Prove $\bigcup _{k=1}^n I_k$ is in ideal

Question

Let $n$ ideals such that $I_1 \subset I_2 \subset \cdots \subset I_n$. Prove that $J=\bigcup _{k=1}^n I_k$ is also an ideal.

We need to show three things:

1. $0\in J$. Trivial...
2. $a,b \in J \implies a+b\in J$. Lets assume $a,b \in J$. Therefore, $a\in I_x$ and $b\in I_y$ for some $x,y$. WLOG, $I_x \subset I_y$. Therefore, $a+b\in I_y$ because $a\in I_x \subset R$.
3. $a\in R, j\in J \implies aj,ja \in J$.
   Again, since $j \in J$ then there's $I_x$ such that $j \in I_x$, so $aj \in I_x$ and therefore $aj \in J$.

Am I right? Should I change something in my proof?

Answer 1

I object to the end of the second line; you should be invoking that $a \in I_y$ rather than $a \in R$. But as pointed out in the other answer, $I_n = J$ and this is all rather unnecessary.

It is necessary in the case of an infinite chain of ideals $I_0 \subset I_1 \subset \dots$. You, then, have given a correct proof that $J = \bigcup_{k=1}^\infty I_k$.

Answer 2

Note: It is an increasing chain of ideals. So,

$$\bigcup_{k=1}^{n} I_k=I_n$$

and $I_n$ is itself an ideal. So there's really nothing to prove.