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Embedding a ring in another ring

I've shown that Z[3]\mathbb{Z}[\sqrt{-3}] isn't a UFD. Now I have to show how we can embed it in a principal ideal domain RR such that it has index 2, i.e. such that the additive subgroup R/Z[3]R / \mathbb{Z}[\sqrt{-3}] has order 2.

Now, the wording of the question makes me think that we don't need to actually find such an RR explicitly. Clearly we need

R/Z[3]={Z[3], r+Z[3]}R/\mathbb{Z}[\sqrt{-3}] = \{ \mathbb{Z}[\sqrt{-3}],\ r + \mathbb{Z}[\sqrt{-3}] \} for some rRr \in R

so every element of RR is of the form a+b3a+b\sqrt{-3} or a+b3+ra+b\sqrt{-3}+r for a,bZa,b \in \mathbb{Z} and rr as above. But I need to justify why RR is a principal ideal domain and I'm not entirely sure how I'd go about this.
(edited 12 years ago)
I think it's easier to just find the ring R. We don't really know any principal ideal domains that aren't Euclidean domains, so it's a safe bet our R will be a Euclidean domain. And that 3\sqrt{-3} is a big giveaway.
Reply 2
Original post by Glutamic Acid
I think it's easier to just find the ring R. We don't really know any principal ideal domains that aren't Euclidean domains, so it's a safe bet our R will be a Euclidean domain. And that 3\sqrt{-3} is a big giveaway.


I was tempted to use Z[1+32]\mathbb{Z} \left[ \dfrac{1+\sqrt{-3}}{2} \right] but for some reason I didn't follow it through... but this seems to work; thanks.
Reply 3
*ringception
Reply 4
Original post by nuodai
I've shown that Z[3]\mathbb{Z}[\sqrt{-3}] isn't a UFD. Now I have to show how we can embed it in a principal ideal domain RR such that it has index 2, i.e. such that the additive subgroup R/Z[3]R / \mathbb{Z}[\sqrt{-3}] has order 2.

Now, the wording of the question makes me think that we don't need to actually find such an RR explicitly.


I'm not sure it can be done, but I like your way of thinking! I imagine the idea is to look at the ideals of Z[3]\mathbb{Z}[\sqrt{-3}] which are not principal and adjoin an element which makes it so. For example, (2,1+3)(2, -1 + \sqrt{-3}) is not principal because those two things don't have a common divisor. But if we add in, say, 1+32\dfrac{-1 + \sqrt{-3}}{2} to our ring, we do have a common divisor (and in fact they become associate).

Unfortunately, I'm not sure such a process will terminate in finitely many steps, nor am I certain whether it's always possible to do it so that each extension has finite index. Any integral domain A embeds into its field of fractions K, and K is a PID. Unfortunately the ideals of K are very boring and worse, typically the index [K : A] is infinite. The naïve idea is to consider the set of all subrings of K containing A which are PIDs, and take their intersection, but the intersection of two PIDs is not in general a PID: for example, Q(X)[Y]\mathbb{Q}(X)[Y] and Q(Y)[X]\mathbb{Q}(Y)[X] are both Euclidean domains (hence PIDs!) but their intersection is Q[X,Y]\mathbb{Q}[X, Y], which is not a PID.
Original post by nuodai
I was tempted to use Z[1+32]\mathbb{Z} \left[ \dfrac{1+\sqrt{-3}}{2} \right] but for some reason I didn't follow it through... but this seems to work; thanks.


This does work. My supervision partner used Z[32] \mathbb{Z}\left[ \frac{\sqrt{-3}}{2}\right] , which seemed to work as well.
Reply 6
Original post by Daniel Freedman
This does work. My supervision partner used Z[32] \mathbb{Z}\left[ \frac{\sqrt{-3}}{2}\right] , which seemed to work as well.


I was hesitant to use that because (32)2=34∉Z\left( \dfrac{\sqrt{-3}}{2} \right)^2 = \dfrac{-3}{4} \not \in \mathbb{Z} (and hence all powers of this are also in the ring), which adds an element of complication. Whereas (1+32)2=1+32=1+1+32\left( \dfrac{1+\sqrt{-3}}{2} \right)^2 = \dfrac{-1+\sqrt{-3}}{2} = 1 + \dfrac{1+\sqrt{-3}}{2}, which simplifies things quite a bit.
Reply 7
Original post by Daniel Freedman
This does work. My supervision partner used Z[32] \mathbb{Z}\left[ \frac{\sqrt{-3}}{2}\right] , which seemed to work as well.


That doesn't work. Z[3]\mathbb{Z}[\sqrt{-3}] does not embed in Z[3/2]\mathbb{Z}[\sqrt{-3}/2] as a subring of finite index, as nuodai has pointed out.

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