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I need your help to solve this question

A subset $X \subset k^n$ ($k$ a field) is called algebraic if there exist polynomials $f_1, \dots, f_m \in k[t_1,\dots,t_n]$ such that

$$X = \{x \in K^n | f_1(x) = ...=f_m(x)=0\}.$$

The coordinate ring $k[X]$ of $X$ is the ring of all functions $f: X\to k$ that can be represented by some polynomial. That is, there exists a polynomial $g \in k[t_1,\dots,t_n]$ such that $f(x) = g(x)$ for all $x \in X$.

1- Show that for two different points $x,y \in X$ there exists $f \in k[X]$ with $f(x) \neq f(y)$.

2- Let $k=\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$ and $X \subset k^2$ be the set $\{(0,1),(1,0)\}$. Is $X$ algebraic? Determine its coordinate ring.

3- Let $g_i$ in $N$ be a sequence of polynomials and define $Y:= \{ x \in k^n | g_i (x) = 0~\forall~i\}$. Show that $Y$ is algebraic set.

Thanks

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    1) isnt true. 2) how about $x+y+1$. for 3), take the product of the $g_i$2011-02-23
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    what is $N$? @yoyo: I think your hint on 3 is wrong.2011-02-23
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    @yoyo: the two points $x$ and $y$ differ in at least one coordinate, so the corresponding coordinate function $t_i$ separates $x$ and $y$.2011-02-23
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    for 3) take the ideal $(g_i)$ generated by the $g_i$. As the ring of polynomials is Noetherian you only need finitely many, say the first n. Then $Y=\{x \in k^n : g_1(x)=\ldots=g_n(x)=0\}$.2011-02-23
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    Thank you could you please explain (3)and (1)? I appreciate your help ...2011-02-26

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