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

Aneeka Hunt 2021-09-21 Answered

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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Expert Answer

Adnaan Franks
Answered 2021-09-22 Author has 6708 answers

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

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