# Let f_1,\dots ,f_r be complex polynomials in the variables x_1,\dots,x_n let V be the variety of their common zeros

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 $$\overline{R}=R/I$$. I to the ring $$\mathbb R$$ of continuous, complex-valued functions on V.

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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 $$\phi (f(x_{1}...x_{n})) = f.$$, 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 $$\displaystyle{\frac{{\overline{\phi}:{C}{\left[{x}_{{{1}}}\ldots{x}_{{{n}}}\right]}}}{{{I}\rightarrow{C}}}}$$.