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“: Assume R is a field. Then we saw here that 0 and R are the only ideals of R. Since a maximal ideal can’t be R itself, it must be that the maximal ideal is the zero ideal.

“\Leftarrow“: Assume that the zero ideal of R is a maximal ideal. Since it is possible to have more than one maximal ideal, we need to show that it is not the case here. Assume there is an ideal I in R such that it doesn’t contain the zero ideal. By definition of an ideal, zero is always contained in an ideal. So that is a contradiction. This means that R has two ideals: the zero ideal and R itself. But that is already the definition of a field, so R is a field.

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