The easiest way to prove this is to use the definitions of a prime ideal, maximal ideal, and P.I.D. In the proof below, we want to show that P is either a maximal ideal or a P.I.D.

Prove that every prime ideal in P.I.D. is a maximal ideal

Proof: Let P = (p) and M = (m) be a prime ideal and a maximal ideal of the principal ideal domain R, respectively. Take for the assumption that P is contained in M. Then there exists an element r \in R such that p = rm. Since P is a prime ideal, we have that either r \in P or m \in P. If m \in P, then P is the maximal ideal. If r \in P, then r = px for some x \in R. Then p = rm = pxm which means that xm = 1, so this implies that m is a unit. So P = R.

Therefore, we have that P = M) or P = R).

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Proof.

“\Rightarrow“: Let (x) be a prime ideal of R[x]. Then we have seen here that R[x]/(x) \cong R is an integral domain.

“\Leftarrow“: Given R an integral domain. Now take the polynomial ring R[x] and the principal ideal (x) of R[x]. Then R[x]/(x) \cong R. As we have seen here, we see that (x) is a prime ideal of R[x].

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Proof. It is important to use the definition of a prime ideal in this proof.

“\Rightarrow“: Assume that P is a prime ideal of R. Then P \neq R and when ab \in P, then either a \in P or b \in P. Take the quotient ring of R by P, that is, R/P. Since P is a prime ideal, we know firstly that R/P \neq \overline{0}.

Let \overline{r} = r + P \in R/P. Then \overline{r} = 0 in R/P if and only if r \in P. Now take the elements \overline{a} = a + P and \overline{b} = b + P in R/P. Take by assumption that \overline{a}\overline{b} = 0. Then either \overline{a} = \overline{0} or \overline{b} = \overline{0} by the definition of a prime ideal. This means that R/P has no zero divisors, and we saw earlier that R/P \neq \overline{0}, which is the definition of an integral domain.

“\Leftarrow“: repeat the proof above in the opposite direction.

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