Three quotient-ring isomorphism questions I need some help with the following isomorphisms.

Let $R$ be a commutative ring with ideals $I,J$ such that $I \cap J = \{ 0\}$. Then 


*

*$I+J \cong I \times J$ 

*$(I+J)/J \cong I$

*$(R/I)/\bar{J} \cong R/(I+J) \quad \text{where} \quad  \bar{J}=\{x+I \in R/I: x \in J \}$



For the first item $\theta: I \times J \ \rightarrow I+J: \ (x,y) \mapsto x+y$ is clearly a surjective homomorphism. It's injective because $I \cap J = \{ 0\}$.
For the second item, the mapping $\eta \ : \ I+J \rightarrow I \ : \ x+y \mapsto x$ is well-defined by item 1, surjective, and the kernel equals $0+J$. Now we can use the first isomorphism theorem.
For the third item, I tried to find a define a mapping:
$$\phi: \quad R/I \ \rightarrow \ R/(I+J) \quad : \quad x + I \ \mapsto \ x+I+J $$
And I tried to show that $\bar{J}$ is the kernel, but it didn't totally feel okay because I got confused. Is the following correct?
$$ x \in \ker(\phi) \ \iff \ x+ I \in I+J \iff x+I \in \bar{J} $$
I would appreciate it if you could tell me if I made mistakes. Could you provide me a little information about the third item? It seems like a blur to me.
 A: The line

$ x \in \ker(\phi) \ \iff \ x+ I \in I+J \iff x+I \in \bar{J} $

is wrong, because $x+I$ could be principally no element of $I+J$ since $I+J$ is an ideal which contains elements of $R$, and $x+I$ is a left coset of an ideal and hence also a set of elements of $R$. You could write instead

$ x + I \in \ker(\phi) \iff  x \in I+J \iff x+I \in (I+J)/I = \overline{J}.$

Are you familiar which the third isomorphism theorem of rings?

If $I \subseteq J$ are ideals of $R$, then $J/I$ is an ideal of $R/I$ and $(R/I)/(J/I) \cong R/J$.

Note, that in your case $\overline{J} = (I+J)/I$.
A: The justification for why $ker(\phi) = \bar{J}$ is faulty, as Dune correctly points out.
So let's drill down on what the definition of each of these things really is.
Let $x + I$ be an element in $ker(\phi)$ this means that $\phi(x + I) = x + I + J = I + J$ in $R/(I + J)$.
From $x + I + J = I + J$ we get that $x\in I + J$. This means we can write $x$ as $x' + i$ for some $i\in I$ and $x' \in J$.
Note that the coset $x + I = x' + i + I = x' + I$. This satisfies the definition of $\bar{J}$. Therefore $x + I \in \bar{J}$.
