This question follows from this basic theorem:
Let $I$ be an ideal of ring $R$. The map $\phi:R\to R/I$ defined as
$\phi(r)=r+I$ is a ring homomorphism of $R$ onto $R/I$ with kernel
$I$.
I've tried to prove the last part of the theorem, but stumbled in understanding the basic thing on the identity element of $R/I$.
I suspect that, since the definition of multiplication of two cosets in $R/I$ defined as
$$(r+I)(s+I)=rs+I$$
then the identity element should be $1+I$?
But, when trying to prove that last part of the theorem I got:
$\ker{\phi}=\{x\in R:\phi(x)=1+I\}$, can I claim from here that $\ker{\phi}=I$?
Thank you.
Best Answer
There are two identity elements at play: the additive identity $0_{R/I} = 0+ I = I$ and the multiplicative identity $1_{R/I} = 1+I$. The kernel of $\phi$ is the set of elements of $R$ that map to the additive identity. Therefore it is precisely $I$.