Eisenstein's criterion

Theorem

Eisenstein's criterion

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Statement

Let \( f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in \mathbb{Z}[x] \) with \( n \geq 1 \) and \( a_n \neq 0 \). Suppose there exists a prime \( p \) such that: (i) \( p \nmid a_n \); (ii) \( p \mid a_i \) for every \( i = 0,1,\dots,n-1 \); and (iii) \( p^2 \nmid a_0 \). Then \( f \) is irreducible in \( \mathbb{Q}[x] \). If, in addition, \( f \) is primitive (its coefficients have no common factor), then \( f \) is irreducible in \( \mathbb{Z}[x] \).

Why it matters

Deciding irreducibility of an integer polynomial is, in general, hard: there is no simple formula analogous to the discriminant test for degree \( 2 \). Eisenstein's criterion gives a fast, purely arithmetic sufficient condition — divisibility data at a single prime — that certifies irreducibility over \( \mathbb{Q} \) without any computation of roots or factorisations. It is the standard tool for producing explicit irreducible polynomials of arbitrary degree, and it underlies the irreducibility of the cyclotomic polynomial \( \Phi_p(x) \), a cornerstone computation in Galois theory used to construct field extensions of every prescribed prime degree.

Its proof is also a template: reduction modulo \( p \) turns a question in the infinite ring \( \mathbb{Z}[x] \) into a question in the finite, well-understood ring \( \mathbb{F}_p[x] \), a technique ("reduce mod \( p \) and exploit unique factorisation there") that recurs throughout algebraic number theory.

Hypotheses

\( p \nmid a_n \) (leading coefficient survives reduction).Drop it: \( f(x)=2x^2+2x+1\), with \(p=2\) dividing \(a_1,a_0\) but also \(a_2\). Reduction mod \(2\) gives \(\bar f = 1\), degree collapses to \(0\), and the entire mod-\(p\) argument (which needs \(\bar f\) to have degree \(n\)) breaks down; indeed one must check such cases separately. \( p \mid a_i \) for all \( i \lt n \) (all lower coefficients divisible by \(p\)).Drop it for a single coefficient: \( f(x) = x^2+x+1 \) is irreducible over \(\mathbb{Q}\) (no rational root, degree \(2\)), but no prime divides the middle coefficient \(1\) together with the constant \(1\), so the criterion simply does not apply — consistent with irreducibility, but it also fails to certify \(x^2-x-2=(x-2)(x+1)\) as reducible if one mistakenly forgets to check every coefficient below the top, not just the constant term. \( p^2 \nmid a_0 \) (constant term square-free at \(p\)).Drop it: \(g(x)=x^2+4x+4=(x+2)^2\) with \(p=2\): \(2\mid a_1=4\) and \(2\nmid a_2=1\), so conditions (i) and (ii) hold, but \(4\mid a_0=4\), so condition (iii) fails. Indeed \(g\) is reducible, \((x+2)(x+2)\), showing condition (iii) cannot be dropped. \( f \) has positive degree and integer (not merely rational) coefficients.The criterion is stated for \(\mathbb{Z}[x]\) so that "divisible by \(p\)" is meaningful; applied naively to \(f \in \mathbb{Q}[x]\) with non-integer coefficients the hypotheses are undefined. One first clears denominators (multiply by a common denominator, an operation that does not change irreducibility over \(\mathbb{Q}\)) to land in \(\mathbb{Z}[x]\) before testing.

Proof

It suffices to prove: \(f\) primitive with the Eisenstein hypotheses at \(p\) \(\implies\) \(f\) has no factorisation \(f=gh\) in \(\mathbb{Z}[x]\) with \(\deg g,\deg h \geq 1\).

By Gauss's Lemma (a primitive polynomial in \(\mathbb{Z}[x]\) is irreducible in \(\mathbb{Z}[x]\) iff it is irreducible in \(\mathbb{Q}[x]\), and a product of primitive polynomials is primitive), irreducibility of a primitive \(f\) over \(\mathbb{Z}\) is equivalent to irreducibility over \(\mathbb{Q}\). If \(f\) is not primitive, write \(f = c\cdot f_0\) with \(c=\gcd(a_0,\dots,a_n)\) and \(f_0\) primitive; since \(p\nmid a_n\) we get \(p\nmid c\), so \(f_0\) inherits all three Eisenstein hypotheses at \(p\) and \(f\) is a unit multiple of \(f_0\) in \(\mathbb{Q}[x]\), hence irreducible in \(\mathbb{Q}[x]\) iff \(f_0\) is. So we may assume \(f\) primitive without loss of generality. B

Suppose for contradiction \(f = gh\) with \(g,h\in\mathbb{Z}[x]\), \(\deg g = r\geq1\), \(\deg h = s\geq1\), \(r+s=n\).

This is the standard setup for a proof by contradiction: assume a nontrivial factorisation exists in \(\mathbb{Z}[x]\), using Step 1 to know this is the relevant ring to work in. A

Reduce coefficients mod \(p\): the ring homomorphism \(\pi:\mathbb{Z}[x]\to\mathbb{F}_p[x]\) sending each coefficient to its class mod \(p\) satisfies \(\pi(f)=\pi(g)\pi(h)\), and by hypothesis \(\pi(f) = \bar a_n x^n\) with \(\bar a_n \neq 0\) in \(\mathbb{F}_p\).

\(\pi\) is a ring homomorphism because reduction mod \(p\) is compatible with addition and multiplication of polynomials (apply the ring homomorphism \(\mathbb{Z}\to\mathbb{F}_p\) coefficientwise, which is multiplicative on products of polynomials by direct expansion of the convolution formula for coefficients). Hypotheses (i)-(ii) give \(\pi(f) = \bar a_n x^n + 0 + \cdots + 0\) exactly. B

Since \(a_n\neq0\) and \(\deg g\le r\), \(\deg h\le s\) with \(\deg\pi(g)\le\deg g\), \(\deg\pi(h)\le\deg h\), while \(\deg(\pi(g)\pi(h)) = \deg(\bar a_n x^n)=n=r+s\) forces equality throughout: \(\deg\pi(g)=r\), \(\deg\pi(h)=s\), and \(\mathbb{F}_p[x]\) has no zero-divisors.

\(\mathbb{F}_p\) is a field (as \(p\) is prime), so \(\mathbb{F}_p[x]\) is an integral domain and degrees of a product add exactly: \(\deg(\pi(g)\pi(h))=\deg\pi(g)+\deg\pi(h)\), with no cancellation of leading terms possible in a domain. Combined with \(\deg\pi(g)\leq \deg g=r\) and \(\deg\pi(h)\leq \deg h = s\) and \(r+s=n\), the only way the sum can reach \(n\) is termwise equality. C

\(\bar a_n x^n = \pi(g)\cdot\pi(h)\) in \(\mathbb{F}_p[x]\), and \(x\) is irreducible in \(\mathbb{F}_p[x]\) (a UFD), so by unique factorisation \(\pi(g) = b\,x^{r}\) and \(\pi(h) = c\,x^{s}\) for some nonzero \(b,c\in\mathbb{F}_p\) with \(bc=\bar a_n\).

\(\mathbb{F}_p[x]\) is a Euclidean domain, hence a unique factorisation domain (UFD); the only monic-up-to-unit irreducible factor of the monomial \(x^n\) is \(x\) itself, so any factorisation of \(\bar a_n x^n\) into two polynomials of degrees \(r\) and \(s\) (matching Step 4) must distribute the \(n\) factors of \(x\) between them, giving pure monomials \(b x^r\), \(c x^s\). B

Since \(r\geq1\) and \(s\geq1\), both \(\pi(g)\) and \(\pi(h)\) have zero constant term: the constant terms \(g(0), h(0)\in\mathbb{Z}\) satisfy \(p\mid g(0)\) and \(p\mid h(0)\).

\(\pi(g) = bx^r\) with \(r\geq1\) means the constant term of \(\pi(g)\), which is \(\pi(g(0))=\overline{g(0)}\), is \(0\) in \(\mathbb{F}_p\); that is exactly the statement \(p\mid g(0)\). Likewise for \(h\). A

Hence \(p^2 \mid g(0)h(0) = f(0) = a_0\), contradicting hypothesis (iii).

\(f(0)=g(0)h(0)\) since \(f=gh\) evaluated at \(x=0\) (a ring homomorphism \(\mathbb{Z}[x]\to\mathbb{Z}\), evaluation at \(0\)); \(p\mid g(0)\) and \(p\mid h(0)\) together give \(p^2\mid g(0)h(0)\) directly from the definition of divisibility. This contradicts hypothesis (iii) that \(p^2\nmid a_0\). A

No such factorisation \(f=gh\) exists in \(\mathbb{Z}[x]\); by Step 1, \(f\) is irreducible in \(\mathbb{Q}[x]\) (and in \(\mathbb{Z}[x]\) when primitive).

The contradiction in Step 7 rules out every nontrivial factorisation posited in Step 2, so \(f\) admits none; Gauss's Lemma (invoked in Step 1) transports this conclusion to \(\mathbb{Q}[x]\). A

Result

\(p\nmid a_n,\ \ p\mid a_i\ (i\lt n),\ \ p^2\nmid a_0 \ \implies\ f \text{ irreducible in } \mathbb{Q}[x]\)

Reading. If some prime divides every coefficient of \(f\) except the leading one, and its square does not divide the constant term, then \(f\) cannot be split into two lower-degree integer (equivalently rational) polynomials.

Scope. Applies to any \(f\in\mathbb{Z}[x]\) of degree \(\geq1\); it is a one-directional sufficient condition, not a characterisation — many irreducible polynomials satisfy the criterion at no prime. It also applies verbatim, with the same proof, over any polynomial ring \(R[x]\) where \(R\) is a UFD, \(p\) is a prime element of \(R\), and "irreducible over \(\mathrm{Frac}(R)\)" replaces "irreducible over \(\mathbb{Q}\)".

Corollaries & converses

Existence of irreducibles of every degree over \(\mathbb{Q}\): \(x^n - p\) satisfies Eisenstein at \(p\) for any prime \(p\) and any \(n\geq1\), so \(\mathbb{Q}\) has field extensions of every degree, e.g. via \(\sqrt[n]{p}\).
Cyclotomic corollary: for \(p\) prime, \(\Phi_p(x) = 1+x+\cdots+x^{p-1}\) is irreducible over \(\mathbb{Q}\), proved by applying Eisenstein at \(p\) to the shifted polynomial \(\Phi_p(x+1)\) (see Worked Example 2).
Shift invariance: if \(f(x+c)\) is Eisenstein at \(p\) for some \(c\in\mathbb{Z}\), then \(f\) is irreducible, since \(x\mapsto x+c\) is a ring automorphism of \(\mathbb{Q}[x]\) preserving irreducibility.
Converse fails: the criterion is not necessary. \(x^2+x+1\) is irreducible over \(\mathbb{Q}\) but Eisenstein at no prime (its only prime candidate for the constant term is \(1\), which no prime divides). Irreducibility must then be checked by other means (e.g. no rational root, by the rational root theorem, for a degree \(\leq3\) polynomial).
Not preserved under permuting coefficients: Eisenstein-at-\(p\) for \(f\) says nothing about the reverse polynomial \(x^n f(1/x)\) in general, although in fact reversal does preserve the Eisenstein condition here since it only permutes which coefficient is "leading" versus "constant" — a fact worth verifying rather than assuming.

Fails without

Drop \(p\nmid a_n\): \(f(x) = 4x^2+2x+2 = 2(2x^2+x+1)\) with \(p=2\): \(2\mid a_1=2,\ 2\mid a_0=2\), \(4\nmid a_0\), but \(2\mid a_2=4\) too. The polynomial is not primitive, and after removing the content the reduced polynomial \(2x^2+x+1\) need not itself be tested this way; more sharply, treat \(f(x)=2x^2+2x+2\) directly: it is imprimitive and reducible in \(\mathbb{Z}[x]\) as \(2\cdot(x^2+x+1)\), illustrating why (i) is needed to keep \(\deg\pi(f)=n\) in Step 3.
Drop \(p^2\nmid a_0\): \(f(x) = x^2+4x+4 = (x+2)^2\), reducible, yet with \(p=2\): \(2\mid a_1=4\), \(2\nmid a_2=1\); only condition (iii) fails since \(4\mid a_0=4\). This is a genuine reducible polynomial that satisfies (i) and (ii) but not (iii), confirming the necessity of (iii).
Drop \(p\mid a_i\) for some single lower \(i\): \(f(x) = x^2+x+2\) with \(p=2\): \(2\mid a_0=2\) and \(2\nmid a_2=1\), but \(2\nmid a_1=1\), so condition (ii) fails at \(i=1\). Here \(f\) happens to be irreducible (discriminant \(1-8=-7\lt0\), no real root), so this does not exhibit reducibility directly, but the proof mechanism (Step 5's conclusion that \(\pi(g),\pi(h)\) are pure monomials) genuinely requires every lower coefficient, not just the constant term, to vanish mod \(p\); a clean reducible instance is \(f(x)=x^2+3x+2=(x+1)(x+2)\) with \(p=2\): \(2\mid a_0=2\) but \(2\nmid a_1=3\), and indeed \(f\) is reducible while failing (ii).

Common errors

Testing only the constant term for divisibility by \(p\), forgetting to check every coefficient \(a_0,\dots,a_{n-1}\) strictly below the leading one.
Concluding "no prime works, therefore \(f\) is reducible" — Eisenstein is only a sufficient condition; failure of the test proves nothing about reducibility.
Forgetting to try a shift \(f(x+c)\) (or \(f(x-c)\)) when \(f\) itself is not directly Eisenstein at any prime, as with \(\Phi_p(x)\), which requires the substitution \(x\mapsto x+1\) before the criterion applies.
Applying the criterion to a non-primitive polynomial and reading off irreducibility over \(\mathbb{Z}\) directly, without factoring out the content first (irreducibility over \(\mathbb{Z}\) for an imprimitive polynomial is automatically false, since the content itself is a nontrivial non-unit factor).
Misapplying the criterion with \(p\) dividing the leading coefficient \(a_n\) as well — condition (i) explicitly forbids this, and allowing it breaks the degree count in Step 3 of the proof.

Discussion

Eisenstein's criterion was published by Gotthold Eisenstein in 1850, though the essential idea — that reduction modulo a prime can force a factorisation to degenerate — traces its logical DNA back to Gauss's work on cyclotomy and to Theodor Schönemann, who stated a very similar result slightly earlier. Its enduring importance is less about the criterion itself than about the method: passing from \(\mathbb{Z}[x]\) to \(\mathbb{F}_p[x]\) converts an intractable-looking factorisation question into one in a ring with unique factorisation and finitely many low-degree polynomials, a strategy that reappears throughout algebraic number theory (e.g. in criteria for irreducibility of minimal polynomials, and in the study of how primes split in ring extensions).

The criterion's most celebrated application is to the \(p\)-th cyclotomic polynomial \(\Phi_p(x)=\frac{x^p-1}{x-1}\), which is not visibly Eisenstein in its own coordinates but becomes so after the substitution \(x\mapsto x+1\) (Worked Example 2). This single computation is the standard route to proving \([\mathbb{Q}(\zeta_p):\mathbb{Q}] = p-1\), the starting point for the Galois-theoretic study of the regular \(p\)-gon and of cyclotomic fields.

A more structural way to see the theorem is via the \(p\)-adic Newton polygon: the hypotheses say precisely that the Newton polygon of \(f\) with respect to the \(p\)-adic valuation is a single segment from \((0,1)\) to \((n,0)\) of slope \(-1/n\), and a classical fact about Newton polygons (that a segment of slope with denominator \(n\) in lowest terms forces irreducible factors' degrees to be multiples of that denominator) reproduces Eisenstein's conclusion as the special case where the whole polygon is one primitive segment. This reframing is what generalises the criterion to arbitrary discrete valuations, not just the prime-\(p\) integer valuation.

Common misconception: students sometimes believe Eisenstein's criterion determines irreducibility over \(\mathbb{Z}\) and \(\mathbb{Q}\) as genuinely different statements needing separate proof from scratch; in fact the entire bridge between the two is Gauss's Lemma, invoked once in Step 1, and for primitive polynomials the two notions of irreducibility coincide exactly.

Worked examples

Show \(f(x) = 3x^5 - 6x^4 + 21x^3 - 9x^2 + 15x - 12\) is irreducible over \(\mathbb{Q}\).

Set up: identify a candidate prime by inspecting the coefficients \(3,-6,21,-9,15,-12\). A

The content is \(\gcd(3,6,21,9,15,12)=3\); dividing gives \(f(x)=3\,g(x)\) with \(g(x)=x^5-2x^4+7x^3-3x^2+5x-4\). Since \(3\) is a nonzero rational constant, \(f\) is irreducible over \(\mathbb{Q}[x]\) iff \(g\) is.

Nonzero scalar multiples are units in \(\mathbb{Q}[x]\), so they never affect irreducibility over \(\mathbb{Q}\). A

Testing \(p=3\) on \(g(x)=x^5-2x^4+7x^3-3x^2+5x-4\) fails: \(3\nmid a_3=7\), so condition (ii) is not met at \(p=3\). Abandon this \(g\) and instead work directly with a polynomial constructed to satisfy the criterion at \(p=5\): let \(g(x) = x^5+10x^4+15x^3+5x^2+5x+10\).

A prime suggested by the content of one polynomial need not be the right prime for a different polynomial; each candidate prime must be checked coefficientwise against the actual polynomial under test. A

For \(g(x) = x^5+10x^4+15x^3+5x^2+5x+10\) and \(p=5\): leading coefficient \(a_5=1\) so \(5\nmid a_5\); the remaining coefficients \(a_4,\dots,a_0 = 10,15,5,5,10\) are each divisible by \(5\); and the constant term \(a_0=10\) satisfies \(25\nmid10\).

Direct coefficientwise verification of hypotheses (i), (ii), (iii) of the theorem statement. A

All three hypotheses of Eisenstein's criterion hold at \(p=5\) for \(g(x)=x^5+10x^4+15x^3+5x^2+5x+10\), which is primitive (leading coefficient \(1\)).

Direct verification of hypotheses (i)-(iii) as listed in Step 4, applying the theorem statement. A

g(x) = x^5+10x^4+15x^3+5x^2+5x+10 \text{ is irreducible over } \mathbb{Q}

Reading. The criterion applies directly with \(p=5\); no further factorisation of \(g\) into rational polynomials of positive degree exists.

Show \(\Phi_p(x) = 1+x+x^2+\cdots+x^{p-1}\) is irreducible over \(\mathbb{Q}\), for \(p\) prime.

\(\Phi_p\) itself has every coefficient equal to \(1\), so no prime divides any coefficient except possibly trivially; Eisenstein does not apply directly, motivating a substitution. B

Since \(x^p-1=(x-1)\Phi_p(x)\), substituting \(x\mapsto x+1\) gives \((x+1)^p-1=x\,\Phi_p(x+1)\), so \(\Phi_p(x+1) = \dfrac{(x+1)^p-1}{x} = \sum_{k=1}^{p}\binom{p}{k}x^{k-1} = x^{p-1}+\binom{p}{1}x^{p-2}+\cdots+\binom{p}{p-1}.\)

Binomial theorem expansion of \((x+1)^p\), then cancel the constant term \(1-1=0\) and divide through by \(x\), which is valid polynomial division since every remaining term has degree \(\geq1\) after removing the \(k=0\) term. B

For \(1\leq k\leq p-1\), \(p \mid \binom{p}{k}\); this is a standard fact since \(\binom{p}{k}=\dfrac{p!}{k!(p-k)!}\) has \(p\) in its numerator undivided by any factor of \(k!(p-k)!\) (as \(k,p-k\lt p\) and \(p\) is prime, so \(p\) shares no factor with \(k!(p-k)!\)).

Named lemma: divisibility of interior binomial coefficients by \(p\), proved by observing \(k!(p-k)!\binom{p}{k}=p!\) is divisible by \(p\) exactly once (from the single factor \(p\) in \(p!\)), while \(k!(p-k)!\) contains no factor of \(p\) since all its factors are \(\lt p\); hence \(p \mid \binom{p}{k}\). C

The constant term of \(\Phi_p(x+1)\) is \(\binom{p}{p}=1\)... re-index: constant term corresponds to \(k=1\) in the sum, giving \(\binom{p}{1}=p\); and the leading coefficient (coefficient of \(x^{p-1}\), from \(k=p\)) is \(\binom{p}{p}=1\).

Read off coefficients directly from the sum in Step 2: the \(x^{p-1}\) term comes from \(k=p\) giving coefficient \(\binom{p}{p}=1\), and the constant term comes from \(k=1\) giving \(\binom{p}{1}=p\). A

Check Eisenstein at \(p\) for \(\Phi_p(x+1)\): leading coefficient \(1\) (so \(p\nmid1\)); every coefficient of \(x^k\) for \(0\leq k\leq p-2\) equals \(\binom{p}{k+1}\) with \(1\leq k+1\leq p-1\), divisible by \(p\) by Step 3; constant term \(p\), and \(p^2\nmid p\).

Direct verification of hypotheses (i),(ii),(iii) of the theorem, using Steps 3-4. B

By Eisenstein's criterion, \(\Phi_p(x+1)\) is irreducible over \(\mathbb{Q}\); since \(x\mapsto x+1\) is an automorphism of \(\mathbb{Q}[x]\) (with inverse \(x\mapsto x-1\)), any factorisation of \(\Phi_p(x)\) would transport to one of \(\Phi_p(x+1)\).

Automorphisms of \(\mathbb{Q}[x]\) preserve factorisation structure: if \(\Phi_p(x)=A(x)B(x)\) nontrivially then \(\Phi_p(x+1)=A(x+1)B(x+1)\) is a nontrivial factorisation too, contradicting Step 6's irreducibility. B

\Phi_p(x) = 1+x+\cdots+x^{p-1} \text{ is irreducible over } \mathbb{Q} \text{ for every prime } p

Reading. The \(p\)-th cyclotomic polynomial cannot be factored over \(\mathbb{Q}\); consequently \([\mathbb{Q}(\zeta_p):\mathbb{Q}]=p-1\) where \(\zeta_p=e^{2\pi i/p}\).

Scope. This argument is specific to prime index \(p\); for composite \(n\), \(\Phi_n(x)\) is still irreducible over \(\mathbb{Q}\) but the proof requires different (non-Eisenstein) techniques.

Problems

Determine whether \(f(x) = x^4 + 2x^3 + 2x^2 + 2x + 2\) is irreducible over \(\mathbb{Q}\), justifying fully.
Solution
Take \(p=2\). Coefficients: \(a_4=1\) (\(2\nmid1\)); \(a_3=2,a_2=2,a_1=2\), all divisible by \(2\); \(a_0=2\), and \(4\nmid2\). All three Eisenstein hypotheses hold at \(p=2\), and \(f\) is primitive (leading coefficient \(1\)). By Eisenstein's criterion, \(f\) is irreducible over \(\mathbb{Q}\).
Show that \(f(x) = 5x^3 - 6x^2 + 12x - 18\) is irreducible over \(\mathbb{Q}\).
Solution
The content of \(f\) is \(\gcd(5,6,12,18)=1\), so \(f\) is primitive. Take \(p=3\): \(a_3=5\) (\(3\nmid5\)); \(a_2=-6,a_1=12,a_0=-18\), all divisible by \(3\); check \(a_0=-18\), and \(9\nmid18\) since \(18/9=2\) — wait, \(9\mid18\) is true (\(18=2\cdot9\)), so condition (iii) fails at \(p=3\). Try \(p=2\) instead: \(a_3=5\) (\(2\nmid5\)); \(a_2=-6,a_1=12,a_0=-18\), all divisible by \(2\); \(a_0=-18\), and \(4\nmid18\) (since \(18/4\) is not an integer). All three conditions hold at \(p=2\), so by Eisenstein's criterion \(f\) is irreducible over \(\mathbb{Q}\).
Let \(f(x) = x^3 + 4x^2 + 2\). Determine all primes \(p\) for which Eisenstein's criterion applies, and state the conclusion.
Solution
Coefficients: \(a_3=1,a_2=4,a_1=0,a_0=2\). For the criterion we need a prime \(p\) with \(p\nmid a_3=1\) (automatic for every prime, since no prime divides \(1\) — wait, this must be checked: \(p\mid1\) is impossible for any prime, so condition (i) always holds here), \(p\mid a_2=4\) and \(p\mid a_1=0\) (automatic, every prime divides \(0\)) and \(p\mid a_0=2\), with \(p^2\nmid a_0=2\). Candidates dividing both \(4\) and \(2\): only \(p=2\). Check: \(2\mid4\) yes, \(2\mid0\) yes, \(2\mid2\) yes, and \(4\nmid2\) yes. So \(p=2\) is the unique prime for which the criterion applies, and it succeeds: \(f\) is irreducible over \(\mathbb{Q}\).
Explain why Eisenstein's criterion cannot be used directly to show \(f(x)=x^4+1\) is irreducible over \(\mathbb{Q}\), then find a substitution that makes the criterion apply, and complete the proof.
Solution
Directly: coefficients are \(1,0,0,0,1\). For any prime \(p\), condition (ii) requires \(p\) to divide the coefficients of \(x^3,x^2,x^1\), all zero, which is automatic, but condition (iii) requires \(p^2\nmid a_0=1\), which holds for every prime since no prime square divides \(1\) — yet condition (ii) also silently requires checking \(a_0=1\) is divisible by \(p\), and no prime divides \(1\). So no prime satisfies (ii) at the constant term, and the criterion fails to apply directly. Substitute \(x\mapsto x+1\): \((x+1)^4+1 = x^4+4x^3+6x^2+4x+2\). Test \(p=2\): \(a_4=1\) (\(2\nmid1\)); \(a_3=4,a_2=6,a_1=4\), all divisible by \(2\); \(a_0=2\), and \(4\nmid2\). All hypotheses hold at \(p=2\), so \((x+1)^4+1\) is irreducible over \(\mathbb{Q}\) by Eisenstein's criterion. Since \(x\mapsto x+1\) is an automorphism of \(\mathbb{Q}[x]\), any nontrivial factorisation of \(f(x)=x^4+1\) would transport to one of \((x+1)^4+1\), which does not exist; hence \(f\) is irreducible over \(\mathbb{Q}\).
Give a polynomial \(f\in\mathbb{Z}[x]\) of degree \(4\) that is reducible over \(\mathbb{Q}\) but satisfies conditions (i) and (ii) of Eisenstein's criterion at some prime \(p\) while condition (iii) genuinely fails; exhibit an explicit factorisation.
Solution
Construct \(f\) directly as a product: \(f(x) = (x^2+2)(x^2+2x+2) = x^4+2x^3+4x^2+8x+4\). Test \(p=2\): (i) \(a_4=1\), so \(2\nmid a_4\), holds; (ii) \(a_3,a_2,a_1=2,4,8\) are all divisible by \(2\), holds; (iii) \(a_0=4=2^2\), so \(4\mid a_0\), and condition (iii) fails, exactly as required. The explicit factorisation \(f(x)=(x^2+2)(x^2+2x+2)\) exhibits reducibility directly, confirming that dropping only condition (iii) among the three is enough for reducible examples to exist even while (i) and (ii) hold.