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 $\cup_{i \in \mathcal{I}} Y_i = \mathcal{Z}\left(\cap_{i \in \mathcal{I}} T_i \right)$?
Why is the infinite union of algebraic sets not necessarily algebraic?
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7Try $n=1$ and $k=\mathbb{C}$... – 2012-09-08
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0@t.b. In that case the algebraic sets are precisely the finite subsets of $\mathbb{C}$. Is that it? – 2012-09-08
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1well, yes, and ... ? – 2012-09-08
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2You realize that since single points are algebraic sets, what you claim would imply that _all_ sets are algebraic? – 2012-09-08
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0@MarcvanLeeuwen: Interesting...Yes i see this implication. – 2012-09-08
1 Answers
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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0Could you please clarify what you mean by "$\left\{X\right\} \cap \left\{X-1 \right\}=\emptyset$ defines all of $\mathbb{C}$"? – 2012-09-08
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0Sure: 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. – 2012-09-08
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0Ok, but then you are saying that the ideal generated by the empty set is the zero ideal? I don't understand this :-) – 2012-09-08
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0Yeah, 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). – 2012-09-08
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0Thanks, 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 $- \infty$. – 2012-09-08
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0Suppose $f$ is a non-zero polynomial in the intersection. Let $n$ be the degree of $f$. Then each of $X-1, X-2, \dots, X-(n+1)$ must divide $f$, and since they are pairwise relatively prime, their product must divide $f$. But a polynomial of degree $n+1$ cannot divide a polynomial of degree $n$. – 2012-09-08