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Disclaimer: I am a research mathematician, but not an algebraic geometer, and so I don't know if this is a good question. I welcome advice for improving it and/or better tags.

Let $K$ be an infinite field (I'm imagining $K=\mathbb{C}$).

Let $P_1, \dots, P_n$ be polynomials in $K[t_1,\dots, t_k, x_1,\dots,x_m]$. In the examples I have in mind, these polynomials all have total degree $\leq 2$, so if assuming this makes this question easier to answer, feel free to do so.

If we specialize $(t_1,\dots,t_k)$ to a point $t$ in $\mathbb{A}^k$, then the system of equations $P_1=\dots=P_n = 0$ cuts out a variety $V_t$ in $\mathbb{A}^m$, which we would expect to be of dimension $\max\{0,m-n\}$. For $l = 1, 2, \dots,$ define $$S_l = \left\{t \in \mathbb{A}^k ~\big|~ \dim(V_t) \geq \max\{0,m-n+l\}\right\}.$$

  1. What kind of set is $S_l$? Is it algebraic? If so (or if not!), is there an algorithm to describe it?
  2. (Assuming this question makes sense) is there a general upper bound in terms of $l$ on the dimension of $S_l$?

Example: take $m=n$ and $k=n^2$, and for $i=1,\dots,n$ let $P_i = t_{i,1}x_1 + \dots + t_{i,n}x_n.$ Then $S_l$ is the locus where the matrix $(t_{ij})$ has rank $\leq n-l$, and this set is cut out by the vanishing of some determinants.

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  • $\begingroup$ I recommend that you read Exercises 3.19 and 3.22 in Chapter II of Hartshorne's "Algebraic geometry". $\endgroup$ – Jason Starr Jan 9 '18 at 8:26
  • $\begingroup$ @Jason a friend pointed me to this post as well: mathoverflow.net/questions/193/… So I guess this answers in the affirmative that $S_l$ is algebraic, but is there a hope of a dimension bound as in my second question? $\endgroup$ – Bobby Grizzard Jan 9 '18 at 23:38
  • $\begingroup$ I recommend that you read Exercises 3.19 and 3.22 in Chapter II of Hartshorne's "Algebraic geometry". $\endgroup$ – Jason Starr Jan 9 '18 at 23:51
  • $\begingroup$ The assumptions you made are consistent with $P_i$ being identically zero. You can't expect an upper bound without assuming something more then this. $\endgroup$ – Alex Gavrilov Jan 10 '18 at 3:27

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