Suppose $R$ is a commutative ring with unity and $I_i$, $i=1,...,n$ are ideals such that $\cap_i I_i=\{0\}$. The quotient rings $R/I_1,...,R/I_n$ are Noetherian. Show that $R$ is Noetherian.
My teacher claims that from the given it is obvious that $R/I_i$ are Noether R-modules. I haven't seen why that is obvious. Is it just because the $R/I_i$'s are Noether-$R/I_i$ modules by definition of Noetherian rings and since for some ring homomorphism, the $R/I_i$'s are ismorphic to the image of that mapping which is a subset in R? If I am able to show this I have a proof of what stands above.
Consider the homomorphism ring $$\phi : R \to \bigoplus_{j=1}^{n}R/I_j$$ $$r \mapsto (r+I_1, \ldots , r + I_n )$$
We see that $\ker(\phi) = \cap_j I_j = 0 $ , and this implies that $R$ is isomorphic to a submodule of $\bigoplus_{j=1}^{n}R/I_j$.
A finite direct sum of noetherian modules is noetherian, thus $\bigoplus_{j=1}^{n}R/I_j$ is noetherian, and then also $R$ is noetherian as a submodule of a noetherian module.