# Why is the infinite union of algebraic sets not necessarily algebraic?

Let $\left\{Y_i\right\}_{i \in \mathcal{I}}$, where $\mathcal{I}$ is infinite, be a family of algebraic sets of $k^n$, where $k$ is an algebraically closed field. Then $Y_i = \mathcal{Z}(T_i)$, i.e. each $Y_i$ is the zeros of some subset $T_i$ of $A=k[x_1,\cdots,x_n]$. Then what is the problem with saying that $\cup_{i \in \mathcal{I}} Y_i = \mathcal{Z}\left(\cap_{i \in \mathcal{I}} T_i \right)$?

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Try $n=1$ and $k=\mathbb{C}$... – t.b. Sep 8 '12 at 17:03
@t.b. In that case the algebraic sets are precisely the finite subsets of $\mathbb{C}$. Is that it? – Manos Sep 8 '12 at 17:16
well, yes, and ... ? – t.b. Sep 8 '12 at 17:18
You realize that since single points are algebraic sets, what you claim would imply that all sets are algebraic? – Marc van Leeuwen Sep 8 '12 at 18:13
@MarcvanLeeuwen: Interesting...Yes i see this implication. – Manos Sep 8 '12 at 18:21

First, even for finite unions, you can't just intersect subsets of equations. For example the point $0$ is the zero set of $X$ and the point $1$ is the zero set of $X-1$. Then $\{X\} \cap \{X - 1\} = \emptyset$, so the intersection of these two sets of equations defines all of $\mathbb{C}$, not just the union of the two points.

To remedy this, use the ideal $I(Y)$ of all polynomials that vanish at the given algebraic set $Y$. Then $\{ 0, 1 \} = \mathcal{Z}((X) \cap (X-1)) = \mathcal{Z}((X (X-1)))$.

Even so, the two sets $\cup_{i \in I} Y_i$ and $\mathcal{Z}(\cap_{i \in I} I(Y_i))$ are not necessarily the same when $I$ is infinite. Suppose $Y_n$ is the single point $n \in \mathbb{N} \subseteq \mathbb{C}$, so that $I(Y_n) = (X - n)$. Then $\cap_{n = 1}^{\infty} I(Y_n) = (0)$ (why?), so $\mathcal{Z}(\cap_{n=1}^{\infty} I(Y_n)) = \mathbb{C}$, not $\cup_{n=1}^{\infty} Y_n = \mathbb{N}$.

In fact, $\mathbb{N}$ is not an algebraic subset of $\mathbb{C}$, so there is no ideal $I$ such that $\mathcal{Z}(I) = \mathbb{N}$. You can prove this by using your knowledge of what all the ideals of $\mathbb{C}[X]$ are.

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Could you please clarify what you mean by "$\left\{X\right\} \cap \left\{X-1 \right\}=\emptyset$ defines all of $\mathbb{C}$"? – Manos Sep 8 '12 at 18:10
Sure: every point satisfies the empty set of equations. So the zero set of the empty set of equations is the whole space. The equivalent ideal-theoretic statement is that the zero set of the ideal $(0)$ is the whole space. – Michael Joyce Sep 8 '12 at 18:15
Ok, but then you are saying that the ideal generated by the empty set is the zero ideal? I don't understand this :-) – Manos Sep 8 '12 at 18:17
Yeah, you have to make this convention for things to be consistent. But the main point is you need to transition away from thinking of "sets of equations" that define algebraic sets (a perfectly fine way of thinking about things at first) and into "ideal of polynomials that vanish on the algebraic set". The correspondence is really geometry - algebra (algebraic sets - ideals) and not geometry - set theory (algebraic sets - sets of equations). – Michael Joyce Sep 8 '12 at 18:21
Thanks, nice explanation. Another question please: any element of $\cap_{n=1}^{\infty} I(Y_n)$ would be a polynomial of infinite degree. But such a polynomial does not exist. How does that lead to the conclusion that $\cap_{n=1}^{\infty} I(Y_n)=(0)$, since e.g. Lang defines the degree of the zero polynomial to be $- \infty$. – Manos Sep 8 '12 at 18:29