This question has been solved.
I have a question about ascending chain between different rings.
Here is my question:
Assume that $f:S \rightarrow R$ be an injective ring homorphism and $I_1\subset I_2\subset\cdots$ is an ascending chain of left $R-$ideals.
Then $I_i$ can also be seen as left $S-$modules.
We assume that $S$ is notherian.
Then we have $I_1\subset I_2\subset\cdots$ is stationary as the $S-$modules chain.
Then can we get that $I_1\subset I_2\subset\cdots$ is stationary as the $R-$ideals chain?
I think since $I_1\subset I_2\subset\cdots$ is stationary as the $S-$modules chain, then we have $I_i=I_{i+1}=\cdots$ as $S-$modules for sufficiently large $i$.
Thus we can also get $I_i=I_{i+1}=\cdots$ as left $R-$ideals.
Is it true?
Any help and references are greatly appreciated.
Thinks!
As @Asigan pointed out in the comment, I assume you mean that the $M_i$'s are submodules of an $R$-module $M$. I will denote by $_{S}M$ the structure of $S$-module induced by restriction of scalars.
To address directly your question, you are right to say that if $M_1\subset M_2\subset\cdots$ is stationary as an $S$-modules chain, then it is stationary also as an $R$-modules chain (simply because the underlying sets are the same and being stationary depends on the sets, not on the structure).
More in general one can say the following.
If $_{S}M$ is noetherian, then also $M$ (as an $R$-module) is noetherian since any $R$-submodule of $M$ is also an $S$-submodule so if you have a chain of $R$-submodules, you can see it as a chain of $S$-submodules, which stabilizes as $_S M$ is noetherian.
The converse is true if $f$ is surjective but not in general. In fact in this casy any $S$-submodule of $M$ is also an $R$-submodule and we can apply the same reasoning.
For a counterexample (where $f$ is also injective) consider $A$ a non-noetherian domain and $K(A)$ the quotient field of $A$. Let $f:A\to K(A)$ be the canonical injection. Then $K(A)$ is obviously noetherian as $K(A)$-module, but not as an $A$-module since it contains $A$ which was not (and if $I\subset A$ is an $A$-submodule of $A$, then it is also an $A$-submodule of $K(A)$).
EDIT (after the OP's comment)
If $A$ is a ring and $S\subset A$ a subring which is noetherian as a ring, it is not true in general that $A$ is a noetherian ring.
For example $S=K$ is a field and $A=K[x_n| n\in \mathbb{N}]$. Then $S$ is a noetherian ring (since it is a field), but $A$ is not a noetherian ring since the ideal $(x_n)_{n\in\mathbb{N}}$ is not finitely generated.