In a PID, every nonzero prime ideal is maximal

In a PID, every nonzero prime ideal is maximal

Proof:

Let R be a PID. Let (x) be a nonzero prime ideal. Suppose (x)\subsetneq (y). Then x=yr for some r\in R.

Note that yr\in(x) implies y\in(x) or r\in(x). Since (x)\neq(y), we have r\in(x), so r=xz for some z\in R. Then, \displaystyle x=yr=yxz which implies 1=yz, thus y is a unit. Hence (y)=R.

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