In Aluffi, he states the following ring isomorphism theorem: $$\frac{R/(a)}{(\bar{b})} \cong R/(a,b)$$
His proof says that this follows from the first isomorphism theorem: $$\frac{R/I}{J/I} \cong R/J$$
and the fact that: $$(\bar{b}) \cong (a,b)/(a)$$
The last statement bothers me. By my reckoning $$(\bar{b}) \cong (b)/(a) \cong (a,b)/(a)$$ because $$(\bar{b}) = Rb+(a) = Rb + Ra $$
What really is important here is that $(a,b)$ is the smallest ring that contains $(a)$ and $(b)$ which is required for the first isomorphism theorem.
And really we have, $$\frac{R/(a)}{(a) + (b)} \cong \frac{R}{(a) + (b)}$$ and $(a) + (b) = (a,b) \cong (\bar{b})$
My question is really whether or not my reasoning is correct here, and if Aluffi's statement is a bit misleading.
Thanks
Unless we are given that $b \,|\, a$ so that $(a) \subseteq (b),$ we cannot say that $\overline{(b)} = (b) / (a).$ Particularly, the Second Isomorphism Theorem does not apply. We must really use that $\overline{(b)} = (b) + (a).$
Given ideals $I$ and $J$ of a ring $R,$ we claim that $$\frac{R/I}{J + I} \cong \frac{R}{I + J}.$$ Consider the map $\varphi : R \to (R / I) / (J + I)$ defined by $\varphi(r) = (r + I) + J + I = r + I + J.$ Given that $r = s,$ we have that $r - s = 0$ is an element of $I + J$ so that $r - s + I + J = 0 + I + J$ and $r + I + J = s + I + J.$ Consequently, it follows that $\varphi$ is well-defined. One can easily see that $\varphi$ is surjective with kernel $I + J,$ hence the claim holds by the First Isomorphism Theorem.
By identifying $I = (a)$ and $J = (b),$ the result you mention is established.