Question about structure of ideals of a sub-ideal

Question

Let $R$ a ring and $I$ be an ideal of $R$. Then, let $I'$ be an ideal of $I$. Is it true that one can always write $I'$ as the intersection $I\cap J$ for some ideal $J$ of $R$?

I already know that taking $J=I'$ does not work in general, but I am not sure how to approach the general case.

Answer 1

Here’s an explicit example.

Let $R$ be the ring of all $2\times 2$ matrices with real coefficients of the form $$\left(\begin{array}{cc} a & b\\ 0 & c \end{array}\right).$$ It is a nice exercise to verify that the ideals of this ring are exactly the following:

1. The zero ideal.
2. The ring $R$.
3. The ideal $J_1$ of all matrices with first column equal to $0$.
4. The ideal $J_2$ of all matrices with second row equal to $0$.
5. The ideal $J_3=J_1\cap J_2$ of all matrices with $(1,1)$ and $(2,2)$ entries equal to $0$.

Now consider the ideal $J_1$; when you multiply two such matrices, you get $$\left(\begin{array}{cc} 0 & b\\ 0 & c\end{array}\right) \left(\begin{array}{cc} 0 & s\\ 0 & t \end{array}\right) = \left(\begin{array}{cc} 0 & bt\\ 0 & ct \end{array}\right).$$ That means that the set of matrices that have first row and first column equal to $0$, that is matrices of the form $$\left(\begin{array}{cc} 0 & 0\\ 0 & c \end{array}\right)$$ are an ideal of $J_1$. However, there is no ideal of $R$ that, intersected with $J_1$, will give this set. Of course, any such intersection would have to be an ideal of $R$, and we already know this set is not an ideal of $R$ (if you believe my list is complete).

Answer 2

No, it is not true. The intersection of ideals of $R$ will be an ideal of $R$ (an easy exercise for you) but you may cook up examples where $I'$ is an ideal of $I$ but not an ideal of $R$. It seems that you already have one, since you know taking $J = I'$ does not work in general. That is, actually, precisely when the statement fails!