Question

Gauss's Lemma on irred. polynomial says,

Let R be a UFD and F its field of fractions. If a polynomial f(x) in R[x] is reducible in F[x], then it is reducible in R[x].

In particular, an integral coefficient polynomial is irreducible in Z iff it is irreducible in Q. For me this tells me something on how the horizontal divisors in the fibration from the arithmetic plane SpecZ[x] to SpecZ intersects the generic fiber: a prime divisor (the divisor defined by the prime ideal (f(x)) in Z[x]) intersect the generic fiber exactly at one point (i.e. the prime ideal (f(x)) in Q[x]) with multiplicity one.

Now here is my question:

Give a ring R, with Frac(R)=F, and a polynomial f(x) in R[x] such that f(x) is reducible in F[x], but is irreducible in R[x].

Of course, R should not be a UFD.

I'd like to see an example for number fields as well as a geometric example (where R is the affine coordinate ring of an open curve or higher dimensional stuff). Thanks

Answer 1

Gauss' Lemma over a domain R is usually taken to be a stronger statement, as follows:

If R is a domain with fraction field F, a polynomial f in R[T] is said to be primitive if the ideal generated by its coefficients is not contained in any proper principal ideal. One says that Gauss' Lemma holds in R if the product of two primitive polynomials is primitive. (This implies that a polynomial which is irreducible over R[T] remains irreducible over F[T].) Say that a domain is a GL-domain if Gauss' Lemma holds.

It is known that this property is intermediate between being a GCD-domain and having irreducible elements be prime (which I call a EL-domain; this is not standard). Here is a relevant MathSciNet review:

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MR0371887 (51 #8104) Arnold, Jimmy T.; Sheldon, Philip B. Integral domains that satisfy Gauss's lemma. Michigan Math. J. 22 (1975), 39--51.

Let $D$ be an integral domain with identity. For a polynomial $f(x)\in D[X]$, the content of $f(X)$, denoted by $A_f$, is the ideal of $D$ generated by the coefficients of $f(X)$. The polynomial $f(x)$ is primitive if no nonunit of $D$ divides each coefficient of $f(X)$ (or equivalently, if $D$ is the $v$-ideal associated with $A_f$). On the other hand, $f(X)$ is superprimitive if $A_f{}^{-1}=D$. The authors study, among other things, the relation between the following four properties on an integral domain: (1) each pair of elements has a greatest common divisor; (2) each primitive polynomial is superprimitive; (3) the product of two primitive polynomials is primitive; (4) each irreducible element is prime. In an integral domain $D$, the implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) hold, while no reverse implication holds in general. On the other hand, the properties (2), (3) and (4) are equivalent in $D[X]$.

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On the other hand, when R is Noetherian, all of these conditions are equivalent, and equivalent to being a UFD: see, e.g., Theorem 17 of

http://math.uga.edu/~pete/factorization.pdf

Thus a Noetherian domain satisfies Gauss' Lemma iff it is a UFD. In particular, such rings must be integrally closed, but this condition is not sufficient: e.g. take the ring of integers of any number field which is not of class number one (for instance Z[\sqrt{-6}]).

Answer 2

For number fields take $R = \mathbb{Z}[ \sqrt{-3} ]$ and $f(t) = t^2 + t + 1$. This polynomial is irreducible over $R[t]$ because the only units of $R$ are $\pm 1$, but $f(t) = \left( t - \frac{-1 + \sqrt{-3}}{2} \right) \left( t - \frac{-1 - \sqrt{-3}}{2} \right)$.

For function fields take $R = \mathbb{C}[x, y]/(x^2 - y^3)$ and $f(t) = t^2 - y$. This polynomial is irreducible over $R[t]$ because $y$ is not a square, but $f(t) = \left( t - \frac{x}{y} \right) \left( t + \frac{x}{y} \right)$.

Gauss's lemma fails for any ring $R$ which is not integrally closed, which is how the above examples are constructed. In the latter case this is because of the singularity at $(0, 0)$ (and in general this means some localization isn't integrally closed). Taking the integral closure resolves singularities for Krull dimension 1 (both of the above examples), although presumably you already knew this.

Answer 3

This is basically a fleshing out of Pete's answer.

Using the formulation of Gauss's lemma in terms of primitive polynomials, in $\mathbb{Z}[\sqrt{-5}]$, the polynomial $2x+(1+\sqrt{-5})$ is primitive, but $$(2x+1+\sqrt{-5})^2 = 4x^2 + 4(1+\sqrt{-5}) x + (-4+2 \sqrt{-5})$$ is divisible by $2$.

Another example, again using the formulation in terms of primitive polynomials: if $R = k[a,b,c,d]/(ad-bc)$ then $(ax+b)$ and $(ax+c)$ are primitive but $$(ax+b)(ax+c) = a (x^2 + (b+c)x + d).$$

In general, if $p$ is irreducible, $p|ab$ but $p$ does not divide $a$ or $b$, then $px+a$ and $px+b$ are primitive but $(px+a)(px+b)$ is divisible by $p$. So Gauss's lemma implies that, if $p$ is irreducible and $p|ab$ then $p|a$ or $p|b$. Pete calls this property EL and shows that, in a Noetherian domain, EL is equivalent to UFD.

I agree with Qiaochu that the formulation of Gauss's lemma which you give should be equivalent to the ring being integrally closed, at least for noetherian rings.

Answer 4

For any commutative ring B and subring A, the following are equivalent:

(1) A is integrally closed in B

(2) If F in A[X] factorizes as F = G.H in B[X] with G and H monic, then G and H are in A[X]

These conditions imply:

(3) If F in A[X] is monic and irreducible, then F remains irreducible in B[X]

When B is an integral domain, all three conditions are equivalent.

Proof: (2) ==> (1) is trivial: if F(b) = 0 for some b in B and F in A[X] monic, then F = (X-b).G in B[X] for some polynomial G (remainder theorem). Then G is monic, so by (2) X-b and G are in A[X]. In particular, b is in A.

(1) ==> (2) is folklore. Take a commutative ring S that contains B, over which G and H can be written as products of linear factors: G = (X-x1)...(X-xn), H = (X-y1)...(X-ym). Then in S the xi and yj are zeroes of the monic polynomial F, so they are integral over A. But the coefficients of G and H are (elementary symmetric) polynomials in the xi and yj, respectively, hence they are again integral over A. As these coefficients are in B, by (1) they must lie in A. (Basically, one can construct such a ring S in the same way as a splitting field for a polynomial over a field is found: first consider S1 := B[X]/(G); then G = (X-x1).G1 in S1[X], where we write x1 := X mod (G) in S1. Then "adjoin another root x2 of G" by passing to S2 := S1[X]/(G1), etc, until we arrive at a ring Sn over which G completely splits into linear factors. Then proceed to adjoin roots for H in the same manner. Note that B remains a subring throughout, i.e. no non-zero element of B will map to 0 in S.)

Condition (3) immediately follows from (2), for if F = G.H in B[X], the leading coefficients of G and H are inverse units of B because F is monic. So we can rewrite this as F = G1.H1 with G1 and H1 monic in B[X].

When A and B are domains, (3) implies (1): if F(b) = 0 with b in B and F in A[X] monic, factor F as F1...Fr with the Fi monic and irreducible in A[X]. (Factoring F as a product of monic polynomials reduces the degree, so eventually we wind up with factors that are irreducible.) Since B is a domain, it follows that Fi(b) = 0 for some i. By (3), Fi is still irreducible in B[X]. But it is divisible by X-b there, and so Fi = X-b. Hence X-b is in A[X] and b belongs to A.

Q.e.d.

(Matthe van der Lee, Amsterdam.)