# how does Macaulay2 computes analytic spread for non-local rings?

Macaulay2 computes analytic spread for R=QQ[a,b,c,d,e,f] which is not a local ring.

In the books like "*Cohen-Macaulay rings* (*Bruns-Herzog*)", "*Integral Closure of Ideals, Rings, and Modules* (*Irena Swanson and Craig Huneke*)" and "*Equimultiplicity and Blowing up* (*M. Herrmann S. Ikeda U. Orbanz*)", analytic spread is defined for local rings. then they reach to an equivalent condition:
"the analytic spread of I is the smallest number of generators of an ideal J such that I is integral over J." See for example see Huneke-Swanson, Proposition 8.3.7.

I have two Questions:

Question1: how does Macaulay2 computes analytic spread for non-local rings?

Question2: if Macaulay2 computes analytic spread for ideal $I=yzR$ in $R=k[x,y,z]/(x)\cap (y,z)$ will this be equal to the analytic spread of ideal $I=yzR$ in $R=k[[x,y,z]]/(x)\cap (y,z)$? (this is an example I don't want analytic spread of this $I$)

thank you for reading this long Question.

## 1 Answer

Actually, you can view the code, http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.7/share/doc/Macaulay2/Macaulay2Doc/html/_code.html, and I believe that it used the variable of ambient ring as the maximal ideal.

My experience is that sometimes its output is a negative number which is not possible. But most of the time, instead computing the dimension of the special fiber ring works fine.

The analytic spread is equal for both rings you mentioned since it is a finitely generated $k$-algebra, where $k = R/$maximal ideal.

In fact, since your ideal is principal, the analytic spread is either $0$ or $1$. If it were $0$, then $yz$ is nilpotent, but it is not (cf. $yz \notin (x))$. Therefore, analytic spread of $I$ is $1$. In other words, it is its own minimal reduction.

• dim specialFiber I. – Youngsu Jan 26 '15 at 17:48