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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.


  • $\begingroup$ 1) isnt true. 2) how about $x+y+1$. for 3), take the product of the $g_i$ $\endgroup$
    – yoyo
    Feb 23, 2011 at 15:46
  • $\begingroup$ what is $N$? @yoyo: I think your hint on 3 is wrong. $\endgroup$
    – Rasmus
    Feb 23, 2011 at 16:05
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    $\begingroup$ @yoyo: the two points $x$ and $y$ differ in at least one coordinate, so the corresponding coordinate function $t_i$ separates $x$ and $y$. $\endgroup$ Feb 23, 2011 at 16:12
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    $\begingroup$ 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 $Y=\{x \in k^n : g_1(x)=\ldots=g_n(x)=0\}$. $\endgroup$ Feb 23, 2011 at 16:14
  • $\begingroup$ Thank you could you please explain (3)and (1)? I appreciate your help ... $\endgroup$
    – saba
    Feb 26, 2011 at 21:57

1 Answer 1


(1) Mariano said it all. Suppose $x=(x_1,x_2,...,x_n)$ and $y=(y_1,y_2,...,y_n)$. Since $y$ is not equal to $x$, $x_i$ is not equal to $y_i$ for some $i$. Take $f=t_i$.

(2) $X$ is algebraic since we can find finite set of polynomials in $K[x,y]$ that vanish on $X$ e.g $xy, x+y+1$. In fact by taking the ideal $I(X)=<xy, x+y+1>$, you get all all the polynomials vanishing on X which is precisely {$0, xy, x+y+1$}. (Convince yourself that this is true by noting that there are at most four polynomials from $X$ to $K$.) The coordinate ring $K(X)=K[x,y]/I(X)$. i.e $K(X)=${$0, 1, x, x+1$}

(3) slackenerny gave you the right idea. You probably need to look-up Hilbert Basis Theorem to know why. This theorem tells you that the ideal generated by the $g_i$'s is finitely generated. Hence, this finite generating set gives you a finite set that vanishes on same set as the infinite set you started with.

By the way,I think you mean a sequence indexed by N in the third question.

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    $\begingroup$ I think this is better left as a comment as opposed to an answer. $\endgroup$
    – Amzoti
    Mar 9, 2013 at 21:00
  • $\begingroup$ @Amzoti: Not anymore. $\endgroup$
    – Hesky Cee
    Apr 10, 2013 at 16:19
  • $\begingroup$ thats the spirit! Good job! Regards $\endgroup$
    – Amzoti
    Apr 10, 2013 at 16:27

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