# Collection of hypersurfaces failing to be a complete intersection

Let $W_d$ be the vector space of degree $d$ homogenous polynomials in $n+1$ variables. If we fix $d_1,\ldots,d_m$, we can consider the locus $Z\subset \mathbb{P}W_{d_1}\times\cdots\mathbb{P}W_{d_m}$ parameterizing $m$-tuples of hypersurfaces that fail to be a complete intersection. Do we know anything about the codimension of $Z$ in $\mathbb{P}W_{d_1}\times\cdots\mathbb{P}W_{d_m}$?

• did you mean "hypersufaces" instead of "hyperplanes"? – Abdelmalek Abdesselam Nov 30 '16 at 21:48
• Do you want lower bounds or upper bounds? If you order the integers so that $d_1 \leq \dots \leq d_m$, then the lower bound that has been used most often is that the codimension is at least as large as the minimum of $\binom{d_1+n}{n}, \binom{d_2+n-1}{n-1},\dots, \binom{d_m + n+1-m}{n+1-m}$. This is the lower bound that is used in my work with Harris and Roth, and Riedl-Yang use a similar bound. – Jason Starr Nov 30 '16 at 22:22
• Maybe I should add: that bound is sharp when $m$ equals $1$ and also when $m$ is arbitrary, yet $d_1=\dots=d_m=1$. – Jason Starr Nov 30 '16 at 22:44
• AbdelmalekAbdesselam: Yes, you're right. I changed it. Jason Starr: Thanks for the reference. In the papers you mentioned, the crude bound obtained by a slow projection onto a plane was enough for their applications. I just didn't know if people have ever tried to work a little harder to do better. – DCT Nov 30 '16 at 22:52
• I think there should be ways to do somewhat better by projecting to a projective space so that the image becomes a hypersurface, and then doing some casework on degrees of components of this hypersurface, etc... I've never tried to do this because it seems possibly fairly ugly and I've never needed a better bound. – dhy Dec 1 '16 at 5:04

I am posting this as an answer, because the comments are getting long. If $m\leq n+1$, the proved lower bound on the codimension equals $$L=L(n,m,d_1,\dots,d_m) = \min\left( \binom{d_1+n}{n}, \binom{d_2+n-1}{n-1},\dots,\binom{d_m+n+1-m}{n+1-m} \right)$$
On the other hand, there is an irreducible component $Z_{\text{main}}$ of $Z\subset W_{d_1}\times \dots \times W_{d_m}$ that parameterizes sequences of polynomials that simultaneously vanish on some (varying) $\ell$-plane, where $\ell =\max(0,n+1-m)$. The codimension of this component is readily computed, and this gives an upper bound, $$U = U(n,m,d_1,\dots,d_m) =$$ $$\binom{d_1+\ell}{\ell} + \dots + \binom{d_m+\ell}{\ell} - (n-\ell)(\ell+1).$$ So if $m\leq n+1$, the upper bound is $$U = \binom{d_1+n+1-m}{n+1-m} + \dots + \binom{d_m+n+1-m}{n+1-m} - (m-1)(n+2-m).$$ When $m$ equals $1$ or when $d_1=\dots=d_m = 1$, then $L$ equals $U$, and this equals the codimension. Of course when $m\geq n+1$, then the true upper bound equals $U$, although $L=1<U$ for $m=n+1$. In "most" cases, speaking only for myself, I would expect that the true codimension equals $U$: the most typical "bad base locus" for a sequence that is not a regular sequence is a linear space.
However, for instance, when $m>1$ and the sequence is $(d_1,\dots,d_{m-1},d_m) = (1,\dots,1,d)$ for $d\geq 2$ and $m\leq n$, the true codimension equals $L= \min(n+3-m,\binom{d+n+1-m}{n+1-m}) = (n+3-m)(n+2-m)/2$, yet $U$ equals $2(m-1)+\binom{d+n+1-m}{n+1-m}$. Here the "bad base locus" is a degree $d$ hypersurface in a $(n+2-m)$-plane. This example illustrates the issue: there may be some family of nonlinear $(n+1-m)$-dimensional varieties $B$, the potential "bad base loci", whose Hilbert functions are necessarily higher than for a $(n+1-m)$-plane, yet that "compensate" for this by having many more moduli (i.e., higher dimension of the corresponding subvariety of the Hilbert scheme of $\mathbb{P}^n$ than the dimension $(m-1)(n+2-m)$ for linear varieties).
Edit. The following paragraph was based on an arithmetic mistake. Another "exceptional" case is when $m=n < 8$ and $(d_1,\dots,d_n)=(2,\dots,2)$. Then the upper bound equals $5n-2$. Yet, when $n<8$, this bound is beat by the codimension of loci of $n$-tuples of quadratic polynomials that are linearly dependent, i.e., $\binom{n+2}{n} - (n-1) = (n^2+n+4)/2$. This equals neither $L$ nor $U$. Presumably there are many other exceptional cases when either $m$ is large compared to $n$, $n$ is small, or there are many repetitions among the integers $(d_1,\dots,d_m)$.