Hello, my name is Michaela Ceria. Today I will present the third part of the joint work with Theomora, Do It Yourself, Böckberger and Janné's basis over effective rings, specialising in particular on an involutive basis. Let us start from some historical milestones.

The works by Janné, which date back to the beginning of the 20th century, gives a preliminary version of Böckberger's theory for the polynomial over a field. Later on, and in particular from 1996, Gert and Blinkel formalised and generalised Janné's work, under the name of Involutive Divisions and Bases, as tools to compute efficient regular basis.

Seiler studied the same topic, pointing to different problems towards solvable polynomial rings and the distinction between strong and weak involutive bases. Well, what is the main idea? Roughly speaking, suppose that you have some polynomial generating your ideal, F1, F2, F3 and F4.

Each of their leading monomials has its own cone, a set containing some of its multiples, chosen according to a main rule, the Involutive Division. In my picture here, I colour the cones of my leading monomials. For example, the cone of X, Y6, consists of all its multiples,

while the cone of X3, Y5, consists only on the multiples that you can obtain by multiplying by powers of X. I want that the set of my multiples of my leading monomials are the union of my cones. This regulates the reduction process.

A monomial can be reduced only with the generating polynomial, or also polynomials, if the cones are not disjoint that may happen, such that the leading monomial cone contains my monomial that I want to reduce, my W. For example, here the cones are disjoint,

W is in the cone of X, Y6, so I will use F4 to reduce it, and W2 is in the cone of X8, Y3, so I will reduce it using F1. Before even giving the formal definition of Involutive Division,

I need to precise that we will work in the setting set up in the first two parts of this talk. In particular, we are not in the cumulative case, and we are not over a field. We will be on some class of neuterial rings, such that the associated graded ring is the ring of cumulative polynomials over a principal ideal ring.

My terms will behave under the twisted Tamari-Weisz-Wenning multiplication. Therefore, when we multiply or divide monomials in principal, we can have some situation in which YX is X2Y,

and also we cannot get rid of coefficients, and this imposes us to define Involutive Divisions over monomials instead of over terms. Here you can see an Involutive Division on the monomials. You get it when, for each monomial, you have a submonoid,

this one here, satisfying these four properties. The multiplication of a monomial U in U by those in the corresponding submonoid gives the cone of U.

Well, here I'm cheating a bit, because the elements in the cone change accordingly to arithmetic set up in the previous part of the talk, if we are in the left, in the right or in the restricted case. Now that I clarified that, I can set up some other terminology.

Take W. W is in the cone of U. Then we call it Involutive Multiple of U, while U is the Involutive Divisor. The element V in the submonoid such that multiplied by U gives W is called Multiplicative for U.

Such V can also be a single variable, and this leads to a partition of variables between those that are multiplicative and those that are not. Moreover, from a partition on the variables into two subsets, we can define a submonoid.

And if the submonoid we get satisfies these conditions of the definition, we can get back an Involutive Division. Finally, when we talk about multiplicative variables for a polynomial, we mean of its leading monomial, with respect to a given term ordering.

Now that we have the main terminology, let us state our questions. 1. How to prove in the principal ideal ring case that locally involutive implies involutive? 2. How to extend the completion again to the principal ideal ring case?

3. How to reformulate in our setting the algorithm by Seiler for computing weak involutive bases? The idea is, we want that the cones of our elements cover all the set of multiples of our generators,

so that we can use the rule imposed by the Involutive Division in order to make reduction. As regards terminology, no worry, definition will come in a few minutes. In the end, we want to compute weak involutive bases, but we can sketch some conjecture also for the strong ones.

Involutive or complete means that the union of the cones is actually the whole set of multiples of my elements in U. But there is a problem, we are not over a field,

we are over a principal ideal ring and therefore in general we have only one of the inclusions. Here it is. The solution of this problem is to force the following relation here on the coefficients.

Now, some definition you need to know. When you take a monomial in U and you multiply it by a variable, you get a prolongation. Such a prolongation is multiplicative or non-multiplicative according to what kind of variable you are multiplying.

Remember indeed that we partition the set of all variables. We come to a keyword for what follows, namely local involutivity. It essentially means that you take a monomial in U and a non-multiplicative prolongation,

and you want to be sure that the prolongation has an involutive divisor, namely it is in the code of some other monomial. The point here is that you are not requiring that all multiples of a monomial in U are in a cone, but you are requiring that only for a multiple by means of one variable.

Now we see that under what condition this actually implies the whole involutivity condition on all multiples. Actually the condition we need for our division is to be continuous.

Namely for each finite set of monomials, we know that there are no repeated terms in a sequence, such that a normal duplicative prolongation of one element in the sequence has an involutive divisor exactly in the next element of the sequence. This means that it cannot happen that, let us say, a normal duplicative prolongation of U4 has U1 as involutive divisor.

This is the crucial thing we needed. In the case we had a continuous division, local involutivity implies involutivity. Therefore my cones cover all multiples.

Let us now give an idea of the proof. It is true, of course, in the classical commutative case over a field. But here we have acquired a messy arithmetic, therefore we should be sure that nothing really bad happens.

So, we have a locally involutive set U, a continuous division and some multiple U times V of U that has an involutive divisor in U. Of course, we may be lucky and a divisor may be U itself.

But I may also be unlucky. In this case, at least, I have divisibility of coefficients and there is a normal duplicative variable of U that is actually a divisor of V and I can find an involutive divisor W1 for the provision.

Moreover, also the coefficient of W1 divides CT. Again, either I am lucky and W was the divisor I wanted or I can repeat again with another variable, finding an analogous W2.

This way I would get a sequence of elements in U satisfying continuity and all dividing the product UV. This has a finite number of divisors. The sequence is finite as well, so in the end, we must have the involutive divisor we are looking for.

Now, what is completion? Completion means an involutive supper set. Finding it is definitely easy. You only have to check whether you can find a normal duplicative prolongation that has no involutive divisor.

Pick the smallest one and add it to the set. Well, now it is time to set up some terminology involving weak involutive bases. Take a set f of our polynomials.

A polynomial p can be reduced modulo f if one of its monomials has the leading monomial of some elements of f as involutive divisor and we can reduce using their coefficients. We have a weak normal form, modulo f, if you cannot reduce anymore.

Computing a weak involutive basis now becomes easy since reduction and normal form are an adaptation of the corresponding notions and algorithms from Buchbergel's theory.

We are getting to involutive bases. We need to know that a set of terms is L auto-reduced if the cons of two distinct elements are disjoint. While a polynomial set is L auto-reduced when it dissolves for the leading

monomial and they do not contain any monomial involutively multiple of a leading monomial. Now, an L auto-reduced set f is weak involutive if and only if for the product of each polynomial by a non-multiplicative variable, the normal form goes to zero.

Finally, we can compute the involutive bases using Muller-Lissen theorem. This theorem takes already care to give our Muller criteria.

Finally, strong involutive bases. For the polynomials over a principal ideal ring, the concept of least common multiple of two monomials exists and it is unique.

If you are over a module, you need that the position of the two monomials is the same, but while it is easy. In a solvable polynomial ring, it is not true that you surely have a least common multiple. Anyway, Pei-Svenig observed that either there is a CZG involving the two monomials and then the least common multiple exists.

Or there is no CZGs and so there are also no confluences to solve and so you don't need it, there is no problem. Anyway, in the solvable polynomial rings, it is possible to decide whether those confluences exist or not.

In the case they exist, you can write the unique least common multiple. We saw the classical theory for involutive completion with coefficients over a field. What we need now is completion where the coefficients are over a principal ideal ring, but luckily it is not so difficult.

We can adapt the classical formulas for Gromlet bases of polynomials with coefficients within a principal ideal ring. Actually, there are two requests. To pass from a weak to a strong Gromlet bases using the rules by BAM.

This means to express the greatest common divisor of the two coefficients using the Bezou identity and adapted to the polynomials. And the second request is to take care of the annihilator of the leading coefficients.

This concludes my presentation and I thank you very very much for your attention. Bye!