Zariski topology: closed set of $\mathbb{A}_F^1$ that contains $\mathbb{Z}$ equals $\mathbb{A}_F^1$

Question

I have a question about a closed set in Zariski topology. I might be overlooking something easy, but I’m stuck.

Let $A$ be a closed set in $\mathbb{A}_F^1$ where $F$ is a field of characteristic zero and $\mathbb{Z} \subset A$. How does it follow that $A= \mathbb{A}_F^1$?

Thanks in advance.

Answer 1

Every polynomial vanishing on each point in $\Bbb Z$ is identically zero, since a non-zero polynomial has only finitely many zeros. The Zariski closure of a subset $A$ is the set of common zeroes of all polynomials which vanish on $A$, so if $\Bbb Z\subset A$, the Zariski closure of $A$ is all of $\Bbb A^1$.

Answer 2

Suppose $S \subseteq \mathbb{F}[x]$, and $A = Z(S)= \{x \in \mathbb{A}^1 | f(x)=0 \, \text{for all}\, f \in S\}$. Since $\mathbb{Z} \subseteq A$, this means that for all $f \in S$, we have $f(x)=0$ for every $x \in \mathbb{Z}$. This is only true if $f = 0$. Thus $S = \{0\}$. Then we have that $Z(0) = A = \mathbb{A}^1$.