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I am working on a problem in Artin's Algebra related to the algebraic geometry talked in Chapter 11. The problem number is 9.2., F.Y.I.

Here goes the problem:

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, and let $I$ be the ideal of the polynomial ring $R = \mathbb{C}\left [ x_1, \dots, x_n \right ]$ they generate. Define a ring homomorphism from the quotient ring $\bar{R} = R/I$ to the ring $\mathcal{R}$ of continuous, complex-valued functions on $V$.

I attempted to use the correspondence theorem w.r.t. the variety of a set of polynomials, i.e. the maximal ideals bijectively correspond to the point in $V$ and we may somehow define the continuous functions there. However I cannot come up with any idea further. Also, the term 'continuous' here seems redundant since I expect the homomorphism will carry polynomials to polynomials.

I appreciate your participation and will be thankful to anything from hints to full solution.

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You can define $\psi\colon\mathbb{C}[x_1,x_2,\dots,x_n]\to\mathcal{R}$ by defining $\psi(g)$, for $g\in\mathbb{C}[x_1,x_2,\dots,x_n]$, as $$ \psi(g)\colon (t_1,\dots,t_n)\in V\mapsto g(t_1,\dots,t_n)\in\mathbb{C} $$ It's easy to see that $\psi(g)$ is continuous and that $\psi$ is a ring homomorphism.

Since $f_1,f_2,\dots,f_r\in\ker\psi$, we have $I\subseteq\ker\psi$. Therefore the homomorphism factors through $\mathbb{C}[x_1,x_2,\dots,x_n]/I$: there exists a unique ring homomorphism $\phi\colon\mathbb{C}[x_1,x_2,\dots,x_n]/I\to\mathcal{R}$ such that $\phi\circ\pi=\psi$, where $\pi\colon\mathbb{C}[x_1,x_2,\dots,x_n]\to\mathbb{C}[x_1,x_2,\dots,x_n]/I$ is the canonical map to the quotient ring.

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  • $\begingroup$ This convinces me more. Marked as accepted solution. Thank you! $\endgroup$ – S. D. ZHU Feb 20 '17 at 14:54
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It's easier than it looks. To each element $h + I$ in $R/I$, associate the function $\phi: V \rightarrow \mathbb{C}$ by setting

$$\phi(t_1, ... , t_n) = h(t_1, ... , t_n)$$

for every $(t_1, ... ,t_n)$. Check that this doesn't depend on the specific coset representative you use.

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  • $\begingroup$ Thank you very much. Seems that I have thought too much... $\endgroup$ – S. D. ZHU Feb 20 '17 at 14:53

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