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I want to understand isomorphisms between quotient rings of a given (noncommutative) ring $R$ (or even an $K$-Algebra) better and if there is some way to connect them to automorphisms of $R$. Now, given an automorphism $$ \phi:R \to R$$ I can mod out any Ideal $I$ in the image and get another ringhomomorphism by composing $\phi$ and the projection onto the quotient ring. Then I can just mod out the kernel of this function and get an isomorphism of quotient rings. Now my question is: If I'm given an isomorphism $$\phi': R/I \to R/J$$ is there some way to construct an automorphism of $R$ that, after going through the process described above, gives me back the isomorphism?

EDIT: Ok, the first answer showed that that's not possible in that generality. What happens if i restrict myself to finitely generated algebras, or even better algebras where every ideal is finitely generated?

A second but somewhat related question: I understand that it's quite hard to determine whether two quotient rings are isomorphic as rings or algebras. Is there any way to at least rule it out? Are there some invariants that are somewhat possible to determine given an, lets say, minimal generating system of the ideals?

Context: I'm interested in the free algebra in $n$ generators over $\mathbb{C}$ and when modding out some Ideals lead to the same quotient algebra over $\mathbb{C}$.

I'm thankful for any comments or answers or corrections!

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Let $R$ be a non-zero commutative ring and $S = \displaystyle \prod_{n=0}^{\infty}R$, then consider the ideals $I=(1,0,0, \dots)$ and $J=(0)$. We have that $S/I \cong S \cong S/J$, but obviously there is no automorphism that sends $I$ to $J$.

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  • $\begingroup$ Thanks for the answer! That shows me that I shouldn't have asked the question in such generality. Maybe I should have restricted myself to finitely generated algebras. $\endgroup$ – Verdruss Oct 13 '17 at 12:58
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An example with finitely generated algebras (in fact, finite rings) would be the following: let $R=\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, let $I=2\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ and let $J = \mathbb{Z}/4\mathbb{Z}\times \{0\}$. Then $$ R/I \cong \mathbb{Z}/2\mathbb{Z} \cong R/J, $$ but no automorphism of $R$ can send $I$ to $J$, since $J$ contains an element of additive order $4$ while $I$ does not.

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    $\begingroup$ In the same spirit but with algebras over a field $k$: $R = k[x,y]/\langle x^4, y^2 \rangle$, $I = \langle x\rangle$, $J = \langle x^2,y\rangle$. Then $R/I \simeq k[x]/\langle x^2\rangle \simeq k[y]/\langle y^2 \rangle \simeq R/J$, but no automorphism of $R$ can send the principal ideal $I$ to the non-principal $J$. $\endgroup$ – Torsten Schoeneberg Oct 18 '17 at 1:07
  • $\begingroup$ Thanks a lot for the answer and the comment @TorstenSchoeneberg! I guess I should start looking for a counterexample in the free algebra case instead of a proof. But maybe it's true there because I have no relations between the variables... But thank you! $\endgroup$ – Verdruss Oct 18 '17 at 10:07

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