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