# Ideals in a ring

• Dec 17th 2008, 03:12 PM
anlys
Ideals in a ring
Let c be in a ring R and let I = {rc | r is in R}
Give an example to show that if R is not a commutative ring, then I need not be an ideal.

Can someone help me this?
• Dec 17th 2008, 03:26 PM
Byun
Quote:

Originally Posted by anlys
Let c be in a ring R and let I = {rc | r is in R}
Give an example to show that if R is not a commutative ring, then I need not be an ideal.

For any $\displaystyle k \in R$ and any $\displaystyle rc \in I$, $\displaystyle k(rc) \in I$.
But $\displaystyle (rc)k$ may not be in $\displaystyle R$.
• Dec 17th 2008, 03:30 PM
anlys
Quote:

Originally Posted by Byun
For any $\displaystyle k \in R$ and any $\displaystyle rc \in I$, $\displaystyle k(rc) \in I$.
But $\displaystyle (rc)k$ may not be in $\displaystyle R$.

Hello Byun,
Thanks for your input. Do you also have a particular example to show this?
• Dec 17th 2008, 04:17 PM
NonCommAlg
Quote:

Originally Posted by anlys
Let c be in a ring R and let I = {rc | r is in R}
Give an example to show that if R is not a commutative ring, then I need not be an ideal.

Can someone help me this?

$\displaystyle R=M_2(\mathbb{Z}), \ c=\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}.$ we have: $\displaystyle I=\{rc: \ r \in R \}=\left \{\begin{bmatrix}a & 0 \\ b & 0 \end{bmatrix}: \ a,b \in \mathbb{Z} \right \}.$ let $\displaystyle s=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}.$ then $\displaystyle cs=s \notin I. \ \Box$