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Suppose $X \subseteq \mathbb{A}^2$ is defined by the equations $f: x^2 + y^2 =1$ and $g: x= 1$. Find the ideal $I_X$ of all the regular functions that vanish on $X$. Is it true that $I_X= (f,g)$?

My attempt : The only common solution of the system composed by $f=0$ and $g=0$ is the point $(1,0)$ and $I_X = \left\{ F \in k[x,y] : F(1,0) =0 \right\}$. So $I_X \neq (f,g)$, because $x^2 - 1 \notin (f,g)$. Is it correct? Is there a more explicit expression of the ideal $I_X$?

Alessio K
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cip
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  • $(g)=(x-1)$. Clearly, $x^2-1$ is contained in that ideal. – Arthur Sep 04 '20 at 08:38
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    Oh, that’s true, I was wrong. Maybe $y \notin (f,g)$, because I can’t find any polynomials such that $y$ is a linear combination of $f$ and $g$ – cip Sep 04 '20 at 09:35

1 Answers1

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$X$ is defined as the zero locus of that two polynomials:

$X:=Z(f,g)$

In this way you get $I_X=I(Z(f,g))=\sqrt{(f,g)}=(x-1,y)$

by Nullstellensatz theorem, if $A$ is algebraically closed.

If you want to prove $\sqrt{(f,g)}=(x-1,y)$, you can observe

$y^2=(x^2+y^2-1)-(x^2-1)=(x^2+y^2-1)-(x-1)(x+1)\in (f,g)$

and so $y\in \sqrt{(f,g)}$, that permit us to say

$(x-1,y)\subseteq \sqrt{(f,g)}$

But $(x-1,y)$ is a maximal ideal of $\mathbb{A}[x,y]$ and so

$(x-1,y)= \sqrt{(f,g)}$

You can observe $Z(x-1,y)=\{(1,0)\}$

Federico Fallucca
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  • Why is $(x -1, y)$ a maximal ideal? – cip Sep 04 '20 at 09:51
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    @cip Because it is the kernel of the surjective map $k[x,y]\to k$ that sends $f$ to $f(1,0)$. As such, $k[x,y]/(x-1,y)$ is isomorphic to the field $k$, and the ideal is therefore maximal. – Arthur Sep 04 '20 at 09:54
  • This is a consequence of the Nullstellenstaz theorem, for which each maximal ideal is in bijection with the points of $\mathbb{A}^2$, the prime ideals are in bijection with the irreducible subvarieties and the radical ideals are in bijection with the subvarieties of $\mathbb{A}^2$ – Federico Fallucca Sep 04 '20 at 09:56