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 Post subject: Isomorphism
PostPosted: 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: Isomorphism
PostPosted: Thu Feb 23, 2017 2:02 am 
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Joined: Tue Nov 10, 2015 8:25 pm
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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: Isomorphism
PostPosted: Fri Feb 24, 2017 3:37 pm 
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Joined: Mon Nov 09, 2015 1:52 pm
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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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