Math 594. Eisenstein criterion and Gauss’ Lemma

Let R be a UFD with fraction field K. In class, we proved the following irreducibility criterion due to Eisenstein, conditional on a lemma of Gauss which we prove in this handout. Recall Eisenstein’s criterion: if f = a_{n}X^{n} + ⋯ + a_{0} ∈ R[X] has positive degree n and π is a prime of R which does not divide a_{n} but does divide a_{i} for all i < n, yet π^{2} ∤ a_{0}, then f is irreducible in K[X].

Before proving Gauss’ Lemma, let’s give one example of Eisenstein’s criterion in action (the trick of “translation”) and one non-example to show how the criterion can fail if we drop primality as a condition on π (recall that in the proof of Eisenstein’s criterion, the role of π being prime was crucial for knowing that R/π is a domain, so (R/π)[X] is a domain).

Example 0.1. Let p be a prime in Z and consider Φ_{p} = X^{p−1} + ⋯ + X + 1 ∈ Z[X]. We claim that this is irreducible in Q[X] (e.g., Φ_{2} = X + 1 and Φ_{3} = X^{2} + X + 1, but for p ≥ 5 it is clear that hands-on manipulation is pointless). To prove Φ_{p} is irreducible in Q[X], we make the change of variable X ⇝ X + 1: it is equivalent to show Φ_{p}(X + 1) is irreducible. But this is a polynomial to which Eisenstein applies in Z for the prime p. Indeed, in F_{p}[X] we have by the magic of characteristic p that

(X − 1)Φ_{p}(X) = X^{p} − 1 = (X − 1)^{p},

so Φ_{p}(X) ≡ (X − 1)^{p−1} mod p, so Φ_{p}(X + 1) ≡ X^{p−1} mod p. This congruence says that the monic degree p−1 polynomial Φ_{p}(X +1) ∈ Z[X] has all lower degree coefficients divisible by p, so we just have to make sure the constant term of Φ_{p}(X + 1) is not divisible by p^{2}. Ah, but the constant term is obtained by specializing the variable to zero, and Φ_{p}(1) = p by inspection.

Example 0.2. To show what goes wrong if we don’t require primality, consider Φ_{5} = X^{4} +X^{3} +X^{2} +X +1 ∈ Z[X]. Applying Eisenstein to Φ_{5}(X + 1) with p = 5 shows irreducibility in Q[X], as we saw above. But consider the ring R = Z[α] where α = (−1 + \sqrt{5})/2. Though α looks “fractional”, in fact it is easy to verify that α^{2} + α − 1 = 0, so in fact α satisfies a monic quadratic with Z coefficients (so R = Z ⊕ Zα). Using some elementary algebraic number theory it can be shown that R is a UFD. But the fraction field K = Q(\sqrt{5}) of R has the property that in K[X] we have the factorization

Φ_{5} = X^{4} + X^{3} + X^{2} + X + 1 = (X^{2} − αX + 1)(X^{2} − (−1 − α)X + 1),

as one checks by using the relation α^{2} = −α + 1.

Thus, Φ_{5} is reducible in K[X] (the factorization even takes place in R[X]), but it is still true in R that Φ_{5}(X + 1) has all lower degree coefficients divisible by 5 (as this is even true in Z) and its constant term Φ_{5}(0) = 5 is not divisible by 5^{2} (one can check that 5/25 = 1/5 does not lie in R). So why didn’t Eisenstein apply to this situation over K with π = 5 ∈ R? What went wrong is that 5 is not prime in R! In fact, \sqrt{5} = 2α + 1 ∈ R has the property that it is a non-unit in R (check!) and its square is 5. Hence, 5 ∈ R admits a non-trivial factorization and thus is not prime.

Theorem 0.3. (Gauss’ Lemma) Let R be a UFD with fraction field K. If f ∈ R[X] has positive degree and f is reducible in K[X], then f = gh with g, h ∈ R[X] having positive degree.

We should give a warning about how careful one has to be concerning factorization statements when the coefficient ring is not a field. For example, consider 2X in Z[X]. Viewing Z[X] as an absract ring in its own right, clearly the element 2X is “reducible” in the sense that it is a product of two non-units (namely, 2 and X). In contrast, 2X ∈ Q[X] is irreducible, since 2 ∈ Q^{×}. The point is that whether or not an element is a unit may change if one passes to a bigger ring (and hence may get more units), so statements of “irreducibility” of an element in R[X] when R is not a field (and hence not all nonzero elements of R are units) must be treated very carefully lest one make oversights.

Proof. (of Theorem 0.3) If f = c · f̃ for some nonzero c ∈ R and some f̃ ∈ R[X], it suffices to treat f̃ instead of f. Thus, by factoring out the greatest common divisor of the coefficients of f (which makes sense since the coefficient ring R is a UFD), we may assume that the coefficients of f have gcd equal to 1. We call such polynomials primitive (e.g., for R = Z, we have that 7X^{2} − 14X + 2 is primitive in Z[X] whereas 6X^{2} − 15 is not).

The key fact we need is that a product of primitives is primitive. To prove it, let g, h ∈ R[X] be such that gh ∈ R[X] is not primitive. We wish to prove that one of g or h is not primitive. The non-primitivity of gh implies that some non-unit c ∈ R divides all coefficients of gh. If π is an irreducible factor of c then π divides all coefficients of gh. Let \overline{R} = R/(π), a domain since π is prime (recall that irreducibles are prime in a UFD). Working in \overline{R}[X], we have \overline{g}\overline{h} = \overline{gh} = 0. But a polynomial ring over a domain is again a domain (why?), so one of \overline{g} or \overline{h} vanishes. This says that π divides all coefficients of g or h, so one of these is non-primitive, as desired.

Say our given non-trivial factorization is f = gh with g, h ∈ K[X] having positive degree. If we write the coefficients of g and h as reduced form fractions (possible since R is a UFD), then we can write g = g̃/d and h = h̃/d′ where d, d′ ∈ R are the nonzero lcm’s of the denominators and g̃, h̃ ∈ R[X] are primitive. If we could show d, d′ ∈ R^{×}, we’d be done. Well, g̃h̃ = dd′f, and the left side is a product of primitives, hence is primitive. Thus, the right side, dd′f, is primitive. This forces the common factor dd′ ∈ R of the coefficients of dd′f to be a unit, so certainly d, d′ ∈ R^{×}.

■

Corollary 0.4. If R is a UFD with fraction field K, then R[X] is a UFD. An element f ∈ R[X] is irreducible if and only if f ∈ R is irreducible (degree 0 case) or f primitive and irreducible in K[X] (positive degree case).

For example, the irreducibles in Z[X] are the ordinary primes, primitive polynomials of positive degree which are irreducible over Q, and negatives of these. The basic principle of the proof is to peel off the UFD property from K[X], using the UFD property of R to control nonzero constant scaling factors which are absorbed as units when working over K. In particular, it must be stressed that the general UFD proof in this corollary uses the known special case when the coefficient ring is a field. That is, to use this corollary with R = k a field (trivially a UFD) to deduce the UFD property of k[X] would be circular reasoning.

Proof. As a warm-up, we begin by verifying the classification of irreducibles in R[X]. Since degrees add when multiplying, it is clear that an irreducible in R is irreducible in R[X] (and a reducible in R is reducible in R[X]; a non-unit of R cannot become a unit in R[X]). Meanwhile, if f ∈ R[X] has positive degree then it is reducible if it is not primitive (e.g., 6X + 15 ∈ Z[X]). If f is primitive, a non-trivial factorization f = ab in R[X] must have a and b of positive degree, so in K[X] we get a non-trivial factorization. Note that conversely if f ∈ R[X] is primitive of positive degree and f is reducible in K[X] then f is also reducible in R[X]; this follows from Gauss’ Lemma! Hence, we conclude that the irreducibles in R[X] are exactly as advertised.

Now we prove that every nonzero non-unit f ∈ R[X] is a product of irreducibles, the factorization being unique up to ordering and unit multiples. For elements of degree 0, the only possible factorizations involve elements of degree 0. Thus, the UFD property of R settles both the existence and uniqueness aspects for degree 0 elements in R[X] (the unit ambiguity is not a problem, since R[X]^{×} = R^{×}). Now consider f ∈ R[X] with positive degree. We use the UFD property of R to write f = cf_{0} with c ∈ R and f_{0} primitive. By Gauss’ Lemma, if f_{0} is not irreducible in K[X] then it has a non-trivial factorization f_{0} = g_{0}h_{0} in R[X] with g_{0} and h_{0} of positive degree, and clearly necessarily primitive (as f_{0} is primitive). Inducting on degrees, we conclude that any positive degree primitive in R[X] is a product of positive degree primitives which are irreducible in K[X] (and hence irreducible in R[X], by Gauss’ Lemma). If we also apply the UFD property of R to the coefficient c, we can break up the factorization f = cf_{0} into f = ∏ p_{i} · ∏ π_{j} where p_{i} ∈ R are irreducibles (there are none of these if f is primitive) and π_{j} ∈ R[X] are positive degree primitive irreducibles (necessarily also irreducible in K[X], by Gauss’ Lemma). This proves the existence of factorization into irreducibles.

For the uniqueness aspect, suppose f ∈ R[X] of positive degree admits two factorizations

∏ p_{i} · ∏ π_{j} = f = ∏ p′_{r} · ∏ π′_{s},

where the p_{i}’s and p′_{r}’s are the degree 0 factors and the π_{j}’s and π′_{s}’s are the positive degree irreducible factors (necessarily irreducible in K[X], from the classification of irreducibles in R[X]). In K[X], the π_{j}’s and π′_{s}’s are irreducible and ∏ π_{j} = u ∏ π′_{s} where u = ∏ p′_{r} · ∏ p_{i}^{−1} ∈ K^{×} is a unit. Hence, by the UFD property of K[X] we conclude that the two collections {π_{j}} and {π′_{s}} are the same up to ordering and K^{×}-multiple. But these collections consist of primitives, and if g, h ∈ R[X] are primitive then g, h are K^{×}-multiples in K[X] if and only if they are R^{×}-multiples in R[X]. To see this, note that if g = ch for some c ∈ K^{×} then we write c = a/a′ with a, a′ ∈ R − {0} having no non-trivial common factor (this can be done since R is a UFD), so then a′g = ah in R[X]. The gcd’s of the collection of R-coefficients on each of the two sides are a′ and a respectively, since g and h are primitive, so since gcd’s are only well-defined up to unit multiple we conclude that a = a′u for u ∈ R^{×}, whence c = a/a′ = u ∈ R^{×} as desired.

We conclude that the two collections {π_{j}} and {π′_{s}} are the same up to ordering and R^{×}-multiple. Cancelling, we see that ∏ p_{i} and ∏ p′_{r} in R are off by unit multiple, so by the UFD property of R we deduce that {p_{i}} and {p′_{r}} are the same up to ordering and R^{×}-multiple. This proves the uniqueness of factorization into irreducibles in R[X] (up to ordering and unit multiples). ■

Corollary 0.5. If R is a UFD, so is R[X_{1}, …, X_{n}] for all n ≥ 1.

Proof. The case n = 1 is the preceding corollary. Since R[X_{1}, …, X_{n}] ≃ (R[X_{1}, …, X_{n−1}])[X_{n}] for n > 1, we may induct on n (note the need to have the previous corollary for arbitrary UFD’s in order to carry out the induction!). ■

As one example, Z[X_{1}, …, X_{n}] is a UFD for all n ≥ 1. It is actually not a PID when n > 1. Likewise, k[X_{1}, …, X_{n}] is a UFD for n ≥ 1, but it is not a PID when n > 1. To give counterexamples to the PID property, one just has to check that if R is a UFD which is not a field and π ∈ R is an irreducible (which exists by the UFD property applied to any non-zero non-unit, such elements existing in R precisely when R is not a field), then (π, X) is a non-principal ideal in R[X]. This is a pleasant exercise.

It is also instructive to chase through the proof of the UFD property in several variables to try to make it effective. For example, follow the method of proof uses above to factor

−6X^{3} + 6X^{2}Y^{2} + 6X^{3}Y − 3XY + 3Y^{3} + 3XY^{2} ∈ Z[X, Y]

into irreducibles.