Are there Zariski-continuous maps between algebraic sets that are not polynomial maps?

Question

Suppose that $S_1$ and $S_2$ are the vanishing sets of a system of polynomial equations in n variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[X_1,\dots,X_n]$) and a system of polynomial equations in m variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[Y_1,\dots,Y_m]$). We can give $S_1$ and $S_2$ the Zariski topology by letting all algebraic subsets of $S_1$ and $S_2$ to be closed.

Surely, if $Y_j\circ f\colon S_1\to\mathbb{k}$ is a polynomial in $\mathbb{k}[X_1,\dots,X_n]/I(S_1)$, for every $j$, then $f$ is continuous in the Zariski topology.

The condition that $Y_j\circ f$ be polynomial, however, seems too strict: as long as the vanishing set of $Y_j\circ f$ were equal to the vanishing set of a polynomial function, $f$ would be continuous, which means that the usual morphisms from $S_1$ to $S_2$ are not the same as the continuous maps between them as (Zariski-)topological spaces.

My question then is of two parts.

1. Is it true that there exist Zarsiki-continuous maps that are not polynomial?
2. If yes, then what is the geometric way of thinking about the extra structure that polynomial maps preserve, i.e. continuous maps squash subsets of $S_1$ in such a way that non-algebraic subsets never get squashed into algebraic ones, but some algebraic ones may get squashed into non-algebraic ones, but polynomials also give what features to the geometric squashing?

Answer 1

If I understand your question correctly, then (1) yes, there are going to be lots of Zariski continuous maps that are not polynomial (although, if one asks that all higher products of the map preserve the Zariski topology on the corresponding sources and targets, this is a much stronger restriction, I think --- does anyone know the details?); and (2) the usual extra structure that one adds is the sheaf of regular functions on source and target; polynomial maps pullback regular functions on the target to regular functions on the source. Thus varieties are naturally thought of as being objects in the category of locally ringed spaces (i.e. spaces equipped with a certain structure sheaf of rings) rather than just in the category of topological spaces.

Answer 2

Consider the affine line and Look for an example there yourself.

Answer 3

Yes, there exists Zariski continuous maps between algebraic varieties that are not polynomials. For example all bijection between $\mathbb A ^1$ and itself is Zariski continuous but not polynomial in general: think of bijection that permutes $0$ and $1$ and leaves other points fixed.

Geometric way of thinking is algebraic subsets must be preserved. Only use polynomials and rational functions.

Answer 4

Just an easy remark. If $k = \mathbb{F}_q$ is the finite field with $q$ elements, then every function $k^n \to k$ is a polynomial map. In fact, for every point $p = (a_1, \dots, a_n) \in k^n$ consider the polynomial $$ f(x_1, \dots, x_n) = (-1)^n \prod_{i=1}^n \prod_{b \in \mathbb{F}_q \setminus \{ a_i \} } (x_i - b); $$ $f(p) = 1$ and $f$ is zero in $k^n \setminus \{ p \}$.