Doing proofs with the variety function and the Zariski topology

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Discussion Overview

The discussion revolves around proving equivalencies involving the variety function and the Zariski topology, specifically focusing on a problem that requires participants to demonstrate the relationship between the intersection and union of sets of polynomials. The scope includes mathematical reasoning and technical explanations related to algebraic geometry.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that if an element x is in the intersection of V(Ai), then x must be in V(Ai) for all i, implying a connection to the union of Ai.
  • Another participant reiterates the definition of V(A) as the common zeroes of polynomials in A, attempting to clarify the equivalency being sought.
  • Some participants express difficulty in rigorously proving the equivalency of the statements, indicating a need for logical notation and set theory concepts.
  • A later reply emphasizes that the phrasing and notation are secondary to the logical structure of the proof, providing a logical argument for why p being in V(∪ Ai) implies p is in V(Ai) for all i.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and the nature of the problem but express differing levels of understanding regarding the proof of equivalency. There is no consensus on the best approach to rigorously demonstrate the equivalency.

Contextual Notes

Some participants note that the proof requires careful use of logic notation and set theory, suggesting that assumptions about the definitions and relationships may be implicit and not fully explored.

Mikaelochi
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TL;DR
Problem is shown in the image
20211020_134055.jpg

I included this image because it is easier than typing it out. Anyway, this is an old problem I need to catch up on. I have a clue as to how to do part a. I could say given an x that is a member of ∩V(Ai) which implies that x is a member of V(Ai) for ∀i. Then we can say ∀i all polynomials are in Ai when the polynomial is equal to zero at x. Apparently this statement is the same as V of the union of Ai. Still a little hazy on that. I don't know how to show the converse is true (which would prove the equivalency). This problem has me quite lost. But I suspect (b), (c), and (d) follow nicely from understanding (a). Any help is greatly appreciated. Thanks!
 
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##V(A)## means the common zeroes of all polynomials in ##A##. The right hand side is the common zeroes of all polynomials in all ##A_i##. The left hand side says the same.
 
martinbn said:
##V(A)## means the common zeroes of all polynomials in ##A##. The right hand side is the common zeroes of all polynomials in all ##A_i##. The left hand side says the same.
I know that, I'm just having some trouble showing the equivalency of problem (a)
 
Mikaelochi said:
I know that, I'm just having some trouble showing the equivalency of problem (a)
But that is the showing of the equivalency!
 
martinbn said:
But that is the showing of the equivalency!
I mean I have to use logic notation and probably some ideas from set theory to rigorously prove this is the case
 
Mikaelochi said:
I mean I have to use logic notation and probably some ideas from set theory to rigorously prove this is the case
What I wrote was rigorous, the notations and the phrasing are unimportant. If you insist just write it with the notations you prefer. Let ##p\in V(\cup A_i)##, then for all ##i## and all ##f\in A_i##, we have that ##f(p)=0##. Thus ##p\in V(A_i)## for all ##i##.
 

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