Prove that for every ideal: $I_1I_2 \subset I_1 \cap I_2 \subset I_1 \subset I_1 +I_2$.

Question

Prove that for every ideal: $I_1I_2 \subset I_1 \cap I_2 \subset I_1 \subset I_1 +I_2$.

My try:

• $I_1 \subset I_1 +I_2$

By definition of an ideal $0\in I$ so:

$I_1 +I_2=\left\{a+b:a\in I_1, b\in I_2\right\}\supset \left\{a+0:a\in I_1, 0\in I_2\right\}=\left\{a:a\in I_1\right\}=I_1$

• $I_1 \cap I_2 \subset I_1$

$I_1 \cap I_2 =\left\{a: a\in I_1, a\in I_2 \right\}$

So the conclusion is real because in $I_1 \cap I_2$ are elements from $I_1$

At this moment I need confirmation if what I've done so far is correct.

Moreover, I don't know how to prove that $I_1I_2 \subset I_1 \cap I_2$, because $I_1I_2=\left\{\sum a_ib_i: a_i\in I_1, b_i \in I_2 \right\}=a_1b_1+...+a_nb_n$ so I think that this conclusion is not real.

Answer 1

Let $a_i\in I_1$ and $b_i\in I_2$ arbitrary.
Ideal means left and right ideal at once.

Then, since $I_1$ is a right ideal, $a_ir\in I_1$ for every element $r$ in the whole ring, in particular $a_ib_i\in I_1$.
Similarly, since $I_2$ is a left ideal, $a_ib_i\in I_2$.

Consequently, $a_ib_i\in I_1\cap I_2$.
And $I_1\cap I_2$ is closed under addition.

An alternative way to write the same: $$I_1I_2\subseteq I_1R\subseteq I_1\\ I_1I_2\subseteq RI_2\subseteq I_2\\ \text{hence } I_1I_2\subseteq I_1\cap I_2\,. $$