Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 7 down vote favorite
1

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?)

share|improve this question
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
1  
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

up vote 6 down vote

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.

share|improve this answer
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.