- #1

Kreizhn

- 743

- 1

## Homework Statement

Let R be a commutative ring, [itex] a, b \in R [/itex] disjoint elements of the ring. Let [itex] (a), (b) [/itex] denote the principal ideals of a and b respectively. If [itex] \bar b \in R/(a) [/itex] is the class of b in the quotient ring, show that

[tex] R/(a)/(\bar b) \cong R/(a,b) [/tex]

## The Attempt at a Solution

This is just an application of the third (second for some people) isomorphism theorem which states that if [itex] I \subseteq J \subseteq R [/itex] are ideals, then [itex] (R/I)/(J/I) \cong R/J [/itex]

What I need to show is that

[tex] (\bar b) = \frac{ (a,b)}{(a)} [/tex]

and this is where I'm having trouble.

Indeed, we know that [itex] \bar b = b + (a) [/itex] in R/(a). So then I figure [itex] (\bar b) = (b+(a)) R [/itex]. If I write this out set-wise

[tex] (b+(a))(R) = \{ (b+ar_1)r_2 : r_1, r_2 \in R \} = \{b r_2 + a r_3 : r_2, r_3 \in R\} = bR + a R = (a,b) [/tex]

But we know, via correspondence theorem, that ideals of [itex] R/(a) [/itex] must have the form [itex] J/(a) [/itex] for some ideal J of R. It seems like I'm close, but something in my reasoning here is wrong.