Ring, ideal, homomorphism

Question

The following is from the book "Algebra" by A. Białynicki-Birula (the translation is mine):

Chapter VII, Theorem 9.1. The difference of two elements belonging to $P_1$ belongs to the kernel of the homomorphism $\phi$ if and only if the mapping $\phi$ takes the same value on these elements.

Chapter VIII. (...). Let a ring $P_1$ and its ideal $I$ be given. We will show the existence of a ring $P_2$ and a homomorphism $\phi$ of the ring $P_1$ onto $P_2$ whose kernel is $I$. For this aim let us first note that the theorem 9.1 in the chapter VII implies that if $\phi_{0}$ is a homomorphism of $P_1$ onto some ring $P_0$, then the mapping $\phi_0$ is constant on kernel cosets and the elements assigned by $\phi_0$ to elements belonging to different cosets are different.

How does the mentioned theorem implies what the author says? From what I understand it must be that we know that the difference of elements from $P_1$ belongs to the kernel of the homomorphism $\phi_0$ - but how?

Answer 1

"The mapping $ϕ_0$ is constant on kernel cosets" means that if $b\in a+Ker\ \phi_0$, then $\phi_0(a)=\phi_0(b)$. This is the "only if" statement of Theorem 9.1.

"The elements assigned by $ϕ_0$ to elements belonging to different cosets are different." means that if $b\notin a+Ker\ \phi_0$, then $\phi_0(a)\neq\phi_0(b)$. The contrapositive of this statement is the "if" statement of Theorem 9.1.

Answer 2

The ring homomorphism $\phi$ is the canonical projection, $\phi:P_1\to P_2\cong P_1/I$.

The author's claim is precisely correct: $\operatorname {ker}\phi=I$ and $r_1+I\ne r_2+I\implies \phi(r_1)\ne\phi(r_2)$. For, in general, $\phi(r)=r+I$.