# Union of closed subschemes with the structure sheaf over it

Elementary commutative algebra fact: for two proper ideals I and J of a commutative ring R, we have $V(IJ)=V(I\cap J)=V(I)\cup V(J)$.

Closed subschemes are related to sheaves of ideals. There is operation of intersection and product between sheaves of ideals, which is similar to the affine case.

I see in many places that the structure sheaf over $V(I)\cup V(J)$ are defined to be $R/I\cap J$ rather than $R/IJ$. Why should the structure sheaf be define in that way?

Because the first one is the right answer in the case of affine varieties, and the second one is not. Indeed, $R/I$, $R/J$ are nilpotent-free implies $R/I\cap J$ is nilpotent-free, but not so for $R/IJ$.

As with all definitions, there is no "proof" that the adopted definition is the right one but only a feeling that it better corresponds to our intuition.

In the case at hand, taking $R/IJ$ as structure sheaf would make union schemes nonreduced for no good reason. For example take $R=k[x,y,z], I=(y,z), J=(x,z)$. Geometrically you are describing the union $U$ of the $x$-axis and the $y$-axis in affine three space $\mathbb A^3_k$ . It should have a reduced structure, correctly described by $I\cap J$, whereas $I.J=(xy,zx, zy, z^2)$ would make the function $z$ nilpotent but not zero on $U$, which feels wrong since $U$ should be a closed subscheme of the plane $z=0$.

A more brutal objection to the idea of defining the union of two subschemes by the product of their ideals is that a subscheme $U$ would practically never be equal to its union with itself, since in general$I\neq I^2$ : we would have (almost always) $$U\neq U \cup U$$

• Thanks for your answer. I guess that the following fact makes the $I\cap J$ better: $R/I\cap J$ is reduced if both $R/I$ and $R/J$ are reduced. We 'love' reduced scheme more than non-reduced scheme. Mar 21, 2010 at 18:51
• Dear 7-adic, it is not just a question of loving reduced schemes. See my answer below for an elaboration. Mar 21, 2010 at 22:03

(i) The union of two closed subschemes should be the smallest closed subscheme containing the two given ones; if you like, an initial object in the category of all closed subschemes containing the given two. In the case when the two given closed subschemes are $V(I)$ and $V(J)$, this is is $V(I\cap J)$. (The operation $V$ interchanges closed sets and ideals, and is order reversing.)

(ii) In function-theoretic terms, a function $f$ (i.e. an element of the ring $A$) should vanish on $V(I) \cup V(J)$ (i.e. lie in the ideal cutting out $V(I)\cup V(J)$) if and only it vanishes on both $V(I)$ and $V(J)$ (i.e. lies in $I$ and $J$), which happens if and only if $f$ lies in $I\cap J$. Thus we are forced to define $V(I)\cup V(J) = V(I\cap J)$, if union is to have anything like its usual meaning.

Emerton explained well that $V(I\cap J)$ is the natural scheme structure; I'd just like to add that if $V(I)$ and $V(J)$ are divisors, it is also reasonable to use the scheme structure $V(IJ)$ on their union, i.e. their sum as divisors. So both versions have their merits.

I'm still learning AG, so I am open to feedback on this answer:

If we think about it in category theoretic terms:

The union of closed subschemes should be a coproduct in the category of closed subschemes of $\operatorname{Spec} R$ (with closed immersions). Thus, in the affine case, you need the product in the category of ideals of $R$ (with inclusions). The product in this category is clearly intersection of ideals, not product of ideals.

• In an ordered set, viewed as a category, the product (resp. coproduct) of two objects is their "inf" (resp. "sup"). Check! You are probably misled by the terminology "product of ideals". Nov 21, 2011 at 19:50

You probably agree that for the intersection, things are easy: the most reasonable definition of $$V(I)$$ and $$V(J)$$ is as the fibred product $$Spec\,R/I\times_{Spec\,R} Spec\,R/J=Spec\,R/I\otimes_R R/J=Spec\,R/(I+J).$$

Now one natural way to view the union $$V(I)\cup V(J)$$ is as the result of glueing $$V(I)$$ and $$V(J)$$ along their intersection, that is, as the pushout of the diagram $$V(I)\leftarrow V(I+J)\rightarrow V(J)$$. And it is known (see e.g. Ferrand, Conducteur, descente et pincement) that such a pushout is representable in the category of schemes by the affine scheme whose function ring is the fibred product. Then you can do the following exercise: check that the 'diagonal' map $$R\to (R/I)\times_{R/(I+J)}(R/J)$$ is surjective with kernel $$I\cap J$$. In other words the sought-for fibred product is $$R/(I\cap J)$$. We end up with the conclusion that this pushout viewpoint leads naturally to the definition of the scheme-theoretic structure on the union being $$Spec\,R/(I\cap J)$$.

If I and J are "coprime" (i.e., if I+J = R) then it's true that the I∩J=IJ, so the two definitions are the same. This is true for example if I and J are distinct primes. To illustrate the kind of thing that happens when there are common factors, consider this silly example: R = ℝ[x,y], I = J = (x). So the union of the two varieties of these ideals is again the y-axis, and the question you need to ask is what do you want as the ring of functions on the union--do you want *R/*I∩J = ℝ[x,y]/(x) or do you want R/IJ = ℝ[x,y]/(x2). the issue is that in the world of algebraic geometry, this choice matters, since an algebraic variety is not only defined by its points, but also it's ring of functions. it seems reasonable that you would want X ∪ X = X for any variety X, so you'd better choose the first choice.

as a side note, the second choice is an example of a "nonreduced" variety, which i guess you wouldn't usually(?) see in a first course on algebraic geometry. it has more functions on it: one "full" dimension plus one "infinitesimal" dimension (i.e., with only linear functions in that direction).

• @maxmoo : It is not true that distinct primes are "coprime": take $I=(x)$ and $J=(y)$ in $k[x,y]$. These are distinct prime ideals, but they are definitely not "coprime". Mar 21, 2010 at 16:16
• hmmm yes that's a good point, better change "This is true" to "It's also true"? Mar 21, 2010 at 16:35
• Still not true: if $\mathfrak{p} \subset \mathfrak{q}$, the intersection and product are different. But even if you rule that out, it's no good: $R=k[x,y,z]/(xz-y^2)$, $\mathfrak{p}=(x,y)$, $\mathfrak{q}=(y,z)$. Then the product is $(xy,xz,yz)$ and the intersection is $(y)$. Mar 21, 2010 at 18:39

Relevant in comparing $IJ$ and $I \cap J$ is Atiyah-MacDonald, exer. 1.13, (iii), at the top of p. 9. It asserts that if $I, J$ are prime then $$IJ \subset rad(IJ) = I \cap J.$$ (Note: prime ideals are radical ideals, as are the intersections of radical ideals.)

Two instructive examples (1): $V(y) \cup V(y)$ as mentioned earlier, where $I = J$. Who wants that the coordinate ring of the x-axis to be to be $K[x,y]/(y^2)$? (2): the variety $V$ which is the union of the $z$-axis ($V_1$) and the $xy$-plane ($V_2$). $I = (x,y)$ and $J = z$ are the corresponding prime ideals. $I \cap J = I J = (zx, zy)$ is the ideal of polynomial functions which vanishes on $V$ corresponding to the decomposition of $V$ into $V_1$ and $V_2$ and the decomposition of $IJ$ into primes.