Question

Show that the projection map pr: \(\mathbb{A}^{n} \rightarrow \mathbb{A}^{r}, n \geq r\), defined by \(\operatorname{pr}\left(a_{1}, \ldots, a_{n}\right)=\) \(\left(a_{1}, \ldots, a_{r}\right)\) is a polynomial map.

Short answer

Answer: Yes, the projection map \(pr: \mathbb{A}^n \rightarrow \mathbb{A}^r\) is a polynomial map. We have shown that it can be expressed as a collection of r polynomial functions, with each function of the form \(f_i(X_1, \ldots, X_n) = X_i\) for \(1 \leq i \leq r\).

Step-by-step solution

Define the projection map

Let's begin by defining the projection map pr formally: $$\operatorname{pr}: \mathbb{A}^n \rightarrow \mathbb{A}^r, \quad \operatorname{pr}(a_1, \ldots, a_n) = (a_1, \ldots, a_r)$$ This map takes a point in \(\mathbb{A}^n\) and outputs a point in \(\mathbb{A}^r\) by preserving its first r components.

Express the map as a collection of functions

Recall that a map is a polynomial map if it can be expressed as a collection of polynomial functions. Let's express the projection map pr as a collection of functions: $$(f_1, f_2, \ldots, f_r): \mathbb{A}^n \rightarrow \mathbb{A}^r$$ where for each \(a = (a_1, \ldots, a_n) \in \mathbb{A}^n\), we have $$\begin{cases} f_1(a) = a_1 \\ f_2(a) = a_2 \\ \ldots \\ f_r(a) = a_r \end{cases}$$

Check if the functions are polynomial functions

A function is a polynomial function if it can be expressed as a polynomial of its arguments. In this case, for each \(1 \leq i \leq r\), we have $$f_i(a) = a_i$$ which can be expressed as a polynomial in the coordinates \((a_1, \ldots, a_n)\) with the polynomial \(X_i\), i.e., $$f_i(X_1, \ldots, X_n) = X_i$$ Thus, each function \(f_i\) is a polynomial function.

Conclude that the map is a polynomial map

As we have shown that the projection map pr can be expressed as a collection of r polynomial functions, we can conclude that the map is a polynomial map.