Doing proofs with the variety function and the Zariski topology

[Mikaelochi]

TL;DR

Problem is shown in the image

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(A_i) which implies that x is a member of V(A_i) for ∀i. Then we can say ∀i all polynomials are in A_i when the polynomial is equal to zero at x. Apparently this statement is the same as V of the union of A_i. 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!

 

[martinbn]

$V(A)$ 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.

 

[Mikaelochi]

$V(A)$ 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)

 

[martinbn]

I know that, I'm just having some trouble showing the equivalency of problem (a)

But that is the showing of the equivalency!

 

[Mikaelochi]

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

 

[martinbn]

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 $i$.