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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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I take it your definition of ring does not require a neutral element for multiplication? Otherwise, $J$ is not a subring unless $J=R$. –  Daan Michiels May 11 '12 at 21:22
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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

1 Answer 1

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

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