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10 votes
1 answer
854 views

Is it a valuation ring?

It is known that a one-dimensional Noetherian local ring is a discrete valuation ring if it is integrally closed. Then, even if it is not Noetherian, would a one-dimensional local ring become a ...
George's user avatar
  • 328
2 votes
1 answer
151 views

For an element in the integral closure of an ideal $I$ - which power is in $I$?

Consider an ideal $I$ in a ring $R$. If $f \in R$ belongs to the integral closure of $I$, then there is $k_0 \geq 0$ such that $f^k \in I^{k-k_0}$ for all $k \geq k_0$. Are there any known upper ...
pinaki's user avatar
  • 5,359
1 vote
1 answer
197 views

Chain of closed irreducible sets on Zariski Riemann spaces

Let $A$ be a domain and $K=\mathrm{Frac}(A)$. The Zariski Riemann space $\mathrm{ZR}(K,A)$ is the set of all valuation rings of $K$ containing $A$. It comes with a natural center map \begin{align}...
nowhere dense's user avatar
0 votes
1 answer
270 views

Valuation ring satisfying either a.c.c. or d.c.c. on prime ideals

If a commutative ring with unity has finite Krull dimension, then it satisfies a.c.c. and d.c.c. on prime ideals. The converse is not true in general, as can be seen from here An infinite dimensional ...
user521337's user avatar
  • 1,209
7 votes
1 answer
2k views

The space of valuations of a function field

Hello, I'm looking for someone who can help me to understand Zariski's theory of valuations. First I outline the theory: we take a field $K$ which is a finitely generated transcendent extension of ...
Xin Nie's user avatar
  • 1,804