Quotients of Principal Ideals: How to Apply the Third Isomorphism Theorem

  • Thread starter Kreizhn
  • Start date
In summary: Z}} to \pi \big|_{(a,b)}: ar_1 + br_2 \mapsto ar_1 +...+ ar_n + br_n \big|_{n \in R} which is identical to the original projection.Case 2Instead, I'll try to make the homomorphism from scratch. Let a typical element of (a,b) be ar_1 + br_2 for r_1,r_2 \in R . I want to map this to an element \bar b r_3 for
  • #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.
 
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  • #2
Hi Kreizhn! :smile:

Can't you apply the first isomorphism theorem. I.e. find a surjection

[tex]f:(a,b)\rightarrow (\overline{b})[/tex]

with kernel (a).
 
  • #3
Thanks for the reply micromass, you really seem to jump on these questions :smile:

Are we allowed to do this though? I mean, (a,b) need not be a ring. I guess we could view (a,b) as a group, use the 1st IT then show it's a ring isomorphism, or maybe we are applying a categorical decomposition of the morphism?

Anyway, assuming that we can do this, I think that my issue is that I'm having trouble seeing what [itex] (\bar b) [/itex] looks like. Let me regurgitate my thinking:

Case 1

Well, we know that the canonical projection [itex] \pi: R \to R/(a) [/itex] has kernel (a). My first thought is that maybe we can use the restriction to (a,b) and apply the fundamental theorem of homomorphisms. Unfortunately, it's not clear to me if the restriction of [itex] \pi [/itex] to (a,b) is surjective on on [itex] (\bar b) [/itex], so I'm hesitant to continue in this fashion.

Case 2

Instead, I'll try to make the homomorphism from scratch. Let a typical element of (a,b) be [itex] ar_1 + br_2 [/itex] for [itex] r_1,r_2 \in R [/itex]. I want to map this to an element [itex] \bar b r_3 [/itex] for some [itex] r_3 \in R [/itex]. Since a,b are fixed, the only thing we can play with is [itex] r_1,r_2,r_3 [/itex]. If we want [itex] \ker f = (a) [/itex] then whenever [itex] r_2 = 0 [/itex] we want [itex] \bar b r_3 = (a) [/itex].

Now, [itex] \bar b r_3 = (b + (a)) r_3 = b r_3 + (a) r_3 = br_3 + (a) [/itex] since (a) is an ideal. So I'm thinking we just set
[itex] f: ar_1 + br_2 \mapsto b r_2 + (a) [/itex]?
This is surjective, but I've only shown that [itex] (a) \subseteq \ker f [/itex]. In particular, if b is a left-zero divisor, it seems like the kernel could be bigger.

Alternatively, I could just be greatly over complicating this.
 
  • #4
Kreizhn said:
Thanks for the reply micromass, you really seem to jump on these questions :smile:

Are we allowed to do this though? I mean, (a,b) need not be a ring. I guess we could view (a,b) as a group, use the 1st IT then show it's a ring isomorphism, or maybe we are applying a categorical decomposition of the morphism?

Very good remark. There are two possible ways to clear this up. Firstly, we can see (a,b) as an R-module and then use the first iso theorem for modules.
However, (a,b) actually is a ring. It doesn't have an identity, but for the rest it is a ring. So you can apply the first isomorphism theorem for rings without identity. (which is true by thesame proof).

Anyway, assuming that we can do this, I think that my issue is that I'm having trouble seeing what [itex] (\bar b) [/itex] looks like. Let me regurgitate my thinking:

Case 1

Well, wer know that the canonical projection [itex] \pi: R \to R/(a) [/itex] has kernel (a). My first thought is that maybe we can use the restriction to (a,b) and apply the fundamental theorem of homomorphisms. Unfortunately, it's not clear to me if the restriction of [itex] \pi [/itex] to (a,b) is surjective on on [itex] (\bar b) [/itex], so I'm hesitant to continue in this fashion.

I like this approach. Basically, we must simply show that there is an element which is being sent to [itex]\overline{b}[/itex]. But isn't b sent to [itex]\overline{b}[/itex].
 
  • #5
Hmm..okay. In that case the canonical projection would map
[tex] \pi \big|_{(a,b)}: ar_1 + br_2 \mapsto ar_1 + br_2 + (a) [/tex]

But this brings something else up. Since (a) is a subgroup of the underlying abelian group of the ring, [itex] ar_1 + (a) = (a) [/itex] right? Does this mean in general that

[tex] aR + bR + (a) = (a) + (b) + (a) = (a) + (b) [/tex]
which would imply that [itex] (a,b) \cong (a,b)/(a) [/itex].
 
  • #6
Kreizhn said:
Hmm..okay. In that case the canonical projection would map
[tex] \pi \big|_{(a,b)}: ar_1 + br_2 \mapsto ar_1 + br_2 + (a) [/tex]

But this brings something else up. Since (a) is a subgroup of the underlying abelian group of the ring, [itex] ar_1 + (a) = (a) [/itex] right? Does this mean in general that

[tex] aR + bR + (a) = (a) + (b) + (a) = (a) + (b) [/tex]

OK, I agree with all of this.

which would imply that [itex] (a,b) \cong (a,b)/(a) [/itex].

Why would that imply that? I don't quite see how this follows. This isn't true of course. Maybe you mean

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

which is true. But note that [itex]\overline{a}=0[/itex] (in R/(a)).
 
  • #7
I think I might see where I screwed up: I messed up the addition. My original thought process was as follows: if we write out elements of (a,b)/(a), they would look like elements of (a,b) with (a) thrown on the end. That is,
[tex] ar_1 + br_2 + (a) [/tex]
so that (a,b)/(a) = aR + bR + (a). Clearly this is where the error happened.

So what did I did wrong? I think this is it, but maybe you could clear it up for me.

When I write [itex] aR + bR + (a) [/itex] this is really addition on the quotient ring [itex] [aR+(a)]+[bR+(a)] [/itex] which is what gives us [itex] (\bar a, \bar b) [/itex] right? But I think where I'm messing up is that the addition signs are technically different. To be more precise,

[tex] [aR\ +_R\ (a)]\ +_{R/(a)}\ [bR\ +_R\ (a)] = (\bar a, \bar b)[/tex]

which is why I can't just blindly do the addition. Does this make sense?
 
  • #8
Kreizhn said:
I think I might see where I screwed up: I messed up the addition. My original thought process was as follows: if we write out elements of (a,b)/(a), they would look like elements of (a,b) with (a) thrown on the end. That is,
[tex] ar_1 + br_2 + (a) [/tex]
so that (a,b)/(a) = aR + bR + (a). Clearly this is where the error happened.

So what did I did wrong? I think this is it, but maybe you could clear it up for me.

When I write [itex] aR + bR + (a) [/itex] this is really addition on the quotient ring [itex] [aR+(a)]+[bR+(a)] [/itex] which is what gives us [itex] (\bar a, \bar b) [/itex] right? But I think where I'm messing up is that the addition signs are technically different. To be more precise,

[tex] [aR\ +_R\ (a)]\ +_{R/(a)}\ [bR\ +_R\ (a)] = (\bar a, \bar b)[/tex]

which is why I can't just blindly do the addition. Does this make sense?

I'm starting to see what you did. The problem is that you can't just write things like bR+(a), without saying what it means.
You can write b+(a) only because it's a notation. The real element is , the equivalence class of b. So b+(a) is only notation. And Rb+(a) is certainly not standard. What do you mean with this. Do you mean

[itex]\{rb+(a)~\vert~r\in R\}[/itex]?

In this case, it is indeed true that

[tex]Rb+(a)=(\overline{b})=(a,b)/a[/tex]
 
  • #9
I just looked over my first post, and I see the mistake I made there as well. Namely, that

[tex] (\bar b) = \bar b R/(a). [/tex]

I forgot the quotient by (a)!
 
  • #10
micromass said:
What do you mean with this. Do you mean

[itex]\{rb+(a)~\vert~r\in R\}[/itex]?

Yeah, that's ultimately what I meant.

Edit: I think the lesson today is that I need to be more careful :smile:
 

Related to Quotients of Principal Ideals: How to Apply the Third Isomorphism Theorem

1. What are quotients of principal ideals?

Quotients of principal ideals are formed by dividing a principal ideal, which is an ideal generated by a single element, by another principal ideal. This results in a new ideal that contains elements that are the quotients of the elements in the original ideals.

2. How are quotients of principal ideals used in algebra?

Quotients of principal ideals are used in algebra to simplify calculations and solve equations involving ideals. They also have applications in algebraic number theory and algebraic geometry.

3. Can quotients of principal ideals be commutative?

Yes, quotients of principal ideals can be commutative. This means that the order in which the principal ideals are divided does not affect the resulting quotient ideal.

4. How are quotients of principal ideals related to quotient rings?

Quotients of principal ideals are closely related to quotient rings. In fact, quotient rings can be viewed as a special case of quotients of principal ideals, where the ideals are generated by a single element.

5. Are there any special properties of quotients of principal ideals?

Yes, quotients of principal ideals have several important properties that make them useful in algebraic structures. For example, they are always finitely generated, and they can be used to classify and study prime ideals in a ring.

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