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If $R$ is a ring, J is an ideal in $R$, and $I$ is an ideal of $J$ (with $J$ considered as a ring), does it follow that $I$ is an ideal of $R$? That is, is $I$ necessarily closed under multiplication by elements of $R$? Surely $I$ is closed under multiplication by elements of $J$ (since $I$ is an ideal of $J$). The "obvious" approach to proof fails so quickly that it must be false, that "an ideal of an ideal is not necessarily an ideal of the big ring". Please provide a counterexample. (Or a proof?)

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Fun fact: for a von Neumann regular ring $R$, being an ideal is transitive in the sense that $I\lhd J\lhd R$ implies $I\lhd R$. – rschwieb May 11 '12 at 22:22
It was very surprising for me to learn that such a result is true for closed ideals of C*-algebras. IIRC one uses approximate identities. – Martin Brandenburg Apr 22 '15 at 18:05

Let $M$ be the matrix ring $M_2(\mathbb Q)$, and let $I=M$ viewed as simply a rational vector space. Consider the abelian group $R=M\oplus I$ and define on it a multiplication such that $$(a,v)\cdot(b,w)=(ab,aw+vb);$$ all products on the right hand side of this definition are good ol' matrix products.

You can easily check that this turns $R$ into a ring and that $I$ is an ideal of $R$ such that the product of any two elements of $I$ is zero in $R$. This has the immediate consequence that any $\mathbb Q$-subspace $J$ of $I$ is an ideal of $I$.

A little work will show, on the other hand, that $I$ does not properly contain any non-zero ideal of $R$.

N.B. I interpreted the word ideal in your question to mean bilateral ideal.

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@rschwieb: Let $K=\left\{\pmatrix{q&0\\0&0}:q\in\Bbb Q\right\}$; $0\oplus K$ isn't an ideal in $R$, since $$\left\langle\pmatrix{1&0\\1&0},v\right\rangle\cdot\left\langle 0,\pmatrix{1&0\\0&0}\right\rangle=\left\langle0,\pmatrix{1&0\\1&0}\right\rangle \notin 0\oplus K\;.$$ – Brian M. Scott May 11 '12 at 22:04
@rschwieb, as an $R$-bimodule $I$ is simple: this is true because this is the same as its being simple as an $R/I$-bimodule (this makes sense because $I^2=0$ so one can turn $I$ into an $R/I$-bimodule), and this statement is equivalent to that of $M$ being a simple algebra. – Mariano Suárez-Alvarez May 11 '12 at 22:11
@BrianM.Scott Ah ok, I had somehow forgotten the left hand entries were matrices and not rationals. – rschwieb May 11 '12 at 22:20
For the OP, I'd like to offer a method of remembering this construction that doesn't involve explaining the formula: the ring may be viewed formally as $R=\{ \pmatrix{a&b\\0&a}\mid a,b\in M \}$, with regular matrix multiplication. – rschwieb May 11 '12 at 23:16

Let us consider the sets$$R=\left\{ \begin{pmatrix} a & b & c\\ d & e & f\\ 0 & 0 & g \end{pmatrix}:a,b,c,d,e,f,g\in\mathbb{Z}\right\}\\J=\left\{\begin{pmatrix} 0&0&a\\0&0&b\\0&0&0\end{pmatrix}:a,b\in \mathbb{Z}\right\}\\I=\left\{\begin{pmatrix}0 &0&a\\0&0&0\\0&0&0\end{pmatrix}:a\in\mathbb{Z}\right\}$$Now we can easily check that $R$ is a ring with respect to usual matrix addition and multiplication and $J$ is an ideal of $R$ and $I$ is an ideal of $J$ but $I$ is not an ideal of $R$.

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I can provide a commutative counterexample: Let $R=\mathbb Q[X]/(X^2)$, and $I=(\bar X)$. Then the ideals of $R$ contained by $I$ are simply $I$ and $0$. But any additive subgroup of $I$ is an ideal of $I$ (since the multiplicative structure on $I$ is the trivial one).

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