MathNugget
Junior Member
- Joined
- Feb 1, 2024
- Messages
- 195
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 [imath]A \simeq^\phi C[/imath], [imath]B \simeq^\psi D[/imath].
Is this true: [imath]\frac{A}{B}\simeq \frac{C}{D}[/imath] ?
For now, I want to work under the assumption that [imath]\psi[/imath] is not necessarily [imath]\phi_{\mid B}[/imath] (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 [imath]\frac{A}{B}[/imath] as [imath]\widehat{a}[/imath], and the elements of [imath]\frac{C}{D}[/imath] as [imath]\overline{a}[/imath].
I want to define the follow function: [imath]\theta: \frac{A}{B} \rightarrow\frac{C}{D}[/imath], [imath]\theta(\widehat{a})=\overline{\phi(a)}[/imath].
I'd check the definition:
Take [imath]a, b \in A[/imath], such that [imath]\widehat{a}=\widehat{b}[/imath]. Then [imath]\widehat{a-b}=\widehat{0}[/imath].
Then [math]\theta(\widehat{a-b})= \theta(\widehat{0})\\ \overline{\phi(a-b)}= \overline{\phi(0)}\\ \overline{\phi(a)-\phi(b)}=\overline{0}\\[/math]So [imath]\overline{\phi(a)}=\overline{\phi(b)}[/imath].
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 [imath]Ker(\theta)=\widehat{0}[/imath].
If [math]\theta(a)=\overline{0}\\ \overline{\phi(a)}=\overline{0}\\ \phi(a)\in D[/math].
It would be easy if [imath]\psi=\phi[/imath] on B... but without that, all I really seem to have is [imath]\exists! b \in B[/imath] so [imath]\psi(b)=\phi(a)[/imath], and I don't know how I'd get that [imath]a \in B[/imath] 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 [imath]\phi: C \rightarrow D[/imath] is also an isomorphism, just because there is an isomorphism from C to D? or maybe the smaller isomorphism [imath]\psi[/imath] can be extended to a [imath]\phi_2[/imath] to replace [imath]\phi[/imath]?
If A, C are rings, B is an ideal in A, D is an ideal in C and [imath]A \simeq^\phi C[/imath], [imath]B \simeq^\psi D[/imath].
Is this true: [imath]\frac{A}{B}\simeq \frac{C}{D}[/imath] ?
For now, I want to work under the assumption that [imath]\psi[/imath] is not necessarily [imath]\phi_{\mid B}[/imath] (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 [imath]\frac{A}{B}[/imath] as [imath]\widehat{a}[/imath], and the elements of [imath]\frac{C}{D}[/imath] as [imath]\overline{a}[/imath].
I want to define the follow function: [imath]\theta: \frac{A}{B} \rightarrow\frac{C}{D}[/imath], [imath]\theta(\widehat{a})=\overline{\phi(a)}[/imath].
I'd check the definition:
Take [imath]a, b \in A[/imath], such that [imath]\widehat{a}=\widehat{b}[/imath]. Then [imath]\widehat{a-b}=\widehat{0}[/imath].
Then [math]\theta(\widehat{a-b})= \theta(\widehat{0})\\ \overline{\phi(a-b)}= \overline{\phi(0)}\\ \overline{\phi(a)-\phi(b)}=\overline{0}\\[/math]So [imath]\overline{\phi(a)}=\overline{\phi(b)}[/imath].
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 [imath]Ker(\theta)=\widehat{0}[/imath].
If [math]\theta(a)=\overline{0}\\ \overline{\phi(a)}=\overline{0}\\ \phi(a)\in D[/math].
It would be easy if [imath]\psi=\phi[/imath] on B... but without that, all I really seem to have is [imath]\exists! b \in B[/imath] so [imath]\psi(b)=\phi(a)[/imath], and I don't know how I'd get that [imath]a \in B[/imath] 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 [imath]\phi: C \rightarrow D[/imath] is also an isomorphism, just because there is an isomorphism from C to D? or maybe the smaller isomorphism [imath]\psi[/imath] can be extended to a [imath]\phi_2[/imath] to replace [imath]\phi[/imath]?