Characterizations of discrete valuation rings

Remarks

The key step is that 2 implies 3. Take $a \in \mathfrak{m}$ nonzero; since $\mathfrak{m}$ is the only nonzero prime, some power $\mathfrak{m}^n$ lies in $(a)$; with $n$ minimal one can pick $b \in \mathfrak{m}^{n-1}$ with $b \notin (a)$. For $y = a/b$, the element $y^{-1}$ is not in $R$, so by integral closedness it is not integral, and then $y^{-1}\mathfrak{m} \not\subseteq \mathfrak{m}$; this forces $y^{-1}\mathfrak{m} = R$, so $\mathfrak{m} = (y)$. The other implications use Nakayama's lemma and the fact that $\bigcap_n \mathfrak{m}^n = 0$.

Condition 4 says that $R$ is a regular local ring of dimension $1$. Example: the local ring of $k[x, y]/(y - x^2)$ at the origin is a discrete valuation ring, with $\mathfrak{m} = (x)$, since $y = x^2$ there. The local ring of $k[x, y]/(y^2 - x^3)$ at the origin is not: its maximal ideal $(x, y)$ needs two generators, and $t = y/x$ satisfies $t^2 = x$, so it is not integrally closed.

The hypotheses matter. Without dimension $1$, the ring $k[\![x, y]\!]$ is a Noetherian integrally closed local domain, but not a discrete valuation ring. Without the Noetherian condition, a valuation ring with value group $\Q$ is an integrally closed local domain of dimension $1$, but its maximal ideal is not principal.

Going deeper

For a student

This result is the local step in the theory of Dedekind domains. A Noetherian domain of dimension $1$ is a Dedekind domain exactly when its localizations at nonzero primes are discrete valuation rings. In geometry, it says that a point on a curve is nonsingular exactly when the local ring there is integrally closed, which is why normalization resolves the singularities of curves. Condition 4 links the theory with regular local rings, which generalise discrete valuation rings to higher dimension.