Regarding an attempt to prove that $\mathbf{A}^1_{k} \backslash \{ 0\}$ is not an algebraic set if $k$ is infinite.

Question

I have seen in different old posts on this site that $X:= \mathbf{A}^1(k) \backslash \{ 0 \}$ is an algebraic set. They give a polynomial map $ V(xy-1) \to X $ by defining $(t,1/t) \mapsto t$, and then they claim that this is an isomorphism. However, the inverse is not a $\textbf{polynomial}$ map. I am not sure about the reasoning after the claim.

However, I actually think that $X$ is not an algebraic set if $k$ is infinite. My question is to check if the following reasoning is right.

Assume the contrary. Note that $X$ is homeomorphic to $X \times \{ 0\}$, thus $X$ is algebraic in $\mathbf{A}^1(k)$ iff $W:= X \times \{0\}$ is algebraic in $\mathbf{A}^2(k)$. Thus by assumption $W = V(f_1,...,f_t)$.

Note that $V(f_1,...,f_t)\cap V(y)$ is an infinite set (namely equal to $W$; $k$ is infinite). Notice that actually $(a,b) \in V(f_1,...,f_t)\cap V(y)$ iff $b = 0$ and $a$ is zero of all the $f_i(x,y)$'s. This can be equivalently stated as looking at zeros of $f_i(x,0)$, because $y$ has to be zero. However, $k[x]$ is a domain so any $\textbf{non-zero}$ polynomial has finitely many zeros, in particular the cardinality of $V(f_1,...,f_t) \cap V(y)$ should be finite. This is can only be the case if $y \mid f_i$ for all $i$, but then $(0,0) \in V(f_1,...,f_t) = W$. This is a contradiction. So $W$ is not algebraic and therefore $X$ cannot be algebraic.

Answer 1

I agree with you that $X=\Bbb{A}^1\setminus \{0\}$ is not an affine algebraic set for $k$ an infinite field. However, as David points out in the comments, this is not an "intrinsic" notion: it depends on what affine space we are considering the set to be a part of. $X$ is isomorphic to the conic $C=V(xy-1)\subset\Bbb{A}^2$. The map $X\to C$ is by $P\mapsto (t(P),1/t(P))$ where $t$ is the coordinate on $X$. This is algebraic because both $t$ and $1/t$ are algebraic functions on $X$. The map $C\to X$ is as you say, $(x,y)\mapsto x$, and it is clearly an inverse to the first map.

This shows that $X$ is not a closed subset of $\Bbb{A}^1$ (in the Zariski topology) but actually is in $\Bbb{A}^2$ with a suitable embedding.

Comment: Over a finite field $k=\Bbb{F}_q$ of characteristic $p$, every point in $\Bbb{A}^1_k\setminus \{0\}$ satisfies the equation $x^{p-1} - 1 =0$ and so actually $\Bbb{A}^1_k\setminus \{0\}$ is an algebraic set in this context.