# Reduced Grobner basis form a regular sequence?

1. Mar 7, 2012

### math2012

Does anyone know if a set of homogeneous polynomials forms a reduced Grobner basis, then they form a regular sequence in the polynomial ring? Any references?

All the references that I have looked at (so far) have not related the two.

If this is not true, can you give me a counterexample?

Thanks.

2. Mar 9, 2012

### mathwonk

isn't there a bound of n for the length of a regular sequence in k[X1,...,.Xn]?

I can see no reason for there to be a bound on the length of a grobner basis for an ideal.

of course i don't know what a grobner basis is, but if it is an ideal basis, then i would think you can cook up an ideal with no basis of n elements. ?????

see page 96, example 1, of Ideals varieties and algorithms, by cox little aND O'SHEA.

there you have a reduced grobner basis consisting of 5 elements in a polynomial ring of three variables. hence no sequence longer than 3 elements can be regular.

oops these are not homogeneous, but i doubt that makes any difference.

e.g. what is the reduced grobner basis for the ideal (x^2, xy, y^2) in k[x,y]?

Again, I know next to nothing about grobner bases, but in my ignorance i see no connection at all between the two concepts.

Last edited: Mar 9, 2012
3. Mar 9, 2012

### mathwonk

Heres another reason this implication is doomed. All ideals have grobner bases, but most ideals, even those with fewer than n generators, do not have a regular sequence as basis. ideals that do, define so called "complete intersection" varieties. Such varieties have nicer properties than completely random varieties. For example they are locally equidimensional.

Hence the union of the z axis and the x,y plane cannot be a complete intersection, since the origin lies on two components of different dimensions.

Thus the ideal (z) intersect (x,y) = (xz,yz), cannot have a regulars sequence as basis.

Even components of the same dim ension cannot meet in a smaller set than expected,

E.g. the union of two planes meeting at a point is not a complete intersection.

Thus the ideal (x,y) intersect (z,w) = (xz, xw, yz, yw) in k[x,y,z,w] does not have a basis of regular elements.

Look in the commutative algebra book of Eisenbud under "complete intersection" for this.

There is one situation however where something positive can be said. I believe he proves there that if a local ring has an ideal with n generators, and if it contains a regular sequence with n elements, then any generating set of n elements is also regular. Hence if an ideal contains a regular sequence of n elements, and if it can be generated by a gorbner basis also having n elements, then that grobner basis is regular.

But I do not know if this makes sense since it is about local rings, but it could be affecting the examples you may have ben looking at. I.e. your examples may have been regular essentially by the accident of looking at very small and simple cases, as happens in books.

i.e. general grobner bases are quite complicated, but the simplest examples may accidentally be regular.

Last edited: Mar 9, 2012
4. Mar 10, 2012

### math2012

So it seems like you did some reading on Grobner basis, but just in case, here's what it means to be a Grobner basis.

A quick background first: let's just say we're in some polynomial ring R with n+N variables (N>0) in some algebraically closed field k and suppose we have imposed some sort of term ordering. By imposing this ordering in a multivariable ring, it makes sense to talk about the leading term, or some people would say the initial term in(f), of a function f.

Defn: an initial ideal in(I) is by definition the ideal ( in(f): f is in I ).

A set G = {f_1, ... , f_n} is a Grobner basis if the ideal generated by the initial term of each f_i equals the initial ideal of I = (f_1, ... , f_n). That is, (in(f_1), ..., in(f_n)) = in(I) [the initial terms of the f_i's generate the initial ideal]. This definition is equivalent to a certain notion using sygyzies but that is not needed here.

In general, a Grobner basis can be quite complicated so suppose we have a minimal or a reduced Grobner basis. That is, we say it is minimal if the initial terms of the f_i's is a minimal set of generators for the initial ideal and we say it is reduced if the coeff of each in(f_i) equals 1 and the initial term in(f_i) of each f_i does not divide any of the terms of f_j for j different from i.

So suppose G is a minimal or a reduced Grobner basis and the minimal set of generators for the initial ideal form a regular sequence [for what I am currently thinking about, we can assume even more than this-- that in(f_i) does not divide in(f_j) for any i different from j. Actually, let's just assume each in(f_i) is an irreducible monomial-- so it is a variable]. It seems plausible that this would imply that G is a regular sequence.

> e.g. what is the reduced grobner basis for the ideal (x^2, xy, y^2) in k[x,y]?

You have an ideal generated by quadratic monomials, so it doesn't really matter whether we impose the monomial ordering x > y or y > x as we will obtain the same grob basis and the same initial ideal. The reduced grob basis for your ideal is G = {x^2, xy, y^2}. Your reduced grob basis equals your minimal grob basis.

So your G is not a regular sequence, thus the need for {in(f_1),...,in(f_n)} forming a regular sequence.

Last edited: Mar 10, 2012
5. Mar 10, 2012

### math2012

> E.g. the union of two planes meeting at a point is not a complete intersection.

What is the ideal for this example? An ideal for the first plane and an ideal for the second plane? Are the planes somehow bent (and embedded in some higher dimensional space) so that they're meeting only at a point?

6. Mar 11, 2012

### math2012

Never mind in regards to this: "What is the ideal for this example?...."

Also, I solved my own original question. Thank you for your time.

7. Mar 12, 2012

### mathwonk

I was hoping {x^2, xy, y^2} was a grobner basis, hence the counterexample you asked for.

a regular sequence has the property that each element cuts down the dimension of every component of the set defined by the previous elements. thus if it has r elements the the set it defines has pure codimension r. that is why it cannot have length longer than the dimension of the polynomial ring.

it is also why it cannot define a set made up of components of diffrent dimensions.

i.e. the definition of regular say that ur cannot be a zero divisor in the ring k[X]/(u1,...ur-1).

but if ur vanished entirely on one of the components defined by (u1,...ur-1), then multiplying it by any function vanishing on the other components but not that one, would give a function vanishing everywhere,

hence usually zero. i.e. it would be a zero divisor in the previous quotient ring.

thats why (xy,xz) (a plane and a line in k^3) is not regular. i.e. in k[x,y]/(xy), we have xz not zero, but (xz)(y) = 0.

so any basis of 3 elements for an ideal in k[x,y], such as (x^2, xy, y^2), cannot be regular.

congratulations on solving your own question.