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Let $F$ be a field, and let $f_{1},\ldots, f_{s}$ be polynomials in $F[x_{1},\ldots, x_{t}]$. Assume that the degree of the polynomials is bounded by $d$, by which I mean, if $\alpha x_{1}^{e_{1}}\cdots x_{t}^{e_{t}}$ is any term in $f_{i}$ (for $i=1,\ldots, s)$, then $e_{1},\ldots, e_{t}\leq d$. Assume also that there is a solution to the polynomial system, so that for some sequence, $(a_{1},\ldots, a_{t})$, of numbers from the algebraic closure of $F$, we have $f_{i}(a_{1},\ldots, a_{t})=0$ for each $i=1,\ldots, s$.

Is there some bound, $D$, depending on $t$ and $d$, such that there exists a solution, $(a_{1},\ldots, a_{t})$, satisfying $[F(a_{1},\ldots,a_{t}):F]\leq D?$ If so, what is that bound, and does it appear in the literature? Many thanks.

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    $\begingroup$ math.stackexchange.com/questions/1801655/… $\endgroup$ May 28, 2016 at 0:14
  • $\begingroup$ The existence of a bound is a formal consequence of "quasi-compactness" and "flattening stratifications". However, I do not know an effective bound. Probably an effective bound can be derived from the theorem about flattening stratifications, but it is likely to be a wild overestimate of the true bound (something like an interated exponential in $d$). $\endgroup$ May 28, 2016 at 1:29
  • $\begingroup$ If the zero set $f_1=f_2=\dots=0$ has dimension zero, then Bezout provides a bound (at least if $F$ is infinite). For the degree restrictions given by OP (degree in each variable) it is natural to consider Bezout in the product of projective spaces $(P^1)^t$. I think this bound is sharp in general, but I don't have an example. It's trivial to reduce to zero-dimensional case by fixing the values of some $x_i$. $\endgroup$ May 28, 2016 at 13:09
  • $\begingroup$ Even if the zero set does not have dimension zero, by existence of the flattening stratification, there are only finitely many possible Hilbert polynomials of zero sets (after homogenizing). The maximum of the degrees (leading coefficients, basically) of those Hilbert polynomials is an upper bound. $\endgroup$ May 28, 2016 at 13:17
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    $\begingroup$ Let $m=\min(s,t)$. Then Bezout theorem gives bound $D\leq m! d^m$. $\endgroup$ May 28, 2016 at 14:16

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Let me indicate this for an infinite field of characteristic zero (just to avoid some separability problems). So, the claim is that if $I$ is an ideal in $k[x_1,\ldots, x_t]=R$ generated by polynomials of degree at most $d$ (I will use the usual notion of degree and if you want to use yours, multiply mine by $t$), and if $I$ is a proper ideal, then it is contained in a maximal ideal with the quotient field whose extension degree over $k$ is at most $d^t$, and this is optimal for general $k$.

As some of the commenters have pointed out, by taking general hyperplane sections, you may assume that $\dim R/I=0$ (and non-zero, which I will not repeat). Of course, $t$ may have become smaller, but does not affect our bound. Now, you can find a general linear combination of the generators of $I$, so that $t$ of them will define a scheme of dimension zero with ideal $J\subset I$. Let $J=(f_1,\ldots, f_t)$ and if $g_1,\ldots ,g_t$ are sufficiently general polynomials of degree at most $d$, then for a parameter $u$, one can assume that $J_u=(f_i+ug_i)$ define a family of finite schemes for `small' $u$, for all $u\neq 0$, $\dim_k R/J_u=d^t$, by Bezout's theorem. Now, pick an irreducible curve $C\subset \mathbb{A}^t\times D$, where $D$ is a suitable open set of the $u$-space which maps onto $D$. Then the map $C\to D$ is onto (if necessary making $D$ a bit smaller, but always not losing $u=0$) and for a general fiber $u=a$, the dimension of the fiber is bounded by $d^t$. But then, the special member also has the smae inequality.
In general, this is optimal. If there are elements $\alpha_i\in \overline{k}$ such that $[k(\alpha_i):k]=d$ and $[k(\alpha_1,\ldots,\alpha_t):k]=d^t$, then take $f_i(x_i)$ be the irreducible polynomial of $\alpha_i$. We see that $(f_1,\ldots, f_t)$ is a maximal ideal and the extension degree of the quotient is $d^t$, proving optimality.

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  • $\begingroup$ The Bezout bound can be sometimes improved. For example, for the degree used by Dillon, the Bernstein-Kushnirenko theorem en.wikipedia.org/wiki/Bernstein%E2%80%93Kushnirenko_theorem gives bounds $t! d^t$ instead of $t^t d^t$ given by Bezout. $\endgroup$ Jun 10, 2016 at 4:31
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    $\begingroup$ The problem is, I may have to take general linear combinations and intersection with hyperplanes to reach the situation I need. For example, if I had $x_1^dx_2^d$ and I intersect with $x_1-x_2$, I end up with say $x_1^{2d}$. $\endgroup$
    – Mohan
    Jun 10, 2016 at 13:05
  • $\begingroup$ That's why it is better to intersect with hyperplanes of the type $x_i=c_i$. Or more geometrically, embed $\mathbb{A}^t$ into a product of projective lines $(P^1)^t$ and combine with a Segre map $(P^1)^t\to P^N$. The chosen by OP notion of degree is the degree induced by this embedding. $\endgroup$ Jun 10, 2016 at 13:51
  • $\begingroup$ @Oleg Eroshkin But, how do I know apriori that the ideal does not contain $P_i(x_i)$, irreducible of large degree for all $i$, in which case intersecting with $x_i=c_i$ will give empty sets. $\endgroup$
    – Mohan
    Jun 10, 2016 at 15:02
  • $\begingroup$ Well, if the ideal contains such polynomials for every $i$, it is zero-dimensional and you don't need to cut. Otherwise, there is an $i$ and a value $c_i$ (assuming that field is infinite) that the intersection is non-empty. $\endgroup$ Jun 10, 2016 at 15:42
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A bound can be found in Lemmas 1.8 and 1.9 in this paper:

https://arxiv.org/abs/1802.10262

The techniques used are only elementary Galois theory.

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This sounds something like an effective Nullstellensatz. You might take a look at the paper of Brownawell:

Brownawell, W. Dale (1987), "Bounds for the degrees in the Nullstellensatz", Ann. of Math. 126 (3): 577–591

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  • $\begingroup$ I don't see a relation. In my understanding (in the simplest form) Nullstellensatz tells that if $f_i$ don't have a common zeros, than there exist $a_1,\dots$ such that $a_1 f_1+\dots$ is a constant. And effective version provides bounds for degree $a_i$. $\endgroup$ May 28, 2016 at 13:00
  • $\begingroup$ @OlegEroshkin The Nullstellensatz is a bit stronger than that. It says that if a polynomial $G$ vanishes on the variety $\{F_1=\cdots=F_s=0\}$, then some power of $G$ is in the ideal generated by $F_1,\ldots,F_s$. Having said that, I think you're right, the Nullstellensatz isn't the right tool for this problem. OTOH, if someone is asking about effective degree bounds, it's useful to know about it, so I think I won't delete the answer (but it certainly doesn't deserve any upvotes!). $\endgroup$ May 28, 2016 at 14:19

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