The Stacks project

Remark 31.16.2. If $(A, \mathfrak m)$ is a Noetherian local normal domain of dimension $\geq 2$ and $U$ is the punctured spectrum of $A$, then $\Gamma (U, \mathcal{O}_ U) = A$. This algebraic version of Hartogs's theorem follows from the fact that $A = \bigcap _{\text{height}(\mathfrak p) = 1} A_\mathfrak p$ we've seen in Algebra, Lemma 10.157.6. Thus in this case $U$ cannot be affine (since it would force $\mathfrak m$ to be a point of $U$). This is often used as the starting point of the proof of Lemma 31.16.1. To reduce the case of a general Noetherian local ring to this case, we first complete (to get a Nagata local ring), then replace $A$ by $A/\mathfrak q$ for a suitable minimal prime, and then normalize. Each of these steps does not change the dimension and we obtain a contradiction. You can skip the completion step, but then the normalization in general is not a Noetherian domain. However, it is still a Krull domain of the same dimension (this is proved using Krull-Akizuki) and one can apply the same argument.