Combining two sets of regular sequence

[naturemath]

This question is in regards to higher dimensional algebraic geometry. The actual problem is very complicated so here is my question which is substantially simplified.
Suppose {f_1,... f_k} is a set of quadratic polynomials and {g_1,...,g_l} is a set of linear polynomials in a polynomial ring R=C[x_1,..., x_M].

Suppose the sets {f_1,... f_k} and {g_1,...,g_l} individually form a regular sequence.

Is the following true: if some f_i and g_j form a regular sequence where f_i is in the first set and g_j is in the second set, then the two sets combined {f_1, ... f_k, g_1,...,g_l} is a set of functions that form a regular sequence? In other words, assume that the [STRIKE]codim[/STRIKE] dim of C[x_1,...x_M] /< f_1,..., f_k> is k while the [STRIKE]codim[/STRIKE] dim of C[x_1,...,x_M]/<g_1,...,g_l> is l, where f_i are homogeneous quadratic polynomials and g_j are linear polynomials.

I would like to believe that the [STRIKE]codim[/STRIKE] dim of C[x_1,...,x_M]/<f_1,...,f_k,g_1,...,g_l> is k+l. So question: doesn't it suffice to assume or prove that the f_i's are not an R-combination of the g_j's?
Is there a nice, relatively simple technique one could use to prove that each f_i couldn't be such a combination?

 

[DonAntonio]

Suppose {f_1,... f_k} is a set of irreducible quadratic polynomials and {g_1,...,g_l} is a set of linear polynomials.

Suppose the sets {f_1,... f_k} and {g_1,...,g_l} individually form a regular sequence.

Is the following true: if some f_i and g_j form a regular sequence where f_i is in the first set and g_j is in the second set, then the two sets combined {f_1, ... f_k, g_1,...,g_l} is a set of functions that form a regular sequence?

As of this moment, this appears to be true. Do you have any books or papers that you could refer me to?

Define "regular sequence". The only similar such object I've heard of is in commutative algebra, quotient modules and stuff, and

I don't think you meant that.

DonAntonio

 

[DonAntonio]

Thanks for your reply but that's what I meant-- the definition from commutative algebra.

I meant each set contains a set of functions so that the codim of the variety cut out by {f_1,..., f_k} is k while the codim of the variety cut out by {g_1, ..., g_l} is l.


I would like to believe that the codim of the variety cut out by {f_1,... f_k, g_1,..., g_l} is k+l.

Then why didn't you provide the ring R (real/complex/whatever functions, or polynomials or...) and the R-module M we're working with?

Anyway, from your OP, it seems to be you'd need some restricting condition on the ring, as being graded or local, as to able to conclude the

sequences remain regular under any permutation of their indexes (if, for example, the given elements are homogeneous, which would

be pretty boring if dealing with quadratic-lineal polynomials).

DonAntonio

 

[naturemath]

One may assume that we're working in a complex polynomial ring with many variables.

Assume that the [STRIKE]codim[/STRIKE] dim of C[x_1,...x_M] /< f_1,..., f_k> is k while the [STRIKE]codim[/STRIKE] dim of C[x_1,...,x_M]/<g_1,...,g_l> is l, where f_i are homogeneous quadratic polynomials and g_j are linear polynomials.

I would like to believe that the [STRIKE]codim[/STRIKE] dim of C[x_1,...,x_M]/<f_1,...,f_k,g_1,...,g_l> is k+l.


If you think this is boring, thanks for your help anyway. Maybe this question will interest someone else.

 

[DonAntonio]

One may assume that we're working in a complex polynomial ring with many variables.

Assume that the codim of C[x_1,...x_M] /< f_1,..., f_k> is k while the codim of C[x_1,...,x_M]/<g_1,...,g_l> is l, where f_i are homogeneous quadratic polynomials and g_j are linear polynomials.

I would like to believe that the codim of C[x_1,...,x_M]/<f_1,...,f_k,g_1,...,g_l> is k+l.


If you think this is boring, thanks for your help anyway. Maybe this question will interest someone else.

Read with care what I wrote in my last post before complaining about what you think I consider boring...

Now, assuming we still can continue to do some work together, let us check whether I'm following you: if we have \,\,R=\mathbb{C}[X_1,...,X_m]\,\,, then

we're in an integral domain and thus we've no non-trivial zero divisors, so for a seq. to be regular we only need to check that

the corresponding quotient rings are non-zero, something that will follow at once if the corresponding ideals are proper (so far so good?).

If so, and only to be sure about notation and naming, we try to answer whether \dim_R R/&lt;f_1,...,f_k&gt; = k\,,\,\dim_R R/&lt;g_1,...,g_l&gt; = l\Longrightarrow \dim_R R/&lt;f_1,...,f_k,g_1,...,g_l&gt;=k+l...?
I mean, I believe that when you say "codimension of a qotient M/N" you actually mean "the dimension of the quotient M/N", which is the same as

"the codimension of N in M"...Perhaps you meant something else, as codimension must always be related to the difference of dimensions

between an algebraic substructure and one of its substructures (algebras, vector spaces, modules and etc.).

Of course, you could actually be talking about the codimension of some quotient M/N wrt to some structure that contains it...but then I can't see what

this structure could possibly be.

In any case, if what I think is what you meant then I think the answer is yes as the only way I can imagine in the

present situation the quotient \,\, R/&lt;f_1,...,f_k,g_1,...,g_l&gt;\,\, is of dimension less than \,\,k+l\,\, is if there's some "collapsing"

between some f_i and some g_j , whcih I can't see how would it be possible as these are pol's of different degrees...

Perhaps a little more formally: if some element in \,\,\{f_1,...,f_k,g_1,...,g_l\}\,\, is an R-combination of the other elements, then

that element was already such a comb. in its own set of elements (f in the f's, g in the g's), which of course is absurd as

then that seq. wouldn't be R-regular...

See if you can extract something worth for your case from the above.

DonAntonio

 

[naturemath]

Thanks DonAntonio.

Yes, we're on the right track.

This is the situation that I would like to avoid (as you have mentioned), that

a g_u + b g_v = f_w, where a and b are in R,

or maybe some linear combination of the g_u's is a linear combination of the f_w's. (*)

How would one avoid (*)? What must one have to assume or prove?
PS. Yes, you are correct-- I meant dim of the quotient ring, or the codim of N in M... Thanks.

 

[DonAntonio]

Thanks DonAntonio.

Yes, we're on the right track.

This is the situation that I would like to avoid (as you have mentioned), that

a g_u + b g_v = f_w, where a and b are in R,

or maybe some linear combination of the g_u's is a linear combination of the f_w's. (*)

How would one avoid (*)? What must one have to assume or prove?



PS. Yes, you are correct-- I meant dim of the quotient ring, or the codim of N in M... Thanks.

Ok, so this is a nice, little bastard exercise, and not boring at all...:>) . I think it is clear we must

impose some restricting conditions on our guys so that we'd be able to conclude * in your post cannot happen, as

we can have, say, \,\,f_i= X_1^2+X_2^2=X_1\cdot X_1+X_2\cdot X_2=X_1g_1+X_2g_2\,\,,\,\,g_1=X_1\,\,,\,\,g_2=X_2...

Now, if all the f_i&#039;s are quadratic and homogeneous , then I think the above problem can be avoided if we require

the g_j&#039;s to be linear pol's with non-zero free coefficients (i.e. linear and non-homogeneous) ...

I'd advice you to try some combinations and examples with 1 and 2 variables (with only one I think is very simple

but, perhaps, it can be misleading).

Another restriction could be, maybe, to require some of the g's (or the f's) NOT to contain some of the variables.

For example, we can require, in two variables for simplicity, that the f's only include one single variable (thus

making them "standard" one-variable quadratics), whereas the g's must contain both variables...That

way, I think, * above won't be possible (but check this thoroughly!)

DonAntonio

 

[naturemath]

You're awesome. =)

Sorry about so many typos-- so thanks for your patience. I will check some easy cases.

 

[DonAntonio]

You're awesome. =)

Sorry about so many typos-- so thanks for your patience. I will check some easy cases.

It's been a pleasure, in spite of some misunderstandings here and there.

Please let me know if you make some progress in this.

DonAntonio