Double Quotienting of a Ring is Isomorphic to Ring Quotient by Sum of Ideals

Question

First let me say, please edit my title if their is a more appropriate one and if this is a duplicate please direct me and close the question. I tried searching for my question, but I don't know if the theorem I am trying to prove has a name - so it was difficult to know what to search.

Also let me say that I have already tried to squeeze this theorem out of one of the isomorphism theorems, but I cant quite see how to get it, so if your hint or answer is `check the isomorphism theorems for rings' I might need more then that. I would like to understand this for personal benefit, but I am kind of limited on time.

I am specifically working with $A$ as a commutative ring with unity, $\mathfrak{a}$ is an ideal of $A$, and I believe $\mathfrak{b}$ is to be taken as an ideal in $A/\mathfrak{a}$, and $\mathfrak{b}'$ is the ideal in $A$ that corresponds to $\mathfrak{b}$. Then I would like to show that

$$A/\mathfrak{a}/\mathfrak{b} \approx A/(\mathfrak{a} + \mathfrak{b}').$$

I tried work through the mechanics of a specific example in full formality for insight, specifically I investigated $$\mathbb{Z}/<12>/<3+<12>>$$ where <12> is my ideal generated by 12 in $\mathbb{Z}$, and <3+ <12>> is my ideal generated by the coset with representative 3 in $\mathbb{Z}/<12>$. Sorry for the horrible notation here. I am aware this is probably unusually formal approach to the "coset" approach to quotienting, But I want to make sure I understand the nuts and bolts before passing off to theorems and the more "homomorphic image" approach to quotienting. Anyways

In this example we get $\mathbb{Z}/<12>/<3+<12>>$ = $$\{ <0+<12>>, <1 + <12>>, <3+<12>> \}$$

Which should be easily isomorphically mapped to $\mathbb{Z}/<12>+<3>$ since that sum of ideals is just <12>+<3> = <3>.

If you have any advice or direction please let me know.

Answer 1

Consider the map $$ \phi\colon R/\def\a{\mathfrak a}\a \to R/(\a + \def\b{\mathfrak b}\b'),\quad r + \a \mapsto r + (\a+\b') $$ which is a ring homomorphism. Note that $$ \phi(r+\a) = 0 \iff r \in \a + \b' \iff r + \a \in \b $$ Hence, $\ker \phi = \b$. As $\phi$ is onto, $$ (R/\a)/\b \cong R/(\a + \b')$$ by the homomorphism theorem for rings.

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Addendum, regarding the comment: Suppose $r \in \a + \b'$, that is, there are $a \in \a$, $b \in \b'$ with $r = a + b$, then $$ r + \a = a+b+\a = b + \a \in \b $$ On the other hand, suppose $r + \a \in \b$, that is $r + \a = b + \a$ for some $b\in \b'$, or $r - b \in \a$, that is $r \in \a + \b'$.

Answer 2

I have already tried to squeeze this theorem out of one of the isomorphism theorems, but I cant quite see how to get it...

Well, it's almost exactly the third isomorphism theorem, so it's a little strange you didn't succeed!

Perhaps the difficulty lay in understanding the relationship of $\mathfrak b$ to $\mathfrak b'$. The corresponding ideal is just an ideal $\mathfrak b'$ of $A$ which contains $\mathfrak a$, such that $\mathfrak b=\frac{\mathfrak b'}{\mathfrak a}$. With that said... is there any reason to write $\mathfrak b$ anymore? Perhaps not.

The third isomorphism theorem says that $\frac{A}{\mathfrak a}/\frac{\mathfrak b'}{\mathfrak a}\cong\frac{A}{\mathfrak b'}$. You shouldn't have to write $\mathfrak a +\mathfrak b'$ because $\mathfrak a+\mathfrak b'=\mathfrak b'$.