I want to know if the following is true and if so, then how to prove it:
Let $x,y$ be distinct elements of a ring $A$ and let $\langle x \rangle $, $\langle x,y \rangle $, etc. be the ideals generated by $\{x\}$, $\{x,y\}$, etc.
Let $\pi_1,\pi_2, p$ be the canonic projections to the quotient rings as shown in the diagram below and let $\pi=\pi_2\circ\pi_1$
I want to show (if true) that $\phi$ is an isomorphism. By the iso. theorem it would suffice to show that $\langle x,y \rangle = \ker(\pi) $ and that $\pi$ is surjective (for then $\phi$'s existance is guaranteed and $A/\ker(\pi)\cong \text{Im}(\phi)$).
I tried this:
Take $z\in \langle x,y \rangle$ then $z=ax+by$ with $a,b\in A$ then $\pi_1(z)=\overline{by}$ then $\pi_2(z)=\overline{\bar{b}\bar{y}}=\overline{\bar{0}}$ thus $z\in \ker(\pi)$.
Take $z\in \ker(\pi)$ then $\overline{\bar{z}}=\overline{\bar{0}}$ then $\bar z = \bar b \bar y$ for some $\bar b\in A/\langle x\rangle$ then $z=ax+b'y$ for some $a\in A, b'\in \bar b$ thus $z\in \langle x,y \rangle$.
Therefore $\ker(\pi)=\langle x, y\rangle$.
Take $w\in A/\langle x\rangle \big{/} \langle \bar y \rangle$ then there exists $v\in A/\langle x\rangle $ such that $w =\overline{v}$ and there exists $u\in A$ such that $\bar u =v$ thus $\pi(u)=\overline{\bar{u}} =w$.
Therefore $\pi$ is surjective.
I am not entirely convinced by this proof, specially by the $\ker(\pi)\subseteq \langle x,y\rangle $ part.
Is what I've done right? Is there another way to prove it?
Thanks in advance for any help whatsoever.
P.S. I donk know why the picture of the diagram is blurry, I uploaded it as a HQ PNG.