I want to understand the proof my textbook provides of the second isomorphism theorem for rings.
Ler $R$ a ring, $I$ and $J$ two ideals of R such that $I \subseteq J$. Then
$(R/I)/(J/I) \simeq R/J$
This theorem seems basically a corollary of the fundamental homomorphism theorem for rings. So it reduces to proving the existence of an epimorphism whose kernel is exaclty $J$; my book choose this $\pi = \pi_2 \circ \pi_1$ from $R$ on $(R/I)/(J/I)$
$R \xrightarrow{\pi_1} R/I \xrightarrow{\pi_2} (R/I)/(J/I)$
This kernel should be
$\ker \pi = \{r \in R | \pi_1(r) \in \ker \pi_2\}$
and till this point I understand but then it continues by writing
$\ker \pi =\{ r \in R | r + I = j +I, j\in J\} = J$
I don't understand the part $\ldots r + I = j +I, j\in J\}$ and how this set is equal to $J$. I know that $r +I \in R/I$ but I don't understand how $j$ is involved.
Best Answer
The author is just writing out the definition of $\ker \pi_2$. After all, $\ker \pi_2$ is defined as those elements $r + I \in R/I$ such that $\pi_2(r + I) = 0$, which happens exactly if $r + I = j + I$ (in $R/I$) for some $j \in J$. Thus, the author's equation $\ker \pi = \{ r \in R \space|\space r + I = j + I, j \in J \}$ is true. Now see if you can show that $r + I = j + I \implies r \in J$.