbb{Q}[x]$ (with inverse $x\mapsto x-1$), and an automorphism carries irreducible elements to irreducible elements (a factorization of $g$ pulls back, via the inverse automorphism, to a factorization of $\Phi_5$ of the same shape). Compute $g(x) = \Phi_5(x+1) = \dfrac{(x+1)^5-1}{(x+1)-1} = \dfrac{(x+1)^5-1}{x}$. Expand:

so

and dividing by $x$:

Now test $p=5$: leading coefficient $a_4=1$, so $5\nmid a_4$ (condition (i)); the remaining coefficients are $5,10,10,5$, all divisible by $5$ (condition (ii)); the constant term is $5$, and $25\nmid 5$ (condition (iii)). All conditions hold, $g$ is monic hence automatically primitive, so by the theorem $g$ is irreducible in $\mathbb{Z}[x]$ and in $\mathbb{Q}[x]$. Pulling back through the automorphism $x\mapsto x-1$, $\Phi_5(x)$ is irreducible in $\mathbb{Q}[x]$. This example illustrates the standard extension of the technique: when $f$ itself is not directly Eisenstein at any prime, a linear substitution $x \mapsto x+c$ (which preserves irreducibility since it is invertible over $\mathbb{Q}$) can produce a shifted polynomial to which the criterion does apply — exactly the mechanism behind the general proof that $\Phi_q(x) = 1+x+\cdots+x^{q-1}$ is irreducible for every prime $q$, via $g(x) = \Phi_q(x+1) = \frac{(x+1)^q-1}{x}$, whose coefficients (aside from the leading $1$) are the binomial coefficients $\binom{q}{1},\dots,\binom{q}{q-1}$, each divisible by $q$ (a standard fact for prime $q$), with constant term $\binom{q}{1}=q$ not divisible by $q^2$.