Lemma 53.19.2. Let $(A, \mathfrak m)$ be a regular local ring of dimension $2$. Let $I \subset \mathfrak m$ be an ideal.
If $A/I$ is reduced, then $I = (0)$, $I = \mathfrak m$, or $I = (f)$ for some nonzero $f \in \mathfrak m$.
If $A/I$ has depth $1$, then $I = (f)$ for some nonzero $f \in \mathfrak m$.
Proof. Assume $I \not= 0$. Write $I = (f_1, \ldots , f_ r)$. As $A$ is a UFD (More on Algebra, Lemma 15.121.2) we can write $f_ i = fg_ i$ where $f$ is the gcd of $f_1, \ldots , f_ r$. Thus the gcd of $g_1, \ldots , g_ r$ is $1$ which means that there is no height $1$ prime ideal over $g_1, \ldots , g_ r$. Then either $(g_1, \ldots , g_ r) = A$ which implies $I = (f)$ or if not, then $\dim (A) = 2$ implies that $V(g_1, \ldots , g_ r) = \{ \mathfrak m\} $, i.e., $\mathfrak m = \sqrt{(g_1, \ldots , g_ r)}$.
Assume $A/I$ reduced, i.e., $I$ radical. If $f$ is a unit, then since $I$ is radical we see that $I = \mathfrak m$. If $f \in \mathfrak m$, then we see that $f^ n$ maps to zero in $A/I$. Hence $f \in I$ by reducedness and we conclude $I = (f)$.
Assume $A/I$ has depth $1$. Then $\mathfrak m$ is not an associated prime of $A/I$. Since the class of $f$ modulo $I$ is annihilated by $g_1, \ldots , g_ r$, this implies that the class of $f$ is zero in $A/I$. Thus $I = (f)$ as desired. $\square$
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