Prove that $ I = \bigcup_{k=0}^\infty I_k$ is an ideal of $R$.

Question

Let $ I_0 \subseteq I_1 \subseteq I_2 \subseteq I_3...$ be a sequence of ideals of a commutative ring $R$. In other words for every $k \in \mathbb{N}$ assume $I_k$ is an ideal of $R$ and that $I_k \subseteq I_{k+1}$ for every $k \ge 0$. Prove that $ I = \bigcup_{k=0}^\infty I_k$ is an ideal of $R$.

How do I go about proving that for all $k$, $I_k$ absorbs products?

Answer 1

I is closed under the action of $R$

Let $x\in I$ and $r\in R$. Then $x\in I_k$ for some $k\in\mathbb{Z}_{\geq 0}$ and thus $rx \in I_k \subset I$, since $I_k$ is an ideal.

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I is closed under addition

Now let $x,y\in I$. We may assume that $x \in I_k$ and $y\in I_j$ with $I_j \subseteq I_k$. Then $x,y\in I_k$ and hence $x+y\in I_k \subset I$ since $I_k$ is an ideal.

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Hence $I$ is an ideal of $R$.

Answer 2

You don't need to prove $I_k$ absorbs products. You are told to assume each $I_k$ is an ideal. So you already know $I_k$ is an ideal, and in particular it absorbs products.