User adam boocher - MathOverflow

What happens to factors of the resultant upon specialization?

2012-06-05T22:05:29Z

Let $f, g$ be two polynomials in $S[t]$ where the coefficient ring is $S = \mathbb{C}[a_1..a_n]$. The resultant of $R(f,g)$ gives some measure as to whether or not $f$ and $g$ share a common factor.

My question is what happens once we set a factor of the resultant equal to zero.

For example, suppose that for some nonconstant polynomials $c, p, q$ (with $p$ and $q$ relatively prime) $f = cp + a_1F$

$g = cq + a_1G$.

Clearly, if we set $a_1 = 0$, then $f$ and $g$ contain a common factor $c$. So $a_1$ divides the resultant. But now if we set $a_1 = 0$, and then cancel the common factor, it is reasonable to next study the resultant of $p$ and $q.$

Question: What is the relationship between $R(p,q)$ and $R(f,g)$?

We first thought that an irreducible factor of $R(p,q)$ must be [(some factor of $R(f,g)$) modulo $a_1$]. This is not true (see example below), however I hope some version of it will be true. The biggest difficulty, is that once we set $a_1$ to zero, we have to divide our polynomials by a common factor, and it's very difficult to say what happens to either the roots of the polynomials, or to the Sylvester matrix - the main tools we have to study resultants.

In even the simplest case: If $R(f,g)$ is a monomial in $S$, is $R(p,q)$ a monomial?

Thanks for your help!

Example:
$f = t*t + at^3;$

$g = t*(t+b) + a(t^2 + 1);$

Then $R(f,g) = a^2(a^3-ba+a+1)$, but upon setting $a=0$, these factors become $0$ and $1$ respectively, whereas the resultant $R(p,q) = R(t,t+b) = b$.

Answer by Adam Boocher for Readings for an honors liberal art math course

2012-02-19T22:49:10Z

I recommend Journey Through Genius. It runs through many different topics, and is focused on proving theorems that were major (but whose proofs are still elementary). For example, it starts with the infinitude of primes, the quadrature of the lune, eventually discussing uncountable sets. I like it because not only does it cover many time periods, but it also conveys well the idea of proof in mathematics - which is a part of math which often is foreign to non-math majors.

The chapters make it quite suitable to skip around as well.

Generic liftings of a regular sequence on the initial ideal

2011-04-03T06:18:22Z

Hi everyone,

I've got a question about explicitly lifting regular sequences. Let $I$ be an ideal in a polynomial ring $S$ with some term order. We'll denote the initial ideal by $in(I)$. It is false in general that a regular sequence on $S/in(I)$ is regular on $S/I$. For example consider $I=(x+y)$, with $x>y$ Then $x+y$ is a regular element on $S/in(I)$ but is not regular on $S/I$. However, $3x-y$ IS a regular element mod $I$.

My question is: Can we can do this in general? i.e. Given a regular sequence on $S/in(I)$, can we obtain a regular sequence on $S/I$ by just replacing all the coefficients in all the elements with generic coefficients?

We know that the depth of $S/in(I)$ is at most the depth of $S/I$, but I haven't actually seen too many proofs of this written down. The ones I've seen first show a bound on Betti numbers and then use the Auslander-Buchsbaum formula. I was wondering if one could prove this fact by answering the question above, and if anyone has a reference. I think one might be able to use a flat family argument. In general it would be nice to have an explicit way of going back and form between regular sequences on $I$ and $in(I)$. Any reference or suggestions would be greatly appreciated.

Thanks so much for your help!

-Adam

Comment by Adam Boocher

2012-07-12T03:22:27Z

I second Thomas' point about the uniqueness. The minimal free resolution of a module is an important invariant of a module and is a finer invariant than say the Hilbert function. Although you can retrieve most numerical information (say codimension or degree) from a nonminimal resolution, it's often easier to compute these from a minimal resolution. For example, the rank of $F_i$ in a minimal free resolution is equal to the dimension of the $k$ vector space $Tor_i(P/I,k)$. Equivalently, if you tensor a minimal free resolution with $k$ all the maps become $0$.

Comment by Adam Boocher

2012-06-11T07:07:53Z

Fernando, have you tried the DGAlgebras package for Macaulay2? Also as an aside - I'm working on "cleaning up" and adding new bits to the homological algebra packages in M2 this summer, and I'd be interested if you have any particular suggestions.

Comment by Adam Boocher

2011-04-03T15:44:26Z

in(I) denotes the initial ideal with respect to the term order. I'll update my post to indicate. Thanks!