Union Over a Totally Ordered Set of Ideals is an Ideal

Question

I am trying to understand the proof of a theorem which uses Zorn's lemma. I understand quite well all parts of the proof except for one point:

Let $R$ be a ring and define $K\doteq \{I\subseteq R \;\vert\; I \textrm{ is an ideal} \}$. Let $T$ be a totally ordered (by inclusion) subset of $K$. Then $$J\doteq\bigcup_{I\in T}I$$ is an ideal.

No explanation was given for this claim, so I imagine it is trivial. However, I cannot seem to see why. I know that the union of ideals in general is not an ideal. By the criteria for a union of ideals to actually be an ideal, we would need that for all $A,B\in T$, there is $C\in T$ such that $A\cup B\subseteq C$. Why is this true? I feel like there is something simple I am overlooking. Thank you for your time.

Answer 1

You are right that for $A,B\in T$, we need $C\in T$ with $A\cup B\subset C$. But we get this automatically because of the total order, since either $A\leq B$ or $B\leq A$ and this order is defined by inclusion.

Answer 2

Well, first you have to see that if $x,y\in J$, then $\exists I'\in T, x,y \in I'$, this is true becasuse T is a totally ordered subset of K. Therefore, $x+y\in I'\subset J$.

Then take $x \in J$, then $\exists I' \in T, x\in I'$, and $I'$ is an ideal, thus you have that $\forall a \in R, ax \in I'$ and $xa \in I'$. So, as $I'\subset J$, $\forall a \in R, ax \in J$ and $xa \in J$ .

And so, you conclude that $J$ is an ideal of $R$.

I have to point that in this particular case the union of ideals is an ideal because $K$ is a totally ordered set.