Question 1.227: Prove that if R is a DVR, then R is a PID. Conversely, if R ......
Prove that if R is a DVR, then R is a PID. Conversely, if R is a PID with a unique maximal ideal, then R is a DVR.Prove that if R is a DVR, then R is a PID. Conversely, if R is a PID with a unique maximal ideal, then R is a DVR.
(\Rightarrow) Let K denote the fraction field of R, and let \nu: K \rightarrow \mathbb{Z} denote the valuation of R. Thus R=\{x \in K: \nu(x) \geq 0\}. First of all, we claim that the set \{y \in R: \nu(y)=0\} is the set of invertible elements of R. Indeed, 0=\nu(1)=\nu(y / y)=\nu(y)+\nu(1 / y) implies that if \nu(y)=0, then \nu(1 / y)=0, hence 1 / y \in R.
Next we show that M=\{y \in R: \nu(y)>0\} is the unique maximal ideal and it is generated by a single element. The maximality of M is clear. To see that M is an ideal let y \in M and z \in R. Then \nu(z y)=\nu(z)+\nu(y)>0, therefore, z y \in M.
Finally, we show that R is a PID. Let I \subset R be an ideal. Let z \in I be an element with \nu(z)=m. Also, let t \in M be an element such that t \in M-M^{2}. In this case \nu(t)=1. Since \nu\left(z t^{-m}\right)=\nu(z)+\nu\left(t^{-m}\right)=m-m=0, we see that z t^{-m} is an invertible element of R. Therefore, z=a t^{m}, where a \in R is a unit. In other words, I is generated by t^{n}, where n is the smallest power such that t^{n} \in I.
(\Leftarrow) Define a valuation \nu on K as follows: if r \in R, then \nu(r)=n; the unique nonnegative integer such that r \in M^{n}-M^{n+1}. It is straightforward to verify that \nu satisfies the axioms of being a valuation.