Posts Tagged ‘PID’

We know that if R is a field and if x is a variable over R, then R[x] is a PID and a non-zero ideal I of R[x] is maximal if and only if I is prime if and only if I is generated by an irreducible element of R[x]. If R is a PID which is not a field, then R[x] could have prime ideals which are not maximal. For example, in \mathbb{Z}[x] the ideal I:=\langle 2 \rangle is prime but not maximal. In this two-part post, we will find prime and maximal ideals of R[x] when R is a PID.

Notation. Throughout this post, R is a PID and R[x] is the polynomial ring in the variable x over R. Given a prime element p \in R, we will denote by \phi_p the natural ring homomorphism R[x] \to R[x]/pR[x].

Definition Let p be a prime element of R. An element f(x) \in R[x] is called irreducible modulo p if \phi_p(f(x)) is irreducible in R[x]/pR[x]. Let \gamma : R \to R/pR be the natural ring homomorphism. Then, since R[x]/pR[x] \cong (R/pR)[x], an element f(x)=\sum_{i=0}^n a_ix^i \in R[x] is irreducible modulo p if and only if \sum_{i=0}^n \gamma(a_i)x^i is irreducible in (R/pR)[x]. Note that R/pR is a field because R is a PID.

Problem 1. Prove that if p \in R is prime and if f(x) \in R[x] is irreducible modulo p, then I:=\langle p, f(x) \rangle is a maximal ideal of R[x]. If f =0, then I is a prime but not a maximal ideal of R[x].

Solution. Clearly I/pR[x]=\phi_p(I)=\langle \phi_p(f(x)) \rangle. So \phi_p(I) is a maximal ideal of R[x]/pR[x] because \phi_p(f(x)) is irreducible in R[x]/pR[x] \cong (R/pR)[x] and R/pR is a field. So I is a maximal ideal of R[x]. If f =0, then I=\langle p \rangle=pR[x] and so R[x]/I \cong (R/pR)[x] is a domain which implies that I is prime. Finally, I= \langle p \rangle is not maximal because, for example, I \subset \langle p,x \rangle \ \Box

Problem 2. Prove that a non-zero ideal I of R[x] is prime if and only if either I= \langle f(x) \rangle for some irreducible element f(x) \in R[x] or I=\langle p, f(x) \rangle for some prime p \in R and some f(x) \in R[x] which is either zero or irreducible modulo p.

Solution. If f(x) \in R[x] is irreducible, then \langle f(x) \rangle is a prime ideal of R[x] because R[x] is a UFD. If f(x)=0 or f(x) is irreducible modulo a prime p \in R, then I=\langle p, f(x) \rangle is a prime ideal of R[x] by Problem 1.
Conversely, suppose that I is a non-zero prime ideal of R[x]. We consider two cases.
Case 1. I \cap R \neq (0) : Let 0 \neq r \in I \cap R. Then r is clearly not a unit because then I wouldn’t be a proper ideal of R[x]. So, since r \in I and I is a prime ideal of R[x], there exists a prime divisor p of r such that p \in I.  So pR[x] \subseteq I and hence \phi_p(I)=I/pR[x] is a prime ideal of R[x]/pR[x] \cong (R/pR)[x]. Thus we have two possibilities. The first possibility is that \phi_p(I)=(0), which gives us I \subseteq \ker \phi_p = pR[x] and therefore I=pR[x]=\langle p \rangle. The second possibility is that \phi_p(I)=\langle \phi_p(f(x)) \rangle= \phi_p(\langle f(x) \rangle) for some irreducible element \phi_p(f(x)) \in R[x]/pR[x], which gives us I=\langle p, f(x) \rangle because \ker \phi_p =pR[x].
Case 2. I \cap R = (0) : Let Q be the field of fractions of R and put J:=IQ[x]. Then J is a non-zero prime ideal of Q[x] because I is a prime ideal of R[x]. Note that J=\{g(x)/r : \ g(x) \in I, \ 0 \neq r \in R \}. So, since Q[x] is a PID, J=q(x)Q[x] for some irreducible element q(x) \in Q[x]. Obviously, we can write q(x)=\alpha f(x), where \alpha \in Q and f(x) \in R[x] is irreducible and the gcd of the coefficients of f(x) is one. Thus J = f(x)Q[x] and, since f(x) \in J, we have f(x) = g(x)/r for some 0 \neq r \in R and g(x) \in I. But then rf(x)=g(x) \in I and so f(x) \in I because I is prime and I \cap R = (0). Hence \langle f(x) \rangle \subseteq I. We will be done if we prove that I \subseteq \langle f(x) \rangle. To prove this, let h(x) \in I \subseteq J=f(x)Q[x]. So h(x)=f(x)q_0(x) for some q_0(x) \in Q[x]. Therefore, since the gcd of the coefficients of f(x) is one, we must have q_0(x) \in R[x] by Gauss’s lemma. Hence h(x) \in \langle f(x) \rangle and the solution is complete. \Box

See the next part here!