Ring theory

[oak jackal]

Let R be a PID, F fraction field of R, A subring of F so we have $R \subseteq A \subseteq F$
prove that A is a PID.

This is the second section of the question, in the first which I think I've done correctly I managed to show that $\frac{a}{b} \in A \to \frac{1}{b}\in A$ and got no clue how to do this part now. any help?

[errant wagon]

mtr123

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[hallow goblet]

Here's to get you started: let I be an ideal in A, then the set of numerators is an ideal in R @oak jackal

so the set of numerators is principal in R, hence generated by some element g. Then show that the ideal generated by g/1 in A is actually I

[oak jackal]

Isn’t the set of numerators is all of R? Or you mean any set of numerators which is an ideal?

I’m a bit confused what is the set of numerators in this context since R has no fractions

[hallow goblet]

all of them have fractions

regardless

I is a set of fractions, the numerators are elements of R

collect all the numerators of the fractions in I, you will find it's an ideal in R

which is by assumption principal

@oak jackal

$$ {r\in R \mid \exists s\in R^*: \frac{r}{s}\in I} $$

show that this set is an ideal in R

there is some technical mumbo jumbo involved here, but recall how field of fractions is obtained from R and you'll be fine

[errant wagon]

aL

[hallow goblet]

here R* is the subset of units in R

[oak jackal]

Ain’t there only a single unit? 🤔

[hallow goblet]

in a ring?!

the subset of units of a ring is a group

you mean "identity" probably

[oak jackal]

Not sure if that’s what you mean, but looking at $R/cap I$ and I , I could claim that $I = Frac(R\cap I)$ because it is generated by it and since $R/cap I$ generated by some g in R since is a pid I’m good to go?

[errant wagon]

mtr123

[hallow goblet]

"because it is generated by it"
generated by what?

"I = Frac(R\cap I)" do you have a proof?

@oak jackal

[oak jackal]

I mean, looking at I ideal in A we can look at I cap R. I cap R is an ideal in R (we proved that in class I think using the isomorphism theorem).
Now we know I cap R is generated by some r in R because r is a PID and it is an ideal in R. Now we could ‘expand’ R cap I from R to A using the same build of the fraction field. That way we get an ideal in A (I think) and still got to prove that this ideal is actually the I we began with

Or I’m messing up and it’s all wrong?

[hallow goblet]

R cap I is an ideal in R, you can directly verify this by definition

"could ‘expand’" that's the idea

but write it down formally

[oak jackal]

I thought about showing $Frac(R \cap I) = I$ by showing $Frac(R \cap I) \subseteq I$ and $I \subseteq Frac(R \cap I)$

[errant wagon]

mtr123

[hallow goblet]

ofc, that always works

[oak jackal]

if only I managed to show that, it sound really simple but I'm not sure what am I missing.

given $x = \frac{a}{b}$ I know that $a,b \in R$ because $I \lhd A \lhd F$ and cuz of the way F was built $a,b \in R$. Now i need to prove they are also in I so I get that they are in $I \cap R$ and in the fraction field of it.
Since $I \lhd A$ and $b \in A$ I get that $\frac{a}{b} \cdot b = a \in I$ but got no clue how to prove that b also in I

[errant wagon]

mtr123

[oak jackal]

and that's just the first direction

[hallow goblet]

given x in where?

why would b in I have to be true?

[oak jackal]

x in I, if no how can I prove that x is also in Frac(R cap I)?

[oak jackal]

Or that’s not even the right way to do it?

[hallow goblet]

I browsed around and found this https://math.stackexchange.com/questions/137876/a-subring-of-the-field-of-fractions-of-a-pid-is-a-pid-as-well

the answer by Frankel is basically what I was talking about

but there is a more detailed answer below

@oak jackal

[hallow goblet]

I don't know about "the" right way, it's typical that showing equalities directly for such problems is difficult

[oak jackal]

Any idea what is S here?

𝑆={multiplicative set of all elements 𝑎∈𝐴 that are units in 𝑅}.

[hallow goblet]

in your notation: the elements in R that are invertible in A

but the S are different in the answers

in what you have quoted S is simply the subset of units in R

as per the construction procedure of Frac R

[oak jackal]

honestly it feels like I should have seen the answer long ago but still got no idea what I'm missing. still trying with that link you gave me

[oak jackal]

also I just saw they gave us a hint to somehow use that if I is ideal in A we should take the generator of I cap R in R

[hallow goblet]

that's been the hint all along

take a generator g for I cap R and show g/1 generates I

if you're wondering about technique you can always take the prototypical setting

R = Z and Frac R = Q

can you prove that an ideal of a subring of Q is principal?

@oak jackal

proof for the general case is not that much different

[oak jackal]

Sorry for the delay again.
I think the definitions here really confuses me.
Like I got to prove now that (g/1) = I so I would try to show that (g/1) is in I and then that I in (g/1)
but I dont even know how to show that g/1 is in I.

well maybe I got it. I showed earlier that if a/b in I then 1/b is in I too but well I don't even know if b itself is in I so I can't claim that b/b = 1 is in I.

and for the other direction I'm even more clueless

[oak jackal]

even when looking at Z and Q, not sure what I'm missing cuz it feel like I'm missing something stupid and you don't want to point it out cuz it gonna reveal it all (which make sense of course)

[oak jackal]

I could show that for any a\b in I, a,b are in <g> because they are both in I and in R

[oak jackal]

Maybe I got it.

Lemma:
if a,b are coprime and $\frac{a}{b} \in A$ then $\frac{1}{b}\in A$

proof:
since they are coprime there are $n,m \in R$ such that $na+mb=1$. now we know that $\frac{mb}{b}, \frac{na}{b} \in A$ and so $\frac{na+mb}{b} \in A$ which is exaclty $\frac{1}{b}$

Now let $I \lhd A$. And we gonna look at $R \cap I$.
since R is a pid we know that there is some single generator g such that $\langle g \rangle = R\cap I$

now we wanna prove $\frac{g}{1}$ is the generator of I.

let $\frac{a}{b} \in I$ we know $a\in R$ therefore there exists some k in R such that kg = a (because g is the generator of R).

now:
$\frac{a}{b} = \frac{k}{b} \cdot \frac{g}{1}$\
therefore for each $\frac{a}{b}$ we there some $\frac{k}{b}$ that help us create it using $\frac{g}{1}$

I think this is fine.?

@hallow goblet

[errant wagon]

mtr123

[oak jackal]

I feel like I got to show that 1/b is also created by g/1?

[hallow goblet]

@oak jackal typo: g is the generator of R cap I

why is a in R cap I? (you write a = kg, which is true if a in I)

You have to be careful with the lemma. Bezout theorem works for PIDs. You have yet to prove A is a PID

you are very close

[hallow goblet]

For subrings of Q the argument goes something as follows.

Suppose Z subset A subset Q. If A=Z then the claim is true, so suppose the containment is proper. Then there exist k in Z such that 1/k in A.
Let S be the multiplicatively closed subset of Z that is generated by all k in Z for which 1/k in A.

Then it immediately follows that A = S^{-1}Z (why?).

Exercise. Show that any ideal of A is of the form S^{-1}I, where I is an ideal of Z.

Now given an ideal J in A, we have that J = S^{-1}I = S^{-1}<g> hence every element in J is of the form (rg) /k, equivalently (r/k) * (g/1), hence J is principal. @oak jackal

[oak jackal]

Assume that lemma from before.

Let $I \lhd in A$ look now at $I \cap R$ in R. We know that $R \cap I \lhd R$ and because R is a pid it is generated by a single element. Denote $g \in R$. \
Let $\frac{a}{b} \in I$ (where gcd(a,b)=1, $b\neq 0$ such $\frac{a}{b}$) exists otherwise we get that I subset R and the question becomes trivial. \
Now I also showed that $\frac{1}{b}\in I$ and since I is ideal and $b\in R\subseteq A$ we get that $\frac{a}{b}\cdot b = a \in I$ \
so we get that $a\in R\cap I$ \
Now since $R\cap I$ is generated by g from earlier we know that there's some $n \in R$ such that $gn=a$.\
so we also get that $n\cdot \frac{1}{b} \in A$ and so $g \cdot (n \cdot \frac{1}{b}) = \frac{gn}{b} = \frac{a}{b}$ and now we are finished cuz we found some $x \in A~ (n\cdot \frac{1}{b})$ such that $gx=\frac{a}{b}$ for some random $\frac{a}{b} \in I$.\

Sorry I thought about this before and I think it works

Hopefully its ok this time

[errant wagon]

mtr123

[hallow goblet]

better @oak jackal but there is still one problem

You used Bezout to prove the lemma. But Bezout applies for PIDs. You don't know yet that A is a PID. So can you prove the lemma without Bezout?

Bezout being this:

[oak jackal]

but a,b are in R and R is a pid so assuming those the same a,b from R and they are coprime it still holds isn't it?

[hallow goblet]

right right, i thought for some reason you applied bezout to A, but no

then your proof is correct @oak jackal

[oak jackal]

Thank you very very much!

For being patient with me and all!