# Why is the arbitrary sum, but not the arbitrary intersection, of ideals an ideal?

Two Questions:

1. What is the definition of an arbitrary sum of ideals? (I.e. arbitrarily infinitely many)

2. Why is the arbitrary sum of ideals an ideal, but not the arbitrary intersection of ideals?

My progress so far:

1. My textbook doesn't have a definition, and I haven't been able to find the definition using Google so far. I have two guesses (i) it is the set of all infinite sums, or (ii) the same as the previous except with the additional condition that cofinitely many summands in each sum be equal to zero. The former seems like the "natural" extension of the definition of sum to an arbitrary index set of ideals, but the latter seems like the "convenient" definition, and I'm not sure which goal (naturalness or convenience) should take priority.

2. I would expect that the arbitrary intersection of ideals is an ideal, because the ideal generated by any set is the arbitrary intersection of all ideals containing that set, and the radical of an ideal is the arbitrary intersection of all prime ideals containing the ideal. Meanwhile, it isn't clear to me that the infinite sum of elements of a ring should even be defined. Usually one has that the product of ideals is included in the intersection, and that usually only the finite product of ideals is an ideal, so in the exceptional cases where the product of ideals equals the intersection of ideals, I can somewhat see how the intersection wouldn't be an ideal. But there are at least two problems with this: if only the finite product of ideals is an ideal, then why isn't the same true for sums of ideals? Why is the multiplication operation in the ring being given a privileged position compared to the addition operation? Second, if the product of ideals equals the intersection of ideals in a certain ring, then why doesn't that just allow one to define arbitrary products of ideals, instead of making it impossible to define anything besides finite intersections of ideals? It seems clear to me that I am fundamentally misunderstanding some aspect of the operations (addition, multiplication, intersections) of ideals.

EDIT: http://planetmath.org/sumofideals Planetmath says that "the sum of ideals is the smallest ideal of the ring containing all of those ideals". So I guess my question could be thought of as -- why isn't the intersection of ideals the smallest ideal of the ring containing all of those ideals? For every other object I know of which is closed under intersections, it is the intersection which has this property. Do the sum and intersection coincide in some sense?

Note: This question is so I can show that the Zariski topology on Spec(R) for a commutative ring with unit is a topology -- I have already shown that finite unions and intersections of Zariski closed sets are Zariski closed, see here, so now the question is how to proceed to the general case. In the finite case, the union of Zariski closed sets corresponds to the intersection of ideals, while the intersection of Zariski closed sets corresponds to the sum of ideals. I would expect the intersection of ideals to be arbitrarily extensible, and the sums to be only finitely defined, but that would make the Zariski closed sets the open sets of a topology, which is clearly incorrect.

• The arbitrary intersection of ideals is an ideal. It is not the smallest ideal containing all ideals considered, but rather the dual : it is the biggest ideal contained in all ideals considered. – Arnaud D. Oct 23 '16 at 12:14
• @ArnaudD. If the arbitrary intersection of ideals is an ideal, does that mean that the arbitrary union of Zariski closed sets (for the Zariski topology of Spec(R)) is again closed? To prevent that result from being true is why I assumed that the arbitrary intersection of ideals must not be an ideal. – Chill2Macht Oct 23 '16 at 12:18
• I think this might help you : math.stackexchange.com/questions/846020/… – Arnaud D. Oct 23 '16 at 12:48
• The explanation is at the end : in the infinite case, union of closed sets does not correspond to intersection of ideals. The reason why the proof you give in your other question does not work is that you cannot take an infinite product of elements in the ring. – Arnaud D. Oct 23 '16 at 12:54

Let $(I_j)_{j\in J}$ be a family of ideals of a commutative ring $R$. Their sum is the set of all finite sums $\sum_{k=1}^n a_k$ such that $a_k\in \bigcup_{j\in j}I_j$ for all $1\leq k\leq n$. This definition is equivalent to the second one you give, and you can check that it is indeed an ideal and that it is the smallest ideal containing all the $I_j$.

For your second question, the intersection $\bigcap_{j\in J} I_j$ is an ideal. Indeed, it is non-empty as $0\in I_j$ for all $j$; and if $a,b\in \bigcap_{j\in J} I_j$ and $r\in R$, then $a,b\in I_j$ for all $j$, so that $a+b\in I_j$ and $ra\in I_j$ for all $j$, and thus $a+b\in \bigcap_{j\in J} I_j$ and $ra\in \bigcap_{j\in J} I_j$.

I finally found a source (see here) which gives the definition: $$\sum_{j \in J} I_j := \left\{ \sum_{j\in J} x_j: x_j \in I_j\ (\forall j)\ \text{and only finitely many }x_j\text{ are nonzero} \right\} .$$

Related to Arnaud D.'s comment, we apparently have the relationship that: $$\sum_{j\in J} I_j = \left\langle \bigcup_{j\in J} I_j \right\rangle$$ where $\langle \rangle$ denotes the ideal generated by a set, the intersection of all ideals containing the set.

Thus the set of all proper ideals $I \subseteq R$ of a ring turns out to not only be a partial order under set inclusion, but actually a lattice, meaning that all supremums (least upper bounds) and infimums (greatest lower bounds) exist/are defined.

The infimum of a set of ideals is the arbitrary intersection of the ideals, while the supremum of ideals is the sum of ideals, since it is the smallest ideal containing all of the ideals. Thus the sum and intersection of ideals are indeed somewhat related operations.

Claim: $$\sum_{j\in J} I_j = \left\langle \bigcup_{j\in J} I_j \right\rangle$$ Proof: Let $x \in \sum_{j\in J} I_j$. Then $x=x_1 + \dots + x_n$, with $x_1 \in I_{j_1}, \dots, x_n \in I_{j_n}$, $j_1, \dots, j_n \in J$. Then since $x_1, \dots, x_n \in \bigcup_{i=1}^n I_{j_i}$, we have that $x_1 + \dots + x_n \in L$ for any ideal $L$ such that $L \supseteq \bigcup_{i=1}^n I_{j_i}$. But of course because $\bigcup_{i=1}^n I_{j_i} \subseteq \bigcup_{j \in J} I_j$, so: $$\bigcup_{i=1}^n I_{j_i} \subseteq \bigcup_{j \in J} I_j \subseteq \bigcap_{J \supseteq \bigcup_{j \in J} I_j} J = \left\langle \bigcup_{j \in J} I_j\right\rangle, \\ L=\left\langle \bigcup_{j \in J} I_j\right\rangle \supseteq \bigcup_{i=1}^n I_{j_i}$$ so $x_1 + \dots + x_n \in \left\langle \bigcup_{j \in J} I_j\right\rangle$.

$x_1 + \dots + x_n \in \sum_{j\in J} I_j$ was arbitrary, we have shown that $\sum_{j\in J} I_j \subseteq \left\langle \bigcup_{j \in J} I_j\right\rangle$.

Since $\sum_{j \in J} I_j$ is an ideal and $\sum_{j \in J} I_j \supseteq \bigcup_{j\in J} I_j$, $$\sum_{j \in J} I_j \supseteq \bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J = \left(\bigcap_{\{J: J \supseteq \bigcup_{j \in J} I _j \} \setminus \{\sum_{j \in J} I_j \}} J\right) \cap \left( \sum_{j \in J} I_j \right).$$ However, $\bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J$ is literally the definition of $\left\langle \bigcup_{j\in J} I_j \right\rangle$, i.e. $\bigcap_{\{J: J \supseteq \bigcup_{j\in J} I_j\}} J = \left\langle \bigcup_{j \in J} I_j \right\rangle$ trivially, we have shown that $$\sum_{j \in J} I_j \supseteq \left\langle \bigcup_{j \in J} I_j \right\rangle$$ In other words, $\left\langle \bigcup_{j \in J} I_j \right\rangle$ is the smallest ideal containing $\bigcup_{j \in J} I_j$, $\sum_{j\in J}I_j$ is an ideal containing the union, so it must contain the smallest ideal containing the union. $\square$