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



"Counter"-example for Gauss's Lemma on irreducible polynomials
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://alpha.math.uga.edu/~pete/factorization.pdf](http://alpha.math.uga.edu/%7Epete/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}]).


