Question

# Let f1,…,fr be complex polynomials in the variables x1,…,xn let V be the variety of their common zeros, and let I be the ideal of the polynomial ring

Multivariable functions

Let f1,…,fr be complex polynomials in the variables x1,…,xn let V be the variety of their common zeros, and let I be the ideal of the polynomial ring $$R=C[x1,…,xn]$$ that they generate. Define a homomorphism from the quotient ring $$\overline{R}=R\ x2F$$; I to the ring RR of continuous, complex-valued functions on V.

2021-06-04

Every polynomial with m variables and complex coefficients can be considered a continuous function $$C" > ©$$. Wecan also restrict its domain to V! Denote $$\displaystyle{C}={V}\rightarrow{C}$$ for better visibility. So we first define a. mapping
$$\displaystyle\phi:{C}{\left[{x}_{{{1}}}\ldots{x}_{{{n}}}\right]}\rightarrow{C}$$
given by $$\phi (f(x_{1}...x_{n})) = f. \text{It is clearly a homomorphism. Now we want to prove that}\ I\subseteq K$$, where $$\displaystyle{K}={k}{e}{r}\phi$$.
Let $$f(x_{1}...x_{n})$$$$\displaystyle\in$$ I. Then for every $$(y1,...,yn)$$ $$\displaystyle\in{V}$$ we have that $$f(y1,...,yn) = 0$$ by the definition of V. But this means that
$$\displaystyle\phi{\left({f{{\left({\left[{x}_{{{1}}}\ldots{x}_{{{n}}}\right]}\right)}}}\right)}={0}{C}$$
(the null function)! Therefore, $$f([x_{1}...x_{n}])$$  $$\displaystyle\in{K}$$, so
$$\displaystyle{I}\subseteq{K}$$
By the Theorem 11.4.2 (a) we conclude that there exists « (unique) homomorphism $$\displaystyle{\frac{{\overline{\phi}:{C}{\left[{x}_{{{1}}}\ldots{x}_{{{n}}}\right]}}}{{{I}\rightarrow{C}}}}$$.