Question 1.225: Suppose R is a Noetherian integral domain. We assume further......
Suppose R is a Noetherian integral domain. We assume further that R is a local ring of Krull dimension 1. Prove that if M is the unique maximal ideal of R and \operatorname{dim}_{R / M} M / M^{2}=1, then R is a DVR.
Since R / M is a field and since \operatorname{dim}_{R / M} M / M^{2}=1, we have an element t \in M such that \bar{t} is the basis vector for M / M^{2}.
We claim that M^{n} / M^{n+1} is an R / M-vector space spanned by the image of t^{n} in M^{n} / M^{n+1}. We prove this by induction on n. n=1 case is given by the hypothesis and it is clear that M^{n} / M^{n+1} is a vector space over R / M. Let y \in M^{n} / M^{n+1} be any element. Then y=x_{1} y_{1}+\cdots+x_{r} y_{r} for some x_{i} \in M, y_{i} \in M^{n-1}. Write x_{i} in the form a_{i} t+p where a_{i} \in R is a unit and p \in M^{2}. Thus, x_{i} y_{i}=a_{i} t y_{i}+p y_{i}. By induction hypothesis y_{i} is of the form c_{i} t^{n-1}+q where c_{i} \in R and q \in M^{n}. Thus, a_{i} t y_{i}+p y_{i}=a_{i} c_{i} t^{n}+a_{i} t q+p y_{i}. Since a_{i} t q and p y_{i} are from M^{n+1}, and since a_{i} c_{i} is a unit, we see that x_{i} y_{i} modulo M^{n+1} is equal to a_{i} c_{i} t^{n}, hence our claim follows.
Next, we show that R is a DVR by exhibiting a valuation for R. By the above discussion we see that any element x of R is of the form a t^{n} modulo M^{n+1} for some unit a \in R. Furthermore, the exponent n is unique. Thus, the function \nu: R \rightarrow \mathbb{Z}_{\geq 0} defined by \nu(x)=n is well-defined. We extend \nu to the fraction field K of R by defining it to be \nu(x / y)=\nu(x)-\nu(y). Well-definedness of the extension is clear. Checking that \nu satisfies the properties of a valuation on K is straightforward.