title: Nagata ring
text: # Nagata ring
  
  In commutative algebra, an N-1 ring is an integral domain $A$ whose integral closure in its quotient field is a finitely generated $A$-module. It is called a Japanese ring (or an N-2 ring) if for every finite extension $L$ of its quotient field $K$, the integral closure of $A$ in $L$ is a finitely generated $A$-module (or equivalently a finite $A$-algebra). A ring is called universally Japanese if every finitely generated integral domain over it is Japanese, and is called a Nagata ring, named for Masayoshi Nagata, or a pseudo-geometric ring if it is Noetherian and universally Japanese (or, which turns out to be the same, if it is Noetherian and all of its quotients by a prime ideal are N-2 rings). A ring is called geometric if it is the local ring of an algebraic variety or a completion of such a local ring (Danilov 2001) harv error: no target: CITEREFDanilov2001 (help), but this concept is not used much.
  
  ## Examples
  
  Fields and rings of polynomials or power series in finitely many indeterminates over fields are examples of Japanese rings. Another important example is a Noetherian integrally closed domain (e.g. a Dedekind domain) having a perfect field of fractions. On the other hand, a principal ideal domain or even a discrete valuation ring is not necessarily Japanese.
  
  Any quasi-excellent ring is a Nagata ring, so in particular almost all Noetherian rings that occur in algebraic geometry are Nagata rings. The first example of a Noetherian domain that is not a Nagata ring was given by Akizuki (1935).
  
  Here is an example of a discrete valuation ring that is not a Japanese ring. Choose a prime $p$ and an infinite degree field extension $K$ of a characteristic $p$ field $k$, such that $K^{p}\subseteq k$. Let the discrete valuation ring $R$ be the ring of formal power series over $K$ whose coefficients generate a finite extension of $k$. If $y$ is any formal power series not in $R$ then the ring $R[y]$ is not an N-1 ring (its integral closure is not a finitely generated module) so $R$ is not a Japanese ring.
  
  If $R$ is the subring of the polynomial ring $k[x_{1},x_{2},...]$ in infinitely many generators generated by the squares and cubes of all generators, and $S$ is obtained from $R$ by adjoining inverses to all elements not in any of the ideals generated by some $x_{n}$, then $S$ is a 1-dimensional Noetherian domain that is not an N-1 ring, in other words its integral closure in its quotient field is not a finitely generated $S$-module. Also $S$ has a cusp singularity at every closed point, so the set of singular points is not closed.
  
  