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 Post subject: IsomorphismPosted: Wed Feb 22, 2017 1:20 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Let $\displaystyle{\left(R,+,\cdot\right)}$ be an associative ring with unity $\displaystyle{1=1_{R}}$

and $\displaystyle{I\,,J}$ double ideals of $\displaystyle{\left(R,+,\cdot\right)}$ . If there exists

$\displaystyle{\phi\in Aut(R)}$ such that $\displaystyle{\phi(I)=J}$, then prove that

$\displaystyle{R/I\cong R/J}$ as rings.

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 Post subject: Re: IsomorphismPosted: Thu Feb 23, 2017 2:02 am
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
Consider the composite map
$R \overset{\varphi}{\to} R \overset{\pi}{\twoheadrightarrow} R/J$
Note that
• $\text{Ker}( \pi \circ \varphi ) = I$, as $\text{Ker}( \pi ) = J$ and $\varphi(I) = J$
• $\text{Im}( \pi \circ \varphi ) = R/J$, as $\pi$ is surjective and $\varphi \in \text{Aut}(R)$
Now, apply the 1st isomorphism theorem to $\pi \circ \varphi$ to obtain the desired result.

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 Post subject: Re: IsomorphismPosted: Fri Feb 24, 2017 3:37 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Here is another idea (similar to the previous one) .

We define $\displaystyle{F:R/I\to R/J}$ by $\displaystyle{F(x+I)=\phi(x)+J}$

If $\displaystyle{x+I=y+I\in R/I}$, then $\displaystyle{x-y\in I}$, so

$\displaystyle{\phi(x-y)\in \phi(I)=J\implies \phi(x)-\phi(y)\in J\implies \phi(x)+J=\phi(y)+J}$ .

Therefore, the map $\displaystyle{F}$ is well defined and it's obviously, ring epimorphism.

Finally, let $\displaystyle{x+I\in\rm{Ker}(F)}$. Then, $\displaystyle{F(x+I)=\phi(x)+J=J}$, which means

that $\displaystyle{\phi(x)\in J=\phi(I)\implies \exists\,a\in I\,,\phi(x)=\phi(a)}$ and since

$\displaystyle{\phi}$ is monomorphism, we get $\displaystyle{x=a\in I\implies x+I=I}$. So,

$\displaystyle{\rm{Ker}(F)=\left\{I\right\}}$ and $\displaystyle{\phi}$ is also monomorphism.

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