Solution verification of this question related to noetheranity This question was asked in my quiz of commutative algebra and I attempted this question. I want to ask here whether my solution is correct or not. (What are the mistakes in the attempt).

Question:   Let $I,J$ be ideals in a ring $A$ such that $I \cap J=\{0\}$. Show that $A$ is noetherian iff $A/I$ and $A/J$ are noetherian.

Attempt: Let $A$ be a noetherian ring $\implies$ Any $A$-module $M$ and any ascending chain of submodules stabilizes or every submodule is finitely generated.
Let $a\in A/I$ and $a'\in A/I$  . $ a-a' \in A/I$. $a= a_x +I $ and $a'=a_y +I$ such that $a_x , a_y \in A$  . $a-a' = a_x - a_y +I  \in A+I$.
$a(a_x +I) = a a_x +I \in A+I$ as $I$ is an ideal. So any $A/I$  submodule of $A/I$ is also a submodule of $A$-module.
So, $A/I$ will be noetherian and similarly, $A/J$ will be noetherian.
Conversly, let $I\cap J= \{0\}$ and $A/I$ and $A/J$ are notherian.
$A /I$ is finitely generated implies that there exists $a_1,..., a_n  $  all in $A$ such that  any $a_x \in A/I$ can be written as  $ a_x= a'_1 a_1 +... + a'_n a_n$.
Similarly there exists $a^*_1 ,..., a^*_m $ and  $a^"_1 ,..., a^"_m \in A$ such that any $a_y \in A/J$ can be written as $a_y = a^"_1 a^*_1 +...+ a^"_m a^*_m$ . To show that $A$ is finitely generated.
Any element $a_x = a_i +I $ such that $a_i \in A$ and $a_y = a_j + J$ such that $a_j \in A$.
Now let $a_i $ and $a_j$ in $A$ be arbitrary. Then $a_i +I = a'_1 a_1 +...+ a_n a'_n$ and $a_j +J= a^"_1 a^*_1 +... + a^"_m a^*_m$ . Now $I \cap J=\{0\}$ implies that $a_i +a_j= a'_1 a_1 +...+ a_n a'_n + a^"_1 a^*_1 +...+ a^"_m a^*_m$ . But $a_i +a_j$ is arbitrary as $a_i$ and $a_j$ are both arbitrary.
So, $A$ is of finite type generated by $\{ a_1,...,a_n ,a^*_1,..., a^*_m\}$.
Kindly let me know how much of it is correct.
Thanks!
 A: There is a canonical exact sequence for any pair of ideals $I,J\le A$:
$$
0 \to I\cap J \to A\to A/I\times A/J \xrightarrow{\varphi} A/(I+J)\to0
$$
The only non-trivial map here is $\varphi$ which is given by
$$
\varphi\colon A/I\times A/J\to A/(I+J),\,(x+I,y+J)\mapsto x-y+(I+J)\,.
$$
See here for more detail.
This immediately gives the result. Indeed, if $I\cap J=0$, then the above reduces to a short exaxct sequence
$$
0 \to A \to A/I\times A/J \to A/(I+J) \to 0
$$
and there is a general theorem that for a short exact sequence of $A$-modules
$$
0 \to M' \to M \to M'' \to 0
$$
we have that $M$ is Noetherian iff $M'$ and $M''$ are Noetherian. Now recall that $A$ is Noetherian as $A$-module iff $A$ is Noetherian as ring given that submodule are precisely ideals.
The crucial observation here is that if $I\cap J=0$, then $A$ embeds into $A/I\times A/J$ along the usual projections. You do not really need the above exact sequence but it is a neat result anyway.
And your attempt precisely tries to do this, though it is written a bit confusingly. However, you should start with an arbitrary submodule and show that it is finitely generated, not with the whole ring. As walkar pointed out whole ring (aka the unit ideal) is always finitely generated, even in non-Noetherian rings.
What you then do is to fix a submodule, that is, an ideal $K\le A$ and consider the ideals it generates $K/I\le A/I$ and $K/J\le A/J$. These are now finitely generated by assumption and you can essentially redo your proof for them. To be more precise: you showed exactly this but only of the trivial case of $K=A$.
