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 $\mathbb{P}W_{d_1}\times\cdots\mathbb{P}W_{d_m}$?

$\begingroup$ did you mean "hypersufaces" instead of "hyperplanes"? $\endgroup$ – Abdelmalek Abdesselam Nov 30 '16 at 21:48

$\begingroup$ 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 $\binom{d_1+n}{n}, \binom{d_2+n1}{n1},\dots, \binom{d_m + n+1m}{n+1m}$. This is the lower bound that is used in my work with Harris and Roth, and RiedlYang use a similar bound. $\endgroup$ – Jason Starr Nov 30 '16 at 22:22

$\begingroup$ Maybe I should add: that bound is sharp when $m$ equals $1$ and also when $m$ is arbitrary, yet $d_1=\dots=d_m=1$. $\endgroup$ – Jason Starr Nov 30 '16 at 22:44

$\begingroup$ 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. $\endgroup$ – DCT Nov 30 '16 at 22:52

1$\begingroup$ 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. $\endgroup$ – 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 $$L=L(n,m,d_1,\dots,d_m) = \min\left( \binom{d_1+n}{n}, \binom{d_2+n1}{n1},\dots,\binom{d_m+n+1m}{n+1m} \right) $$
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 $\ell =\max(0,n+1m)$. 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 $$U = \binom{d_1+n+1m}{n+1m} + \dots + \binom{d_m+n+1m}{n+1m}  (m1)(n+2m).$$ 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 $(d_1,\dots,d_{m1},d_m) = (1,\dots,1,d)$ for $d\geq 2$ and $m\leq n$, the true codimension equals $L= \min(n+3m,\binom{d+n+1m}{n+1m}) = (n+3m)(n+2m)/2$, yet $U$ equals $2(m1)+\binom{d+n+1m}{n+1m}$. Here the "bad base locus" is a degree $d$ hypersurface in a $(n+2m)$plane. This example illustrates the issue: there may be some family of nonlinear $(n+1m)$dimensional varieties $B$, the potential "bad base loci", whose Hilbert functions are necessarily higher than for a $(n+1m)$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 $(m1)(n+2m)$ 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 $5n2$. 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., $\binom{n+2}{n}  (n1) = (n^2+n+4)/2$. 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 $(d_1,\dots,d_m)$.

1$\begingroup$ There should be an explicit formula along these lines for the codimension of the locus where the bad set contains an arbitrary complete intersection one dimension higher, in terms of the degrees of the hypersurfaces cutting out this complete intersection. In theory it should be possible to maximize this, although of course your examples suggest this is nontrivial. But I don't see any reason the worst case should always be related to a complete intersection. $\endgroup$ – Will Sawin Dec 1 '16 at 11:06