Proving that Zariski Topology is actually a topology [duplicate]

I am reading from class notes given by a senior and I was unable to prove this assertion. I think I need help in this:

Define $V_K(A)= ${$a\in K^n |F(a)=0 $ for all $F\in A$}, A is an ideal in a finite type K-algebra A' Prove that this is actually a topology called zariski topology.

Attempt: I have proved all the assertions except of finite union is also in topology.

It is sufficient to prove that for $V_L (A) \cup V_L(B) =V_L(A \cap B)$. In this I have proved that $V_L (A\cap B) \subseteq V_L(A)\cup V_L(B)$ but unable to prove the converse.

So, can you please help with that?

Solution 1:

Hint: it's easier to show $$V(I) \cup V(J) = V(IJ) = V(I \cap J).$$

Edit: You've noted that you've shown $V(I \cap J) \subseteq V(I) \cup V(J)$. As we always have that $IJ \subseteq I \cap J \subseteq I,J$, it follows that (as $V(\cdot)$ is inclusion-reversing) $$V(I),V(J) \subseteq V(I\cap J) \subseteq V(IJ)$$ hence $$V(I) \cup V(J) \subseteq V(I \cap J) \subseteq V(IJ).$$ So with this, you are actually finished. The other method I alluded to above is showing that $$V(I ) \cup V(J) \subseteq V(I \cap J) \subseteq V(IJ) \subseteq V(I) \cup V(J).$$ Which of course proves the hinted statement. The first two inclusions in the above line follow from the argument above, as for the last inclusion, take $p \in V(IJ)$, and assume without loss of generality that $f \not \in V(I)$. Then for all $g \in V(J)$, we have $$(fg)(p) = f(p)g(p) = 0,$$ and as $K$ is a field hence an integral domain, and $f(p) \neq 0$ since $f \not \in V(I)$, we must have that $g(p) = 0$, so $p \in V(J)$. This proves this hint.