This is came from the study of Hartshorne Exercise I.5.13., which asks to show the set of nonnormal points of a variety is a proper closed subset. Below is my trial for answering the question.
If a variety $X$ is affine, then let $A := A(X)$. Using the notation of theorem 3.9, let $A'$ be the integral closure of $A$. Then, by Theorem 3.9, by setting $L=K$, we may assume $A' = A[f_{1},\cdots, f_{l}]$ where $f_{i} = a_{i}/b_{i}$ for some $a_{i},b_{i} \in A$. Thus, $A_{b_{1}, \cdots ,b_{m}} \supseteq A'$ and $A_{b_{1},\cdots, b_{m}} \cong A'_{b_{1},\cdots, b_{m}}$. Since the union of zero locus of $b_{1},\cdots, b_{m}$ is proper (since $A$ is integral domain), for any points $P$ which is outside of the zero locus, its related maximal ideal $\mathfrak{m}_{P}$ do not contain $b_{1},\cdots, b_{m}$, thus the local ring $A_{\mathfrak{m}_{P}} \cong A'_{\mathfrak{m}_{P}}$.
Thus we knows that $A_{\mathfrak{m}_{P}}$ is integrally closed local domain. My question is, does it implies regular? All I found is that regular local ring is integrally closed, however, I do not know whether its converse is true or not.
EDIT: I want to know the answer of questions; if it is true, then it directly gives an answer of the Exercise. (If the answer of the question is true, then $A_{\mathfrak{m}_{P}}$ is regular for all $P$ which is not in the union of zero locus of $b_{1},\cdots, b_{m}$, thus the set of singular points are proper closed set, done.)
Best Answer
No, this is not true. One may consider Hartshorne exercise I.3.17(b): solving this gives that $V(z^2=xy)\subset \Bbb A^3$ is normal, so all of its local rings are integrally closed, but by the Jacobian criteria, this variety is singular at the origin.
Your proof of exercise I.5.13 looks correct to me, assuming you've proven the fact about the union of the zero loci of the $b_i$ carefully elsewhere.