# I Two-sided Prinicipal Ideal - the Noncommutative Case - D&F

1. Aug 20, 2016

### Math Amateur

I am reading Dummit and Foote's book: Abstract Algebra ... ... and am currently focused on Section 7.4 Properties of Ideals ... ...

I have a basic question regarding the generation of a two sided principal ideal in the noncommutative case ...

In Section 7.4 on pages 251-252 Dummit and Foote write the following:

In the above text we read:

" ... ... If $R$ is not commutative, however, the set $\{ ras \ | \ r, s \in R \}$ is not necessarily the two-sided ideal generated by $a$ since it need not be closed under addition (in this case the ideal generated by $a$ is the ideal $RaR$, which consists of all finite sums of elements of the form $ras, r,s \in R$). ... ... "

I must confess I do not understand or follow this argument ... I hope someone can clarify (slowly and clearly ) what it means ... ...

Specifically ... ... why, exactly, is the set $\{ ras \ | \ r, s \in R \}$ not necessarily the two-sided ideal generated by $a$?

The reason given is "since it need not be closed under addition" ... I definitely do not follow this statement ... surely an ideal is closed under addition! ...

... ... and why, exactly does the two-sided ideal generated by a consist of all finite sums of elements of the form $ras, r,s \in R$ ... ... ?

Hope someone can help ... ...

Peter

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To give readers the background and context to the above text from Dummit and Foote, I am providing the introductory page of Section 7.4 as follows ... ...

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• ###### D&F - Intro to Section 7.4 ... ....png
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2. Aug 20, 2016

### Staff: Mentor

In an ideal $\mathcal{I}$ you have $x+y\in \mathcal{I}$ for elements $x\,,\,y\in \mathcal{I}$. This is a basic part of its definition.
If you now have two elements $ras \; (r,s\in R\,; a\in \mathcal{I})$ and $paq \; (p,q \in R\,; a\in \mathcal{I})$ there is - in general - no way to write $ras+paq=uav$ because you cannot pull the factors $r,s,p,q$ on the other side of $a$.
Therefore $\mathcal{I}=(a) = RaR = LC(\{ras \,|\, r,s \in R\}) \supsetneq \{ras \,|\, r,s \in R\}$ is - in general - a proper inclusion. The latter is only a set.

In the commutative case we have
$ras+paq=r(as)+p(aq)=r(sa)+p(qa)=(rs)a+(pq)a=(rs+pq)a$ and all linear combinations are of the form $ua\;(u=rs+pq \in R)$.
However, this calculation is not allowed in non-commutative rings.

3. Aug 20, 2016

### micromass

Staff Emeritus
You should try to find a counterexamples yourself where $\{ras~\vert~r,s\in R\}$ is not an ideal. Think of some simple noncommutative rings.

4. Aug 21, 2016

### Math Amateur

Thanks fresh_42 ... that was most helpful ...

Thanks for suggestion micromass ... will try messing around with 2 by 2 matrices ...

Peter