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I have a conjecture but I'm not sure if it's true. Intuitively, it seems correct, but... here it is:

Conjecture: Let $S$ be a set of polynomials in $n$ variables over $\mathbb{C}^n$ (*see below). Let $Z(S)$ be the zero locus of that system (the "solution", where my polynomials are simultaneously zero). Let $S_1,S_2,...,S_m\subset S$, and let $Z(S_i)$ be the zero locus for $S_i$. Then $Z(S)=\bigcap_{i=1}^mZ(S_i)$.

Is this true? Intuitively, if $Z(S_i)$ is the entire solution for set $S_i$, then when it is intersected with the rest of them what remains is the portion of $Z(S_i)$ on which the other polynomials are also zero. But I don't think that's formal enough to be convincing.

*I'm not sure if this is true for $\mathbb{C}^n$, but I actually want it to be true for a general field $\mathbb{F}^n$. I can handle the latter case if the former is true.

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The intersection of sets translates into a logical conjunction, thus: $$ z\in Z(S) \Leftrightarrow \forall i: z\in Z(S_i) \Leftrightarrow z\in \bigcap_i Z(S_i) \; .$$

And this means that the sets $Z(S)$ and $\bigcap_i Z(S_i)$ are equal.

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  • $\begingroup$ Great, thanks. I have one question on how to formalize this a bit better, though. If I have $n$ variables, but some of the $S_i$ only depend on some proper subset of those variables, I want to also state that the remaining variables may take on any value in $Z(S_i)$ (otherwise the intersection of solutions may be empty if the subsets of variables are disjoint, even if there is a solution to the problem). How would I state that in the conjecture? (which I guess is now a theorem?) $\endgroup$ May 18, 2016 at 13:59
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    $\begingroup$ If $p$ is a polynomial in, say, $\mathbb F[z_1,z_2,\ldots,z_{n-1}]$ (which does not depend on $z_n$), by the natural embedding $\mathbb F[z_1,\ldots, z_{n-1}] \to \mathbb F[z_1,\ldots, z_{n-1},z_n]$ it can be considered a polynomial $\tilde p$ in $\mathbb F[z_1,\ldots, z_{n-1},z_n]$. In this case $Z(\tilde{p})=Z(p)\times \mathbb F$. You might as well say, that $Z(\tilde p)$ is invariant in $\mathbb F^n$ with respect to the operation $\mathbb F \times \mathbb F^n\to \mathbb F^n$, $(z,(z_1,\ldots,z_n))\mapsto (z_1, \ldots, z_{n-1},z_n+z)$. You can generalise to any number of missing $z_i$'s. $\endgroup$ May 18, 2016 at 19:12

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