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I want to prove that

$I(X_1 \cap X_2) = \sqrt{I(X_1)+I(X_2)}$

for algebraic sets $X_1=Z(G_1)$ and $X_2=Z(G_2)$, with $G_1,G_2 \subseteq \mathbb{K}[X_1,\ldots,X_n]$.

Remark: Unfortunately I named the indeterminates $X_1,X_2,\ldots$ but obviously the first ones $X_1,X_2$ are algebraic sets and not indeterminates.

I've already proved $\supseteq$ but I'm struggling with proving $\subseteq$.

The problem is equivalent to prove that, if $g \in I(X_1 \cap X_2)$, then $g^n \in I(X_1)+I(X_2)$ for some $n \in \mathbb{N}$, but the $+$ is bothering me... some hint can definitely put myself in the right way.

Also, I'd like to:

Interpret geometrically what it means to have $I(X_1 \cap X_2) \neq I(X_1)+I(X_2)$,

thing that I don't see at all.

Leafar
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1 Answers1

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Since both $I(X_1\cap X_2)$ and $\sqrt{I(X_1)+I(X_2)}$ are radical ideals, by the Nullstellensatz it is enough to prove that $Z(I(X_1\cap X_2))=Z(\sqrt{I(X_1)+I(X_2)})$. We have $Z(I(X_1\cap X_2))=X_1\cap X_2$ and $Z(\sqrt{I(X_1)+I(X_2)})=Z(I(X_1)+I(X_2))=Z(I(X_1))\cap Z(I(X_2))=X_1\cap X_2$.

The inequality $I(X_1\cap X_2)\neq I(X_1)+ I(X_2)$ is because an algebraic reason: the sum of two radical ideals is not in general a radical ideal.

Edit: In the proof I assumed that the field is algebraically closed. The equality does not hold without this hypothesis. For example, take $\mathbb{K}=\mathbb{R}$, $X_1:\ x^2+y^2=1$, and $X_2:\ x=2$. Then $X_1\cap X_2=\emptyset$, so that $I(X_1\cap X_2)=\mathbb{R}[x,y]$, whereas the ideal $I(X_1)+I(X_2)=(x^2+y^2-1)+(x-2)$ (and hence also its radical) is a proper ideal.

Diego
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  • In the first part I'll try to understand the 2nd equalty from the end. – Leafar Mar 07 '15 at 01:47
  • In the second part I think you have made some mistake with the cap but I understand what you said. However, I'd like to have a geometrical interpretation of ''the sum od two radical ideals is not in general a radical ideal'' – Leafar Mar 07 '15 at 01:50
  • However, I found an answer to the second part in http://math.stackexchange.com/questions/110903/geometrical-interpretation-of-ix-1-cap-x-2-neq-ix-1ix-2-x-i-algebrai so let's focus only on the first part – Leafar Mar 07 '15 at 02:08
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    If $z$ is annihilated by all $I(X_1)+I(X_2)$ then in particular it is annihilated by each $I(X_1)$ and $I(X_2)$. Reciprocally, if $z$ is annihilated by $I(X_1)$ and $I(X_2)$ then it is annihilated all $I(X_1)+I(X_2)$ because the set of polynomials which annihilate $z$ is an ideal. – Diego Mar 07 '15 at 02:18
  • @Diego: Why is it enough to show that the zero-sets of the ideals are equal ? – tj_ Mar 07 '15 at 10:56
  • @tj_ Because of Nullstellensatz, and we can eliminate the square root because they are radical ideals – Leafar Mar 08 '15 at 00:36
  • Yes, but the Nullstellensatz requires an algebraically closed field which isn't assumed in the question. When making additional assumptions in an answer it's good practise, to point them out explicitly and not to introduce them implicitly. – tj_ Mar 08 '15 at 00:59
  • @Diego Thanks for your answer, but I don't understand why $Z(\sqrt{I(X_1)+I(X_2)})=Z(I(X_1)+I(X_2))$, maybe it is obvious and I am missing something – Leafar Mar 08 '15 at 01:16
  • The inclusion $\subseteq$ is clear, but the other one isn't clear for me – Leafar Mar 08 '15 at 01:45
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    Let $z\in Z(I)$ and $f\in\sqrt{I}$. For some $n\in\mathbb{N}$, $f^n\in I$. Then $(f(z))^n=f^n(z)=0$ and therefore $f(z)=0$. – Diego Mar 08 '15 at 14:59
  • @Diego Are you assuming $K$ to be algebraically closed ? It seems you are using $1-1$ correspondence between algebraic sets and radical ideals. – user371231 Aug 02 '20 at 18:32
  • @user371231. Yes. I edited my answer to clarify this point. – Diego Aug 03 '20 at 23:37