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The example is given in the following picture:

enter image description here

Here is the text for clarity

EXAMPLES. If $I$ is a left ideal of a ring $R$, then $I$ is a left $R$-module with $ra(r \varepsilon R,a \varepsilon I)$ being the ordinary product in $R$. In particular, $0$ and $R$ are $R$-modules. Furthermore since $I$ is an additive subgroup of $R$, $r / I$ is an (abelian) group. $R/I$ is an $R$-module with $r(r_1+I) = rr_1 + I$. $R/I$ need not be a ring however unless $I$ is a two sided ideal.

It is not clear for me how "$I$ is a left $R$-module with ra being the ordinary product in $R$", I know that by the definition of an ideal it is a ring and hence an additive abelian group but how the ring action is applied it is not clear for me the details, how multiplication of a ring element is distributed over addition of 2 ideal elements and so on, could anyone clarify this for me please?

Also,I did not understand why "$R/I$ need not be a ring, however, unless $I$ is a two-sided ideal" could anyone explain this for me please?

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    $\begingroup$ With the additive structure it is definitely a group. However, the multiplication of cosets is well-defined if and only if $I$ is a two-sided ideal. $\endgroup$ – Randall Oct 5 '17 at 2:11
  • $\begingroup$ why " the multiplication of cosets is well-defined if and only if II is a two-sided ideal."? @Randall $\endgroup$ – Intuition Oct 7 '17 at 18:16
  • $\begingroup$ Standard result proven in many algebra books. It's a long argument. $\endgroup$ – Randall Oct 7 '17 at 18:19
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The meaning of the sentence seems straightforward so I guess you are looking for a counterexample.

Let $T$ be the right ideal of matrices in $M_2(F)$ with bottom row zero.

$T$ is not closed by multiplication on the left by $A=\begin{bmatrix}0&0\\ 1&0\end{bmatrix}$.

Then multiplication on the cosets isn't well defined since $B=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}\equiv 0$ mod $T$, but $AB\not\equiv 0$ mod $T$.

It is not clear for me how "$I$ is a left $R$-module with $ra$ being the ordinary product in $R$"

The module structure on $R/I$ is given by $r\cdot(a+I):=ra+I$. This is well-defined since $I$ is a left $R$ module. That is, if $r=r'$ and $a+I=a'+I$, we first have that $a-a'\in I$, and therefore $r(a-a')\in I$. This says $ra+I=ra'+I$. Since multiplication in $R$ is well-defined, $ra'=r'a'$ and hence $ra'+I=r'a'+I$. Stringing things together, $ra+I=r'a'+I$. Therefore the map $R\times R/I\to R/I$ is well-defined.

It is an easy matter to show that the rest of the module axioms hold for $R/I$ because they hold for $R$ itself.

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  • $\begingroup$ I have updated my question .... could you please read it if you have time? $\endgroup$ – Intuition Oct 7 '17 at 2:12
  • $\begingroup$ why $R/I$ need not be a ring unless $I$ is a 2 sided ideal? $\endgroup$ – user426277 Oct 30 '17 at 11:13
  • $\begingroup$ @Idonotknow $R/I$ won't be a ring using the natural definition ($(x+I)(y+I)=xy+I$) because this operation will not be well-defined. As proven above. $\endgroup$ – rschwieb Oct 30 '17 at 13:23

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