Examples of (non-normal) unibranched rings?

Question

For a local integral domain $R$ the following are equivalent:

a) The integral closure of $R$ in its fraction field (i.e., the normalization of $R$) is again local.

b) The henselization of $R$ is again an integral domain.

(See the book Anneaux locaux Henséliens by Raynaud, p. 99). Such rings are called unibranched. Clearly, normal local rings are unibranched.

My question is: Are there any further "nice" classes of unibranched rings? All of the few examples of non-normal unibranched rings I have seen so far were very special (the local ring in a certain point of a variety defined explicitly by some equations). I absolutely appreciate this notion but if there are no broader classes of examples, I feel like I'm cheating when I write down a theorem for unibranched rings and claim it's more general than for normal rings.

I would also be interested in "nice" classes of integral domains $R$ which are locally unibranched, i.e., $R_P$ is unibranched for all primes $P$.

Answer 1

Let R be the local ring of point on a an irreducible plane algebraic curve, i.e., $R = (k[X,Y]/(f(X,Y)))_M$ for a maximal ideal $M$ of $k[X,Y]$ containing an irreducible $f \in k[X,Y]$, by translation let us assume $M=(X,Y)$. Then R is unibranch if and only if $f$ is irreducible in $k[[X,Y]]$. This, in turn is equivalent to, there is exactly one valuation ring birationally dominating R. On the other hand R is non-normal if and only if R is not a valuation ring itself. Thus, using the last criterion, R is nonnormal unibranch if and only if there is exactly one valuation ring birationally dominating R and it is different from R.

If we use the other criterion, we need an irreducible polynomial $f$ which, as a power series, is irreducible and has order > 1. For the irreducibility of power series, there are criteria as well as construction procedures using Newton Polygons. See, for example, Abhyankar's papers 'Irreducibility criterion for germs of analytic functions of two complex variable, Advances in Mathematics, vol 74(1989), p. 190-257', 'What is the difference between a parabola and a hyperbola?, Mathematical Intelligencer, vol. 10, Springer,(1988)', or T. C. Kuo, Canadian Jour. of Math vol 47(1995), p. 801-816

Answer 2

You can reverse engineer a zillion examples by starting with the normalization: Let $S$ be a normal local domain. Let $I$ be an ideal of $S$ with $\sqrt{I}=\mathfrak{m}$ and let $A$ be a subring of $S/I$. Take $R = \{ f \in S : [f] \in A \}$, where $[ \ ]$ denotes reduction modulo $I$. Then $\mathrm{Frac}(R) = \mathrm{Frac}(S)$, since $I \subset R$. I claim that $S$ is the normalization of $R$. Clearly, $S$ is a normal subring of $\mathrm{Frac}(R)$ containing $R$. Using that $\sqrt{I}= \mathfrak{m}$, one can show that $S$ is a finite $R$-module, so $S$ integral over $R$. (An earlier version of this answer had a weaker condition on $I$, but then $S$ need not be integral over $R$. For example, take $S=k[x,y]$, $I=(x)$ and $A = k \subset k[x,y]/(x)$. Then $y$ isn't integral over the (nonnoetherian) ring $R$.)

To get the cusp, take $S = k[t]$, $I = (t^2)$ and $A = k \subset S/I = k[t]/t^2$.

Answer 3

Sorry to reply to a very old question, but I thought I should record one interesting class of examples of Krull dimension $1$, (completions of) numerical semigroup rings. Fix $T=\{t_1,...,t_n\}$ a set of natural numbers whose greatest common divisor is $1$. Let's also assume $t_i>1$ for each $i$.

Consider the ring $R=K[x^{t_1},...,x^{t_n}]$, viewed as a subring of $K[x]$. Suppressing the process of localizing at the origin, this is a local ring whose maximal ideal is generated by $(x^{t_1},...,x^{t_n})$, and it is an integral domain as it is a subring of $K[x]$ which is an integral domain.

I claim the normalization of $R$ is $K[x]$, a local ring. To see this, first note that since $\{t_1,\cdots,t_n\}$ has greatest common divisor $1$, there are integers $r_1,\cdots,r_n$ so that $r_1t_1+\cdots+r_nt_n=1$. Then, $(x^{t_i})^{r_i}$ is an element of the fraction field $F$ of $R$ for each $i$, so too is their product $x^{r_1t_1+\cdots+r_nt_n} =x$. Then, $x \in F$ satisfies the monic polynomial $X^{t_1}-x^{t_1}$ in $R[X]$, and thus $x$ is integral over $R$. Consequently, $K[x]$ is inside the normalization of $R$. Further, since $K[x]$ is itself normal, $K[x]$ must be normalization of $R$ as $\operatorname{Frac}(K[x])=F$.

Thus, $R$ is a local domain whose integral closure is local, but $R$ itself is not normal. This shows that many interesting examples of non-normal unibranched rings exist even in dimension $1$.