# Zariski topology, non-empty intersection of open sets

Let $k$ be a field and $X=\mathbb k^n$. A subset $Y \subset X$ is closed if there are $f_1,\ldots,f_m \in k[x_1,.\ldots,x_n]$ such that $Y=\{a \in k^n :f_i(a)=0 \space \forall \space i\}$, we write $Y=V(f_1,\ldots,f_m)$. The Zariski topology is $T=\{U \subset X : X \setminus U=V(f_1,\ldots,f_m) \}$.

I am trying to show that given the Zariski topology on $k^n$, if $U$ and $V$ are non-empty open subsets, then $U \cap V$ is non-empty. I don't know how to prove this so any hints would be greatly appreciated.

• Hint: take their complements and note that the zero polynomial is the unique polynomial having all elements of $k$ as roots. – Ángel Valencia Mar 18 '16 at 3:16
• Are you assuming that the ground field $k$ is algebraically closed (or even just infinite)? – anomaly Mar 18 '16 at 3:32

## 2 Answers

We need $k$ to be an infinite field. If $k$ is finite, since point sets are closed in the Zariski topology, we could find two proper closed sets (the union of all but one point of $X$) whose union is all of $X$.

Suppose $U\cap V = \emptyset$. Then $X = (U\cap V)^{c} = U^{c} \cup V^{c}$.

Since $U$, $V$ are open, $U^{c} = V(I)$, $V^{c} = V(J)$ for some ideals $I,J$ of $k[x_{1},...,x_{n}]$. We have then that $X = V(I) \cup V(J) = V(IJ)$.

$I,J\neq (0)$ since $V(I),V(J)$ are proper subsets of $X$. So $IJ \neq (0)$ also since $k[x_{1},...,x_{n}]$ is an integral domain.

Thus there exists an $f\in IJ\setminus \{0\}$ such that $f$ vanishes at all points of $X$. Note that $f$ must in fact be nonconstant.

This is now where we use the fact that $k$ is infinite. Since $f$ is nonconstant, there exists an $i\in\{1,...,n\}$ such that $f$ has a nonzero term with a nonzero power of $x_{i}$ in it. Suppose for convenience of notation that $i=1$. We can then write $f = g_{m}(x_{2},...,x_{n})x_{1}^{m}+ ...+ g_{0}(x_{2},...,x_{n})$ for some $g_{0},...,g_{m}\in k[x_{2},...,x_{n}]$, $g_{m}\neq 0$, and $m>0$. So there is a $(a_{2},...,a_{n})\in k^{n-1}$ with $g_{m}(a_{2},...,a_{n})\neq 0$.

Then $f(x_{1},a_{2},...,a_{n})\in k[x_{1}]$ is nonzero, so can only have finitely many roots. This is a contradiction since $k$ is infinite, and $f$ must vanish at all points of $X=k^{n}$.

• Usually $k$ is assumed to be algebraically closed; so $k$ must be infinite. – Ángel Valencia Mar 18 '16 at 3:42
• I know, but this was not specified by the op. – catfish Mar 18 '16 at 3:43
• @catfish Thanks for your answer! I have one question to make: you've affirmed that since $X=V(IJ)$ and $IJ \neq 0$, then there exists an $f \in IJ \setminus 0$ such that $f$ vanishes at all points of $X$. I don't see why this is true, could you explain this to me? – user156441 Mar 18 '16 at 3:57
• The reason $f$ vanishes on all of $X$ is because $X = V(IJ)$, and $V(IJ) = \{ x\in k^n$ | $g(x) = 0$ $\forall g\in IJ\}$. So since $f\in IJ$, $f(x) = 0$ for every $x\in V(IJ) = X$. The notation $V(I)$ for an ideal $I$ of $k[x_{1},...x_{n}]$ is more general than the notation you define in your post (of the form $V(f_{1},...,f_{m})$ for some polynomials $f_{1},...,f_{m}$). These notations are actually the same though, since $k[x_{1},..,k_{n}]$ is a Noetherian ring, so all ideals are finitely generated. Does this answer your question? – catfish Mar 18 '16 at 4:18
• Yes it does, thank you. – user156441 Mar 20 '16 at 6:51

Let $U,V$ be two open sets of $k^n$ under Zariski topology. Thus they have associated two families of nonzero polynomials $f_1,...f_m;g_1,...,g_l\in k[x_1,...,x_n]$ such that $$f_i(a)=0\,\forall a\not\in U,\forall i\qquad g_j(b)=0\,\forall b\not\in V,\forall j.$$ Note that the intersection $U\cap V$ has associated the products $f_ig_j$, varying $i,j$. Since $k[x_1,...,x_n]$ is an integer domain, all these products are not zero.

Now, suppose that $U\cap V=\varnothing$. This means that all points of $k^n$ are roots of all $f_ig_j$. But this is a contradiction because this would imply $f_ig_j=0$ for all $i,j$.