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?

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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