Doing proofs with the variety function and the Zariski topology

  • #1
Mikaelochi
40
1
TL;DR Summary
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!
 

Answers and Replies

  • #2
martinbn
Science Advisor
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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.
 
  • #3
Mikaelochi
40
1
##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)
 
  • #4
martinbn
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I know that, I'm just having some trouble showing the equivalency of problem (a)
But that is the showing of the equivalency!
 
  • #5
Mikaelochi
40
1
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
 
  • #6
martinbn
Science Advisor
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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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