If we have n homogeneous polynomials (over algebraically closed field) $f_1\ldots , f_n$ on variables $x_0, \ldots , x_n$ $$ f_i(x_0, \ldots , x_n) = \sum_{j_0,\ldots , j_n} a_{i, j_0, \ldots , j_n} x_0^{j_0}\ldots x_n^{j_n} $$ then the common condition is the condition when there are finitely many solutions of the system $f_1=\ldots = f_n=0$.

What is the codimension of non-common condition on coefficients $a_{i, j_0, \ldots , j_n}$ ? Is it true that it is at least 2?

The origin of this question is of following: there is an article of Shakirov http://arxiv.org/abs/0807.4539 where he is stating his formula on resultants. There is a Poisson's lemma: $Res(f_0, \ldots , f_n)=C\prod_{i=1}^N f_0(a_i)$ where $a_i$ are the common zeroes of $f_1, \ldots , f_n$ (if $f_1, \ldots , f_n$ are in common condition) and C is a non-zero element of the field. Then he is using this formula to state an equation on logarithms of resultants. But if the non-common condition is of codimension 1 then his proof fails: the element $C$ may be a rational function with zeroes and poles lying inside non-common condition.

I've wrote him and he said that the editor of his article (in Russian journal "Functional analysis") also is in misunderstanding with his proof.

The common-condition means that the factor $\Bbbk [x_0, \ldots , x_n]_m/(f_1, \ldots , f_n)_m$ (where the index $_m$ means that it is a homogeneous part of degree $m$) is of maximal possible dimension when $m$ is sufficiently large. This gives us a system of inequalities (some determinants are non equal to zero). But this inequalities can still have a common factor...

If you don't understand my English I'm very-very sorry :)


1 Answer 1


I think the answer is yes, this set has at least codimension $2$.

I suppose by solutions, you mean solutions in $\mathbb P^n$.

So, let $V_i$ be the hypersurface defined in $\mathbb P^n_{x_0,\dots,x_n}\times \mathbb P^N_{a_{i, j_0,\dots,j_n}}$ by $f_i=0$.

I believe it is easy to prove that for these $V_i$ it is true, that if $V=\cap V_i$, then $$ \dim V = N. $$ The point is that each $V_i$ cuts down the dimension by $1$. I also claim that all the $V_i$ and $V$ are irreducible. I think this should be easy to prove, but "I leave this to the reader" because I don't have time to think about it much. (I am fully aware that some reader might point out that this is actually not true, but let's go with assuming it for now. I might come back later to fix this part.)

Now, then the projection $V\to \mathbb P^N$ is a surjective, generically finite morphism between irreducible (projective) varieties of the same dimension. Let $Z\subset \mathbb P^N$ be the "non-common condition set" and $W\subset V$ the pre-image of $Z$ in $V$.

Then it follows, since $W$ is a proper subvariety that $$\dim V > \dim W \geq \dim Z +1$$ (also since the restriction of the projection $W\to Z$ has positive dimensional fibers).

In particular ${\rm codim}_{\mathbb P^N} Z\geq 2$.

  • $\begingroup$ Mmmm! Yes, I think you're completely right! $\endgroup$
    – zroslav
    Nov 29, 2010 at 4:14

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