Recall that a prime $\mathfrak{p}$ is called nonsingular (regular) if the localization at that prime is a regular local ring. If all primes of a ring $R$ are nonsingular, $R$ is called regular. Let $S\subseteq R$ be a ring extension, and let $\mathfrak{q}$ be a prime ideal of $S$. Then $R$ is called regular above $\mathfrak{q}$ if any prime contracting to $\mathfrak{q}$ is nonsingular.

Recall that a ring $R$ is called regular in its fibre over $\mathfrak{q}$ if the scheme-theoretic fibre $R\otimes_S \frac{S_\mathfrak{q}}{\mathfrak{q}S_\mathfrak{q}}=R\otimes_S \kappa(\mathfrak{q})$ is regular.

Are these conditions equivalent? There is an obvious implication that regularity above $\mathfrak{q}$ implies regularity in the fibre, but the converse is not obvious to me. If it's not true in general, is it true when $R$ is a subring of a number field and $S=\mathbf{Z}$ (with $\mathfrak{q}$ a nonzero prime)?

Recall that a prime $\mathfrak{p}$ is called nonsingular (regular) if the localization at that prime is a regular local ring. If all primes of a ring $R$ are nonsingular, $R$ is called regular. Let $S\subseteq R$ be a ring extension, and let $\mathfrak{q}$ be a prime ideal of $S$. Then $R$ is called regular above $\mathfrak{q}$ if any prime contracting to $\mathfrak{q}$ is nonsingular.

Recall that a ring $R$ is said to have regular fibre over $\mathfrak{q}$ if the scheme-theoretic fibre $R\otimes_S \frac{S_\mathfrak{q}}{\mathfrak{q}S_\mathfrak{q}}=R\otimes_S \kappa(\mathfrak{q})$ is regular.

Are these conditions equivalent? There is an obvious implication that regularity above $\mathfrak{q}$ implies regularity of the fibre, but the converse is not obvious to me. If it's not true in general, is it true when $R$ is a subring of a number field and $S=\mathbf{Z}$ (with $\mathfrak{q}$ a nonzero prime)?

Recall that a prime $\mathfrak{p}$ is called nonsingular if the localization at that prime is a regular local ring. If all primes of a ring $R$ are nonsingular, $R$ is called regular. Let $S\subseteq R$ be an integral extension, and let $\mathfrak{q}$ be a prime ideal of $S$. Then $R$ is called regular above $\mathfrak{q}$ if any prime contracting to $\mathfrak{q}$ is nonsingular.

Recall that a ring $R$ is called regular in its fibre over $\mathfrak{q}$ if the scheme-theoretic fibre $R\otimes_S \frac{S_\mathfrak{q}}{\mathfrak{q}S_\mathfrak{q}}=R\otimes_S \kappa(\mathfrak{q})$ is regular.

Are these conditions equivalent? There is an obvious implication that regularity above $\mathfrak{q}$ implies regularity in the fibre, but the converse is not obvious to me. If it's not true in general, is it true when $R$ is a subring of a number field and $S=\mathbf{Z}$ (with $\mathfrak{q}$ a nonzero prime)?

Recall that a prime $\mathfrak{p}$ is called nonsingular (regular) if the localization at that prime is a regular local ring. If all primes of a ring $R$ are nonsingular, $R$ is called regular. Let $S\subseteq R$ be a ring extension, and let $\mathfrak{q}$ be a prime ideal of $S$. Then $R$ is called regular above $\mathfrak{q}$ if any prime contracting to $\mathfrak{q}$ is nonsingular.

Recall that a ring $R$ is called regular in its fibre over $\mathfrak{q}$ if the scheme-theoretic fibre $R\otimes_S \frac{S_\mathfrak{q}}{\mathfrak{q}S_\mathfrak{q}}=R\otimes_S \kappa(\mathfrak{q})$ is regular.

Are these conditions equivalent? There is an obvious implication that regularity above $\mathfrak{q}$ implies regularity in the fibre, but the converse is not obvious to me. If it's not true in general, is it true when $R$ is a subring of a number field and $S=\mathbf{Z}$ (with $\mathfrak{q}$ a nonzero prime)?