# Why is the intersection of algebraic subsets of an algebraic variety again an algebraic subset?

This question is inspired by the Wikipedia article on the Zariski topology: https://en.wikipedia.org/wiki/Zariski_topology

Since I know next to nothing about algebraic geometry, and no advanced abstract algebra, only the typical beginning groups, rings, fields, Galois theory sequence, any answer which "explains it like I am five" would be most appreciated. Again, please do not assume that I am intelligent in any response, it would be most helpful.

Why is the intersection of algebraic subsets of an algebraic variety again an algebraic subset?

I assume that this is true because the Wikipedia article on the Zariski topology states that the closed sets in the Zariski topology are all algebraic subsets of the algebraic variety, and obviously the arbitrary intersection of closed sets is closed in any topological space.

I want to understand this because this seems like the only answer I could find for my previous question: What is the motivation behind the arbitrary union topological axiom? Namely since the arbitrary intersection of closed sets being closed is equivalent to the arbitrary union of open sets being open, and the Zariski topology is apparently neither separable nor first/second-countable.

• Basically, algebraic subsets are defined by polynomial equations. Suppose you have a subset $V_1$ defined by $f_1(x) = f_2(x) = \ldots = f_n(x) = 0$, and $V_2$ defined by $g_1(x) = g_2(x) = \ldots = g_k(x) = 0$. Then clearly their intersection is described by the union of the sets of equation describing each, namely by $f_1(x) = f_2(x) = \ldots = f_n(x) = g_1(x) = g_2(x) = \ldots = g_k(x) = 0$ – xyzzyz Jun 12 '16 at 1:44
• Oh that makes sense, so you could basically say that it follows directly from the definition/propositional logic? – Chill2Macht Jun 12 '16 at 1:47
• I don't think this is going to give you the motivation you're looking for. Most of the spaces in algebraic geometry are noetherian topological spaces, so there is hardly ever a need to take infinite intersections. – Hoot Jun 12 '16 at 1:49
• @Hoot: I don't buy that. While Noetherian is typical, arbitrary intersection is still the 'obvious' thing to do for schemes. – user14972 Jun 12 '16 at 7:43
• @Hurkyl I guess I am tuning my response toward the OP, who seems to be beginning to learn and question the foundations general topology. My feeling was that the prime spectrum of a totally arbitrary ring is going to be tough sell in this case. I"m sure the OP would appreciate it if you gave it a shot, though! – Hoot Jun 12 '16 at 18:31