Exercise 11.1.8

Proof.

(a)

$I+J$ is a subgroup of $(R,+)$. If $t\in I+J$ and $u\in R$, then $t=r+s,r\in I,s\in J$. As $I,J$ are ideals of $R$, $\mathit{ur}\in I,\mathit{us}\in J$, therefore $\mathit{ut}=\mathit{ur}+\mathit{us}\in I+J$, so $I+J$ is an ideal of $R$.

$\bar{I}$ is a subgroup of $R∕J$, image of $I$ by the natural projection $\pi :R\to R∕J$.

Let $\bar{r}=r+J,r\in I$ be an element of $\bar{I}$, and $\bar{s}=s+J,s\in R$ be any element of $R∕J$. Then by definition of the laws in $R∕J$,

$$\bar{s}\bar{r}=(s+J)(r+J)=\mathit{sr}+J,$$

and $\mathit{sr}\in I$, since $s\in R,r\in I$. Therefore $\bar{s}\bar{r}\in \bar{I}$, so $\bar{I}$ is an ideal of $R∕J$.

(b)

If $r+(I+J)=s+(I+J)$, then $r-s\in I+J$, so $r-s=i+j$ for some $i\in I,j\in J$.

Then $(r+J)-(s+J)=(r-s)+J=i+j+J=i+J\in \bar{I}$, since $i\in I$. So $(r+J)+\bar{I}$ depends only of the class of $r$ modulo $I+J$, and the map

$$\phi :\{\begin{array}{ccc}\hfill R∕(I+J)\hfill & \hfill \to \hfill & \hfill (R∕J)∕\bar{I}\hfill \\ \hfill r+(I+J)\hfill & \hfill \mapsto \hfill & \hfill (r+J)+\bar{I}\hfill \end{array}$$

is well defined.

$\phi$ is a ring homomorphism: if $\bar{r}=r+I+J,\bar{s}=s+I+J$, then

$$\begin{array}{llll}\hfill \phi (\bar{r})+\phi (\bar{s})& =((r+J)+\bar{I})+((s+J)+\bar{I})& \hfill & \\ \hfill & =((r+J)+(s+J))+\bar{I}& \hfill & \\ \hfill & =(r+s+J)+\bar{I}& \hfill & \\ \hfill & =\phi (\overline{r+s})& \hfill & \\ \hfill & =\phi (\bar{r}+\bar{s})& \hfill & \end{array}$$

and similarly $\phi (\bar{r})\phi (\bar{s})=\phi (\bar{r}\bar{s})$ (and $\phi (1)=(1+J)+\bar{I}$ is the unit of $(R∕J)∕\bar{I}$).

$\phi$ is injective: if $r+(I+J)\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $(r+J)+\bar{I}=\bar{I}$, therefore $r+J\in \bar{I}$, so $r+J=i+J$ ,where $i\in I$, so $j=r-i\in J$ and $r=i+j\in I+J$, $r+(I+J)=I+J$ is zero in $R∕(I+J)$.

$\phi$ is surjective: any element $y$ of $(R∕J)∕\bar{I}$ is of the form $y=u+\bar{I}$, where $u\in R∕J$ is such as $u=r+J$, where $r\in I$, so $y=(r+J)+\bar{I}=\phi (r+I+J)$.

Conclusion :

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