MHB Why Is the Set {ras | r, s ∈ R} Not a Two-Sided Ideal in Noncommutative Rings?

[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:https://www.physicsforums.com/attachments/5899
https://www.physicsforums.com/attachments/5900In 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======================================================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 ... ...

 

[Opalg]

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 ... ...

You might find it helpful to explore a concrete example. One of the simplest examples of a noncommutative ring is the ring $R$ of $2\times2$ matrices (over the real numbers, say). Suppose you take $a$ to be the matrix $\begin{bmatrix}1&0 \\0&0 \end{bmatrix}.$ If $r = \begin{bmatrix}r_{11}&r_{12} \\r_{21}&r_{22} \end{bmatrix}$ and $s = \begin{bmatrix}s_{11}&s_{12} \\s_{21}&s_{22} \end{bmatrix}$ then you can check that $ras = \begin{bmatrix}r_{11}s_{11}&r_{11}s_{12} \\ r_{21}s_{11}&r_{21}s_{12} \end{bmatrix}.$ Notice that $ras$ has determinant $0$.

Suppose now that we take $b = \begin{bmatrix}0&0 \\1&0 \end{bmatrix}$ and $c = \begin{bmatrix}0&1 \\0&0 \end{bmatrix}$. Then you find that $bac = \begin{bmatrix}0&0 \\0&1 \end{bmatrix}$. If $e$ is the identity matrix $e= \begin{bmatrix}1&0 \\0&1 \end{bmatrix}$ then $a = eae$, and $$eae + bac = \begin{bmatrix}1&0 \\0& 0\end{bmatrix} + \begin{bmatrix}0&0 \\0&1 \end{bmatrix} = \begin{bmatrix}1&0 \\0&1 \end{bmatrix} = e.$$ But the determinant of $e$ is not zero, so $e$ is not of the form $ras.$ So the above equation gives you an example of two matrices of the form $ras$ whose sum is not of that form.

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 $$ras, r,s \in R$$ ... ... ?

That $2\times2$ matrix example illustrates the fact that the set $$\{ ras \ | \ r, s \in R \}$$ can contain two elements whose sum is not in the set. In other words, the set need not be closed under addition and so is not necessarily a two-sided ideal.

But if you allow finite sums of elements of that form then they are closed under addition, because if you add two finite sums then the result is still a finite sum!

 

[Math Amateur]

You might find it helpful to explore a concrete example. One of the simplest examples of a noncommutative ring is the ring $R$ of $2\times2$ matrices (over the real numbers, say). Suppose you take $a$ to be the matrix $\begin{bmatrix}1&0 \\0&0 \end{bmatrix}.$ If $r = \begin{bmatrix}r_{11}&r_{12} \\r_{21}&r_{22} \end{bmatrix}$ and $s = \begin{bmatrix}s_{11}&s_{12} \\s_{21}&s_{22} \end{bmatrix}$ then you can check that $ras = \begin{bmatrix}r_{11}s_{11}&r_{11}s_{12} \\ r_{21}s_{11}&r_{21}s_{12} \end{bmatrix}.$ Notice that $ras$ has determinant $0$.

Suppose now that we take $b = \begin{bmatrix}0&0 \\1&0 \end{bmatrix}$ and $c = \begin{bmatrix}0&1 \\0&0 \end{bmatrix}$. Then you find that $bac = \begin{bmatrix}0&0 \\0&1 \end{bmatrix}$. If $e$ is the identity matrix $e= \begin{bmatrix}1&0 \\0&1 \end{bmatrix}$ then $a = eae$, and $$eae + bac = \begin{bmatrix}1&0 \\0& 0\end{bmatrix} + \begin{bmatrix}0&0 \\0&1 \end{bmatrix} = \begin{bmatrix}1&0 \\0&1 \end{bmatrix} = e.$$ But the determinant of $e$ is not zero, so $e$ is not of the form $ras.$ So the above equation gives you an example of two matrices of the form $ras$ whose sum is not of that form. That $2\times2$ matrix example illustrates the fact that the set $$\{ ras \ | \ r, s \in R \}$$ can contain two elements whose sum is not in the set. In other words, the set need not be closed under addition and so is not necessarily a two-sided ideal.

But if you allow finite sums of elements of that form then they are closed under addition, because if you add two finite sums then the result is still a finite sum!

Thanks Opalg ... most helpful to have an example to clarify the issue ...

Appreciate your help ...

Thanks again,

Peter