It might be silly question though, here are the things bothering me several days.
(The $A$ is a group or ring, plus $B$ is a normal or ideal.)
First) Is this statement true?
Let $A / B$ and $B_i \lhd A _i$ with $A \simeq (A_1 \times A_2)$ and $B \simeq (B_1 \times B_2)$
then, $A / B \simeq (A_1 \times A_2) / (B_1 \times B_2)$
my trial) Since the ring case is similar with the group. I tried as a group case. I took the mapping $\phi : A \to A_1 \times A_2 / B_1 \times B_2$ like a $\phi(a) = (a_1 \bullet B_1, a_2 \bullet B_2)$.
This function is well defined and on to function. So all I have to do is just $ker \phi$ = $B$ But failed it. :(
Second) Is my solution and answer right?(I used first statement)
If not, What is the correct method?
Q) Find the isomorphic ring with $\mathbb{Z_{15}}[x] / \langle 3x^2 + 5x \rangle $
$sol)$
Let $f(x) = 3x^2 + 5x$ and $[f]_n$ be $f(x)(mod n)$
Then $\mathbb{Z_{15}}[x] / \langle 3x^2 + 5x \rangle $ = $\mathbb{Z_{15}}[x] / \langle [f]_{15} \rangle $ for $\langle [f]_{15}(= 3x^2 + 5x) \rangle \lhd\mathbb{Z_{15}}[x]$
Here , $\mathbb{Z_{15}}[x] \simeq \mathbb{Z_{3}}[x] \times \mathbb{Z_{5}}[x]$
Plus By C.R.T, $\langle [f]_{15} \rangle \simeq \langle ([f]_3, [f]_5) \rangle \lhd \mathbb{Z_{3}}[x] \times \mathbb{Z_{5}}[x] $
By the way, Since $\mathbb{Z_{15}}[x] (\simeq \mathbb{Z_{3}}[x] \times \mathbb{Z_{5}}[x])$ is a ring with unity, $\langle ([f]_3, [f]_5) \rangle = \langle [f]_3 \rangle \times \langle [f]_5) \rangle$
So, ($\mathbb{Z_{3}}[x] \times \mathbb{Z_{5}}[x]$) / $\langle ([f]_3, [f]_5) \rangle $ $\simeq$ $(\mathbb{Z_{3}}[x] / \langle [f]_3 \rangle) \times $$(\mathbb{Z_{5}}[x] / \langle [f]_5 \rangle $)
By the first statement and $\langle [f]_{15} \rangle \simeq \langle ([f]_3, [f]_5) \rangle = \langle [f]_3 \rangle \times \langle [f]_5) \rangle$
Hence $\mathbb{Z_{15}}[x] / \langle [f]_{15} \rangle $ $\simeq$ $(\mathbb{Z_{3}}[x] / \langle [f]_3 \rangle) \times $$(\mathbb{Z_{5}}[x] / \langle [f]_5 \rangle) $
Therefore, $(\mathbb{Z_{3}}[x] / \langle 5x \rangle) \times $$(\mathbb{Z_{5}}[x] / \langle 3x^2 \rangle $ ) $\simeq$ $\mathbb{Z_{3}} \times \mathbb{Z_{5}} \times \mathbb{Z_{5}}$
Thanks.
There is a fairly big problem here, which is that you seem to be confusing $$\frac{A_1\times A_2}{B_1\times B_2}$$ with $$\frac{A_1}{B_1}\times\frac{A_2}{B_2}.$$ You write the first, but your proposed function $\phi$ actually maps to the second. They are not the same thing, though under some circumstances they are isomorphic.
The second problem is that you only have isomorphisms, not identities. That matters.
Allow me to explain.
First, if $B_1\triangleleft A_1$ and $B_2\triangleleft A_2$, then it is indeed the case that $B_1\times B_2\triangleleft A_1\times A_2$, and that $$\frac{A_1\times A_2}{B_1\times B_2}\cong \frac{A_1}{B_1}\times\frac{A_2}{B_2}.$$ To prove this, define a map $f\colon A_1\times A_2$ to $\frac{A_1}{B_1}\times \frac{A_2}{B_2}$ by $f(a_1,a_2) = (a_1B_1,a_2B_2)$. This is a surjective homomorphism (easy to check, or you can invoke the universal property of the product from maps in $\frac{A_1}{B_1}$ and into $\frac{A_2}{B_2}$). The kernel is precisely $B_1\times B_2$, which gives the isomorphism. This is your attempt with $\phi$, but it is defined into the wrong group. To verify that this is the kernel, note that $$\begin{align*} (a_1,a_2)\in\mathrm{ker}(f) &\iff (a_1B_1,a_2B_2) = (e,e)\\ &\iff a_1B_1=eB_1\text{ and }a_2B_2=eB_2\\ &\iff a_1\in B_1\text{ and }a_2\in B_2\\ &\iff (a_1,a_2)\in B_1\times B_2. \end{align*}$$
However, what you have is just an isomorphism, and that is not good enough. That is: you are asking whether $N\triangleleft A_1\times A_2$, $N\cong B_1\times B_2$ implies that $\frac{A_1\times A_2}{N}\cong \frac{A_1\times A_2}{B_1\times B_2}$. The answer is "no".
For example, take $A_1=C_6$ the cyclic group of order $6$ generated by $x$, $A_2=C_4$, the cyclic group of order $4$ generated by $y$, $B_1=\langle x^2\rangle$ (which is cyclic of order $3$), $B_2=\langle y^2\rangle$ (which is cyclic of order $2$). Then $\frac{A_1\times A_2}{B_1\times B_2}\cong C_2\times C_2$, the Klein $4$-group.
On the other hand, $B_1\times B_2 \cong C_3\times C_2\cong C_6$. So if we take $N=A_1\times\{e\}$, then $N\cong B_1\times B_2$, but $\frac{A_1\times A_2}{N} = \frac{A_1}{A_1}\times \frac{A_2}{\{e\}}\cong A_2\cong C_4$, the cyclic group of order $4$. So the two quotients are not isomorphic, even though we are moding out by isomorphic groups.
Now, if you had equality between $A$ and $A_1\times A_2$ and between $B$ and $B_1\times B_2$, then you would be fine. Or if the isomorphism between $B$ and $B_1\times B_2$ was the restriction of the isomorphism between $A$ and $A_1\times A_2$, so that you have truly the same groups being quotiented out, then you would be okay. But as given, it just doesn't work.
Note that my examples are also rings and ideals: you have $\mathbb{Z}_6\times\mathbb{Z}_4$, and the ideals $3\mathbb{Z}_6$ and $2\mathbb{Z}_2$ for $B_1\times B_2$, and the ideal $\mathbb{Z}_6\times\{0\}$ for $N$. We have that $\mathbb{Z}_2\times\mathbb{Z}_3\cong\mathbb{Z}_6$ as rings (Chinese Remainder Theorem), and $\mathbb{Z}_6\cong\mathbb{Z}_6\times\{0\}$, so that you have the required isomorphisms. But the group structures of $(\mathbb{Z}_6\times\mathbb{Z}_4)/(3\mathbb{Z}_6\times 2\mathbb{Z}_4)$ and of $(\mathbb{Z}_6\times\mathbb{Z}_4)/(\mathbb{Z}_6\times\{0\})$ are different, so the two rings cannot be isomorphic.