Does $A/I \simeq A$ implies $I =0$? 
Let $A$ be a unital algebra over a field, and $I$ an ideal of $A$. If there is an algebra isomorphism $A/I\simeq A$, does it implies that $I=0$? If no, then for what type of algebras can this be true?

I'm not an expert on algebras, so I cannot make the question precise, but its related to my work and I need to answer this particular question. Any help is appreciated.
 A: The isomorphism theorem tells you that if you have a map $\varphi : A \to B$ of $k$-algebras, then $A/\ker \varphi$ is isomorphic to the image of $\varphi$.
So your question can be reduced to asking: Is there an endomorphism $A \to A$ which is surjective but not injective.
(Why is this equivalent to your question?)
Can you now solve this? Consider examples such as $A = k[x_1, x_2, \ldots]$ or $A = \prod_{i \in \Bbb N} k$.
A: This is false, even for finitely generated algebras. For non-finitely generated algebras, examples such as an infinite direct product of copies of a given ring are easy to construct.
It is a theorem of Jacques Lewin from 1967, like for groups, that a finitely generated residually finite ring is Hopfian, i.e., never isomorphic to a proper quotient of itself. Residually finite here means that the intersection of all subrings of finite index is $0$. This leaves open the option that a non-residually finite ring might be non-Hopfian.
There are examples of finitely generated groups that are not Hopfian (and thus not residually finite either). If $G$ is such a group then $kG$ is also non-Hopfian for a field $k$. (This appears to be well known, see the first line of Section 5 of Orzech-Ribes' Residual Finiteness and the Hopf Property in Rings.)
