becomes reducible, but in the former example R is already integrally closed. General version Let be a commutative ring. If is a polynomial in , then we write for the ideal of generated by all the coefficients of ; it is called the content of . Note that for each in . The next proposition states a more substantial property. A polynomial is said to be primitive if is the unit ideal . When (or more generally when is a Bézout domain), this agrees with the usual definition of a primitive polynomial. (But if is only a UFD, this definition is inconsistent with the definition of primitivity in #Statements for unique factorization domains.) Proof: This is easy using the fact that implies Proof: () First note that the gcd of the coefficients of is 1 since, otherwise, we can factor out some element from the coefficients of to write , contradicting the irreducibility of . Next, suppose for some non-constant polynomials in . Then, for some , the polynomial has coefficients in and so, by factoring out the gcd of the coefficients, we write . Do the same for and we can write for some . Now, let for some . Then . From this, using the proposition, we get: . That is, divides . Thus, and then the factorization constitutes a contradiction to the irreducibility of . () If is irreducible over , then either it is irreducible over or it contains a constant polynomial as a factor, the second possibility is ruled out by the assumption. Proof of the proposition: Clearly, . If is a prime ideal containing , then modulo . Since is a polynomial ring over an integral domain and thus is an integral domain, this implies either or modulo . Hence, either or is contained in . Since is the intersection of all prime ideals that contain and the choice of was arbitrary, . We now prove the "moreover" part. Factoring out the gcd's from the coefficients, we can write and where the gcds of the coefficients of are both 1. Clearly, it is enough to prove the assertion when are replaced by ; thus, we assume the gcd's of the coefficients of are both 1. The rest of the proof is easy and transparent if is a unique factorization domain; thus we give the proof in that case here (and see for the proof