I want to show the following:
Let $I_{1},...,I_{n}$ be ideals of a commutative ring (with 1) $A$ such that $\bigcap_{i=1}^{n} I_{i}$ is the zero ideal. If each quotient $A/I_{i}$ is a Notherian ring show that $A$ is a Notherian ring.
My work:
Let $f: A \rightarrow A/I_{1} \times A/I_{2} \times .... \times A/I_{n}$ be the map defined by $f(a)=(a+I_{1},a+I_{2},...,a+I_{n})$. Then $A$ embeds in $A/I_{1} \times A/I_{2} \times .... \times A/I_{n}$. Since the direct sum of Noetherian modules is Noetherian and every submodule of a Noetherian module is Noetherian the result follows.
Is the above OK?