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Is it true that any nonempty open set is dense in the Zariski topology on $\mathbb{A}^n$? I'm pretty sure it is, but I can't think of a proof! Could someone possibly point me in the right direction? Many thanks!

Note: I am not asking about the Euclidean topology at all!

3nondatur
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Edward Hughes
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3 Answers3

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As an exercise, let's reduce everything to statements about polynomials. Every open set contains a basic open set $U$, which is the complement of the zero set of some nonzero polynomial $f$, so it suffices to show that these are Zariski dense. The Zariski closure of a set is the intersection of the zero sets of all polynomials vanishing on it. This is equal to $\mathbb{A}^n$ if and only if any polynomial vanishing on $U$ vanishes on $\mathbb{A}^n$. Thus the claim is equivalent to the following statement about polynomials:

Suppose a polynomial $g$ has the property that if $f(x) \neq 0$, then $g(x) = 0$. Then $g(x) = 0$ for all $x$.

But the condition is equivalent to the claim that $f(x) g(x) = 0$ for all $x$. Can you finish the problem from here? (Note that you need to assume $k$ infinite.)

Community
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Qiaochu Yuan
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    Hi,

    I was trying to prove it. I cleared my conception before the highlighted line. Would you please give me the proof from that highlighted line?

     – Ri-Li Jul 19 '19 at 01:17
  • @Qiaochu Yuen Hi, I am having a bit of troubles understanding "The Zariski closure of a set is the intersection of the zero sets of all polynomials vanishing on it." Let $U = \bigcup_{i=1}^nD(f_i)$, then $\overline{U}=\overline{\bigcup_{i=1}^nD(f_i)}=\bigcup_{i=1}^n\overline{D(f_i)}=\bigcup_{i=1}^n\overline{\mathbb{A}k^n\setminus V(f_i)}=\bigcup{i=1}^n(\mathbb{A}k^n\setminus int(V(f_i)))=\mathbb{A}_k^n\setminus \bigcap{i=1}^n int(V(f_i))$. So it is equal to $\mathbb{A}_k^n$ iff these polynomials have no common roots? – Ishigami Oct 18 '24 at 17:42
  • @Ri-Li Let me try to explain what Yuan has highlighted. Denote by $Z(f)$ the zero set of a polynomial $f$ in $\mathbb{A}^n$. Assuming $U$ is a basic open set, we can write $U=Z(f)^{C} = \mathbb{A}^n \ Z(f)$ for some non-zero $f$. If $ \overline{U} \subseteq Z(g)$, then we get $Z(f)^C \subseteq Z(g)$. Since every point $a \in \mathbb{A}^n$ lies either in $Z(f)$ or in $Z(f)^C \subseteq Z(g)$, we obtain $\mathbb{A}^n \subseteq Z(fg)$. Now prove that this can happens only when $fg$ is zero polynomial (provided $k$ is infinite). Thus $g=0$ and $Z(g) = \mathbb{A}^n$. – NumDio Jun 13 '25 at 10:47
  • Continuing earlier comment : Since $\overline{U} = \cap_{g \ \text{vanishing at} \ U} Z(g)$, we are done. – NumDio Jun 13 '25 at 10:47
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We have the following proposition:

Proposition: Let $X$ be a non-empty topological space. The following are equivalent:

  1. If we can write $X$ in the form $X=F\cup G$, where $F$ and $G$ are closed, then $X=F$ or $X=G$.
  2. If $U,V$ are two open sets of $X$ and $U\cap V=\emptyset$, then either $U$ or $V$ is empty.
  3. Any non-empty open set of $X$ is dense.

When one (hence any) of these conditions is satisfied, we say $X$ is irreducible.

Now, all that remains to show is that $\mathbb{A}^n$ is irreducible, which follows from the fact that its affine algebra is an integral domain.

M Turgeon
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  • As an exercise in topology, you can try to prove the equivalence. The first two being equivalent is easy to see; the last two being equivalent is a good exercise in translation. – M Turgeon Aug 06 '12 at 21:34
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A non-empty topological space $X$ is said to be irreducible if every non-empty open subset is dense.
Now there is a nice criterion for an affine scheme $X=Spec(A)$ to be irreducible (in the Zariski topology, of course): $$X \;\text {is irreducible} \iff A_{red} =A/Nil(A) \;\text {is a domain} $$ So certainly if $k$ is a field, affine space $\mathbb A^n_k=\text {Spec}(k[T_1,...,T_n])$ is irreducible: every non-empty open subset $U \subset \mathbb A^n_k$ is dense.

Edit: An elementary point of view
Since $\mathbb A^n(k)=k^n$ has a basis of its topology given by the $D(f)=\lbrace P\in k^n\mid f(P)\neq 0\rbrace \subset k^n \; (f\in k[T_1,...,t_n])$, it is enough for proving irreducibility of $\mathbb A^n(k) $ to see that $D(f)\cap D(g)\neq \emptyset$ as soon as $D(f), D(g)\neq \emptyset$.
Equivalently to see that $V(f)\cup V(g)=V(fg)\neq \mathbb A^n(k)$ (where $V(f)=\mathbb A^n(k)\setminus D(f)$ etc.).
But this is indeed true as soon as the field $k$ is infinite: no non-zero polynomial can vanish at all points of $k^n$.
Note carefully however that if $k$ is finite, $k^n$ is discrete and thus $\mathbb A^n(k)$ is not irreducible.

[I have sneakily turned from $\mathbb A^n_k$ to $ \mathbb A^n(k) $ in the elementary treatment. In the scheme-theoretic point of view $ \mathbb A^n_k $ is irreducible even for finite $k$, as seen above the Edit]

Georges Elencwajg
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