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Let X and Y be two varieties in affine n-space such that $X\cap Y=\emptyset$. Let $K[X]=K[X_1,...,X_n]/I(X)$ be the coordinate ring of X.

I have managed to convince myself that $X\cup Y$ is an affine variety with an ideal of annihilators $I(X \cup Y)=I(X) \cap I(Y)$, and that, since X and Y are disjoint, that $I(X) \cap I(Y) =I(X)I(Y)$.

Am I correct here?

Also, is it true that $K[X \cup Y]=K[X] \times K[Y]$? I want to use the Chinese remainder theorem here I think but I don't see how to guarantee that $I(X)+I(Y)=K[X_1,...,X_n]$

user3131035
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  • $I(X)+I(Y)=K[X_1,\cdots,X_n]$ is correct. Use the fact that $X\cap Y=\emptyset$ and find its associated ideal. – Rachmaninoff Nov 18 '14 at 02:17
  • Is it as simple as saying $I(X)+I(Y)=I(X \cap Y)=I(\emptyset)=K[X_1,...,X_n]$, where the first equality is by general properties of $I$ and the final equality is using Hilbert's nullstellensatz? – user3131035 Nov 18 '14 at 02:22
  • The general properties of $I$ boil down to showing in this case $\sqrt{IX+IY}=I(X\cap Y)$. And yes, the second equality is by the Nullstellensatz. – Rachmaninoff Nov 18 '14 at 02:54
  • This might be helpful to see why your first equality doesnt hold in general http://math.stackexchange.com/questions/110903/geometrical-interpretation-of-ix-1-cap-x-2-neq-ix-1ix-2-x-i-algebrai – Rachmaninoff Nov 18 '14 at 02:59
  • Ok, so $I(X \cap Y)=\sqrt{I(X)+I(Y)}=K[X_1,X_2,...,X_n]$, but this doesn't imply that $I(X)+I(Y)=K[X_1,X_2,...,X_n]$ because $I(X)+I(Y)$ might not be radical. So I'm still stuck. (PS: Thanks for your patience!) – user3131035 Nov 18 '14 at 17:26
  • Try to prove the following: $1\in \sqrt{I} \iff 1\in {I} $. – Rachmaninoff Nov 19 '14 at 01:33

1 Answers1

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Even though the question has been more or less answered by Rachmaninoff in the comments, I would like to provide an alternative approach to the idea:

Let $k$ be an algebraically closed field and let $Z_1,Z_2$ be classical affine $k$-varieties. Then the disjoint union $Z_1\amalg Z_2$ (the coproduct in the category of spaces with $k$-valued functions¹) is also classical affine. Indeed, denote $A_i=\Gamma(Z_i,\mathcal{O}_{Z_i})$ to the coordinate ring and define $A=A_1\times A_2$. Then $A$ is a reduced $k$-algebra of finite type (because so is $A_i$). Hence, $A$ is the coordinate ring of a classical affine variety $Z$. For $i=1,2$, define $e_i\in A$ as $e_1=(1,0)$, $e_2=(0,1)$. Then $Z=D(e_1)\amalg D(e_2)$ (see for instance here). By Proposition 3.32 of Milne's notes, we have that $D(e_i)$ is affine and with coordinate ring $A_{e_i}$. By the isomorphism $$(S_1\times S_2)^{-1}(R_1\times R_2)\cong S_1^{-1}R_1\times S_2^{-1}R_2$$ (where $R_i$ are rings and $S_i\subset R_i$ is a multiplicative subset; see this), it follows $A_{e_i}\cong A_i$. By the equivalence of categories between affine varieties and reduced $k$-algebras of finite type, $D(e_i)\cong Z_i$.

This constitutes an alternative proof of the isomorphism $K[X \cup Y]=K[X] \times K[Y]$ of your question: if $X,Y\subset\mathbb{A}_k^n$ are algebraic sets that are disjoint as sets, then it is easy to verify that the affine variety their union gives rise to is the disjoint union as spaces with functions of $X$ and $Y$.

(Note that the infinite disjoint union of affine algebraic varieties is never affine for it is not quasi-compact.)

¹ For the terminology, see first chapter of Görtz, Wedhorn, Algebraic Geometry I: Schemes, 2nd edition. In his notes, Milne calls a “(morphism of) $k$-ringed space(s)” to what Görtz, Wedhorn (and me here) call “a (morphism of) space(s) with functions.” You can also read about the sheaf-theoretic approach to classical algebraic geometry in the notes of Gathmann and Ellingsrud, Ottem.

Elías Guisado Villalgordo
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