isomorphisms question (with rings and ideals)

[MathNugget]

This time it's a hobby question, as I really like algebra.

If A, C are rings, B is an ideal in A, D is an ideal in C and $A \simeq^\phi C$, $B \simeq^\psi D$.

Is this true: $\frac{A}{B}\simeq \frac{C}{D}$ ?
For now, I want to work under the assumption that $\psi$ is not necessarily $\phi_{\mid B}$ (phi restricted to B). I'd like to see if this is true (can't seem to find it online, I've only seen the isomorphism theorems for rings on wikipedia).

I'd like to note the elements of $\frac{A}{B}$ as $\widehat{a}$, and the elements of $\frac{C}{D}$ as $\overline{a}$.
I want to define the follow function: $\theta: \frac{A}{B} \rightarrow\frac{C}{D}$, $\theta(\widehat{a})=\overline{\phi(a)}$.
I'd check the definition:
Take $a, b \in A$, such that $\widehat{a}=\widehat{b}$. Then $\widehat{a-b}=\widehat{0}$.
Then $$\theta(\widehat{a-b})= \theta(\widehat{0})\\ \overline{\phi(a-b)}= \overline{\phi(0)}\\ \overline{\phi(a)-\phi(b)}=\overline{0}\\$$So $\overline{\phi(a)}=\overline{\phi(b)}$.
I guess correct definition is checked, and morphism seems pretty much a given.

I am not sure how I would go about injectivity and surjectivity.
Let's try to prove $Ker(\theta)=\widehat{0}$.
If $$\theta(a)=\overline{0}\\ \overline{\phi(a)}=\overline{0}\\ \phi(a)\in D$$.
It would be easy if $\psi=\phi$ on B... but without that, all I really seem to have is $\exists! b \in B$ so $\psi(b)=\phi(a)$, and I don't know how I'd get that $a \in B$ to have injectivity.

I figure , as a counterexample, a permutation could move elements from B outside of D, and elements from outside of B into D (if we consider the case A=C and B=D). But then, these are clearly isomorphic (since it's the same thing). Are the premises insufficient, or can I somehow prove that maybe $\phi: C \rightarrow D$ is also an isomorphism, just because there is an isomorphism from C to D? or maybe the smaller isomorphism $\psi$ can be extended to a $\phi_2$ to replace $\phi$?

 

[blamocur]

Unless I am missing something the statement you are trying to prove is not correct. Consider $A = \mathbb Z,\;\;\;B=\mathbb Z,\;\;\;B=2\mathbb Z,\;\;\;C=3\mathbb Z$

 

[MathNugget]

Unless I am missing something the statement you are trying to prove is not correct. Consider $A = \mathbb Z,\;\;\;B=\mathbb Z,\;\;\;B=2\mathbb Z,\;\;\;C=3\mathbb Z$

Well, it does look like you have 2 B's and no D .
Besides that...would we have $\phi: 2\mathbb{Z} \rightarrow 3\mathbb{Z}\\ \phi(x)=\frac{3x}{2}$ as 1 of the isomorphisms?
I think it's not working like that: $\phi(4ab)=\phi(2a)\phi(2b)\\ 6ab=3a3b$

Could you be a bit more specific about the morphisms?

 

[MathNugget]

By the way, in the last part of my post

Are the premises insufficient, or can I somehow prove that maybe $\phi: C \rightarrow D$ is also an isomorphism, just because there is an isomorphism from C to D? or maybe the smaller isomorphism $\psi$ can be extended to a $\phi_2$ to replace $\phi$?

Of course it was actually supposed to say B instead of C, like down here

Are the premises insufficient, or can I somehow prove that maybe $\phi: B \rightarrow D$ is also an isomorphism, just because there is an isomorphism from B to D? or maybe the smaller isomorphism $\psi$ can be extended to a $\phi_2$ to replace $\phi$?

 

[blamocur]

Well, it does look like you have 2 B's and no D .
Besides that...would we have $\phi: 2\mathbb{Z} \rightarrow 3\mathbb{Z}\\ \phi(x)=\frac{3x}{2}$ as 1 of the isomorphisms?
I think it's not working like that: $\phi(4ab)=\phi(2a)\phi(2b)\\ 6ab=3a3b$

Could you be a bit more specific about the morphisms?

You got me here -- sorry for disinformation You are right: 2Z and 3Z are isomorphic as additive groups, but not as rings.

 

[blamocur]

$\overline{\phi(a-b)} = \overline{\phi(0)}$

I don't see this as proven that $\overline{\phi(a-b)} = \overline{\phi(0)}$. Essentially you need to prove that if $x\in B$ then $\phi(x) \in D$, but I don't see this proven in your post.

 

[blamocur]

It looks to me that in your post #1 you are trying to prove that $\phi(B) = D$ -- am I right?

 

[MathNugget]

It looks to me that in your post #1 you are trying to prove that $\phi(B) = D$ -- am I right?

That would be 1 way to solve my question. Or finding another isomorphism using $\phi$ and $\psi$ ...
I am starting to think there's not enough info to say $\phi(B)=\psi(B)$, or $\phi(B)\subseteq D$

 

[blamocur]

That would be 1 way to solve my question.

I believe this is not true in the general case because it would imply that any two isomorphic ideals must be identical.

 

[blamocur]

I believe this is not true in the general case because it would imply that any two isomorphic ideals must be identical.

As a kind of closure/summary after one week: below is a counterexample, i.e. a ring [imath]A\ with two isomorphic ideals \(B[/imath] and $D$ which are not identical. This means that identical automorphism $\phi : A \rightarrow A$ does not map $B$ to $D$.

Ring $A = \mathbb Z^2 = \mathbb Z\times \mathbb Z$ in which $(a,b)+(c,d) = (a+c,b+d)$ and $(a,b)\cdot(c,d) = (a\cdot c, b\cdot d)$.
Ideal $B = 2\mathbb Z \times \mathbb Z$, i.e. the first coordinate is even, and $C=\mathbb Z \times 2\mathbb Z$, i.e. the second coordinate is even.
Note that this is not a counterexample to your original statement because both $A/B$ and $A/D$ are isomorphic to $\mathbb Z_2 = \mathbb Z/2\mathbb Z$, but it shows that $\phi$ does not necessarily map $B$ to $D$.

 

[MathNugget]

As a kind of closure/summary after one week: below is a counterexample, i.e. a ring [imath]A\ with two isomorphic ideals \(B[/imath] and $D$ which are not identical. This means that identical automorphism $\phi : A \rightarrow A$ does not map $B$ to $D$.

Ring $A = \mathbb Z^2 = \mathbb Z\times \mathbb Z$ in which $(a,b)+(c,d) = (a+c,b+d)$ and $(a,b)\cdot(c,d) = (a\cdot c, b\cdot d)$.
Ideal $B = 2\mathbb Z \times \mathbb Z$, i.e. the first coordinate is even, and $C=\mathbb Z \times 2\mathbb Z$, i.e. the second coordinate is even.
Note that this is not a counterexample to your original statement because both $A/B$ and $A/D$ are isomorphic to $\mathbb Z_2 = \mathbb Z/2\mathbb Z$, but it shows that $\phi$ does not necessarily map $B$ to $D$.

Thank you. I see now, you're right. I suppose a counterexample is probably very complicated, or the 2 functions have to be creatively assembled to make the/an isomorphism...

 

[MathNugget]

I guess I'll try a different approach (some background information to this problem).
I am trying to do isomorphisms of the type $\frac{\frac{A}{B}}{\frac{C}{D}}$ (A is a ring, C is a subring of A, B is ideal in A, D is ideal in C, the whole factorization makes sense) BUT at least for me, it's quite troublesome to write an isomorphism from a class of elements factored by another class of elements...

so I am looking forward for this kind of result:
if $\frac{A}{B}\simeq X$ and $\frac{C}{D}\simeq Y$, and $\frac{X}{Y}$ makes sense (you can factor by Y, meaning Y is an ideal of X), then I am trying to prove $\frac{\frac{A}{B}}{\frac{C}{D}} \simeq \frac{X}{Y}$
I am familiar with these formulas, but the closest there is the 'simplification' that happens with stronger conditions: $\frac{\frac{A}{B}}{\frac{C}{D}} \simeq \frac{A}{D}$ (but then I'd need to have B = D, among other things (which I don't).

 

[blamocur]

A is a ring, C is a subring of A

Just a reminder that you cannot factor over a subring, only over an ideal. An exercise: why? And I am not sure that C/D is an ideal in A/B.
Otherwise I feel too rusty to be much help with this problem

 

[MathNugget]

Ok, a concrete example:
Here p is a prime (in $\mathbb{Z}$), and the polynomial is in $\mathbb{Z}[X]$

I try to prove
$\frac{\mathbb{Z}_p[X]}{(X^2+tX+q)\mathbb{Z}_p[X]}\simeq \frac{(\mathbb{Z}[X]/p\mathbb{Z}[X])}{((X^2+tX+q, p)\mathbb{Z}[X]/p\mathbb{Z}[X])}$

The isomorphism formula $\frac{\frac{R}{I}}{\frac{J}{I}}\simeq \frac{R}{J}$, when $I\subseteq J \subseteq R$ doesn't really help here.

So I was hoping there's another isomorphism rule out there... maybe like factoring the 'numerator' and the 'denominator' and it would give the same thing. (it sure looks like this time it applies like that).
I am trying to avoid writing a morphism from classes of elements to different classes of elements...

 

[MathNugget]

I don't know how legal the source is, this is the original paper (the link sends you to a researchgate.net page, and you have to scroll down for the article). There's more isomorphisms there, and none have any arguments as to why they work .The rest of the paper is weirdly simplistic, but the necessary isomorphisms are quite troublesome...

 

[MathNugget]

I suppose, after careful consideration, I didn't understand too well what I needed to prove.
Here's the updated question:
Say, we have A a ring, B an ideal (they represent the factorization ring and ideal respectively, from above, but I think we can simplify it like this for the sake of me making less mistakes).
if I find a isomorphism $\phi: A \rightarrow X$ (X is a ring).
Is it true that $\frac{A}{B}\simeq \frac{X}{\phi(B)}$? I actually think this last claim of mine is true; I can prove $\phi(B)$ is an ideal of X, at least.

For x, y in $\phi(B)$, $\phi^{-1}(a), \phi^{-1}(b)$ are elements of B, which is an ideal, I add them there to get an element of B, and return to $\phi(B)$. Similar for multiplication.

Normally, I'd then have to prove correct definition of function, bijection, and morphism properties, right? But I think morphism at least is by default, given $\phi(B)$ is isomorphism, once I prove correct definition and bijection.
I'll add after this comment another where I try to prove these 2 things...

 

[blamocur]

I agree that $\frac{A}{B} \simeq \frac{X}{\psi(B)}$, and don't expect the proof to be difficult.

 

[MathNugget]

I agree that $\frac{A}{B} \simeq \frac{X}{\psi(B)}$, and don't expect the proof to be difficult.

Just for practice, I'll try to write the rest of the proof.

$\overline{x} \in \frac{X}{\phi(B)}$, and $\widehat{x} \in \frac{A}{B}$ are the equivalence classes.

say, $\hat{a}=\hat{b}$ in $\frac{A}{B}$. Then $a-b \in B$, so $\phi(a)-\phi(b)=\phi(a-b)\in \phi(B)$, therefore $\overline{\phi(a)}=\overline{\phi(b)}$.
It is surjective (for an element $\overline{x} \in \frac{X}{\phi(B)}$ there is an element $\widehat{\phi^{-1}(x)} \in \frac{A}{B}$ with $\phi(...)=\overline{x}$), and a morphism (from it being an isomorphism and correct definition, I guess?), and

$\overline{x}=\overline{0} \Leftrightarrow x \in \phi(B) \Leftrightarrow \phi^{-1}(x)\in B \Leftrightarrow \widehat{\phi^{-1}(x)}=\widehat{0}$ shows that $Ker(\phi)=B$.


Thanks for help, by the way!!

 

[blamocur]

Just for practice, I'll try to write the rest of the proof.

$\overline{x} \in \frac{X}{\phi(B)}$, and $\widehat{x} \in \frac{A}{B}$ are the equivalence classes.

say, $\hat{a}=\hat{b}$ in $\frac{A}{B}$. Then $a-b \in B$, so $\phi(a)-\phi(b)=\phi(a-b)\in \phi(B)$, therefore $\overline{\phi(a)}=\overline{\phi(b)}$.
It is surjective (for an element $\overline{x} \in \frac{X}{\phi(B)}$ there is an element $\widehat{\phi^{-1}(x)} \in \frac{A}{B}$ with $\phi(...)=\overline{x}$), and a morphism (from it being an isomorphism and correct definition, I guess?), and

$\overline{x}=\overline{0} \Leftrightarrow x \in \phi(B) \Leftrightarrow \phi^{-1}(x)\in B \Leftrightarrow \widehat{\phi^{-1}(x)}=\widehat{0}$ shows that $Ker(\phi)=B$.


Thanks for help, by the way!!

Looks good to me.
You are welcome.