Can you find a domain where ax+by=1 has a solution

Josh Sizemore 2022-01-02 Answered
Can you find a domain where \(\displaystyle{a}{x}+{b}{y}={1}\) has a solution for all a and b relatively prime, but which is not a PID?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

sirpsta3u
Answered 2022-01-03 Author has 4830 answers
Step 1
Define a domain \(\displaystyle{R}_{{{0}}}\) as follows. Take a field K, adjoin an indeterminate \(\displaystyle{x}_{{{0}}}\), and localize at \(\displaystyle{\left({x}_{{{0}}}\right)}\) (that is, adjoin inverses to everything not a multiple of \(\displaystyle{x}_{{{0}}}\)).
\(\displaystyle{R}_{{{0}}}\) has all its ideals principal and linearly ordered: \(\displaystyle{\left({x}_{{{0}}}\right)}\) contains \(\displaystyle{\left({{x}_{{{0}}}^{{{2}}}}\right)}\) contains \(\displaystyle{\left({{x}_{{{0}}}^{{{3}}}}\right)}\)
Now given \(\displaystyle{R}_{{{i}}}\) define \(\displaystyle{R}_{{{i}+{1}}}\) inductively: Adjoin an indeterminate \(\displaystyle{x}_{{{i}+{1}}}\), so we have \(\displaystyle{R}_{{{i}}}{\left[{x}_{{{i}+{1}}}\right]}\). Quotient by \(\displaystyle{\left({{x}_{{{i}+{1}}}^{{{2}}}}-{x}_{{{i}}}\right)}\). Finally, localize at the prime ideal \(\displaystyle{\left({x}_{{{i}+{1}}}\right)}\)
This effectively just gives us one more principal ideal containing all the principal ideals from \(\displaystyle{R}_{{{i}}}:{\left({x}_{{{i}+{1}}}\right)}\) contains \(\displaystyle{\left({{x}_{{{i}+{1}}}^{{{2}}}}={\left({x}_{{{i}}}\right)}\right.}\) contains \(\displaystyle{\left({{x}_{{{i}}}^{{{2}}}}\right)}\)
Now let R be the union of all the \(\displaystyle{R}_{{{i}}}\), and it's obvious that any finitely generated ideal is principal, but there's a non-fg one generated by all the \(\displaystyle{x}_{{{i}}}\).
Not exactly what you’re looking for?
Ask My Question
0
 
Debbie Moore
Answered 2022-01-04 Author has 5720 answers
The easiest example I know is the ring of all algebraic integers (roots of monic polynomials with integer coefficients).
As noted, it is a Bezout domain, so every finitely generated ideal is principal, and in particular for every two algebraic integers a and b there exist algebraic integers \(\displaystyle\alpha\) and \(\displaystyle\beta\) such that \(\displaystyle\alpha{a}+\beta{b}={d}\), where d is a gcd for a and b. However, the ideal
\(\displaystyle{\left({2},\ {2}^{{{\frac{{{1}}}{{{2}}}}}},\ {2}^{{{\frac{{{1}}}{{{4}}}}}},\ {2}^{{{\frac{{{1}}}{{{8}}}}}},\ \cdots,{2}^{{{\frac{{{1}}}{{{2}^{{{n}}}}}}}},\cdots\right)}\)
is not principal, so the ring is not a PID.
0
karton
Answered 2022-01-09 Author has 9103 answers

Step 1
Given an integral domain R, we say that \(a,\ b\in R\) are coprime if gcd(a,b) exists and \(gcd(a,\ b)=1\).
On the other hand, we say that a and b are comaximal if there are \(x,\ y\in R\) such that \(ax+by=1\).
It's easy to see that comaximal \(\Rightarrow\) coprime, but the other implication isn't necessarily true. Domains where coprime \(\Rightarrow\) comaximal were called by Cohn Pre-Bézout domains.
As the names suggest, these aren't necessarily Bézout domains, because we only have the "Bézout relationship" for coprime elements.
But, it turns out that we can use Pre-Bézout domains to characterize Bézout domains among the class of GCD domains. More exactly, it's true the following.
Step 2
Theorem: Let R be an integral domain. TFAE:
i) R is a Bézout domain.
ii) R is a GCD Pre-Bézout domain.
Proof: i) \(\Rightarrow\) ii) It's immediate.
ii) \(\Rightarrow\) i)
Let \(a,\ b\in R\)
WLOG, we can suppose that \(a\ne0\ne b\)
As R is a GCD domain, then \(d=gcd(a,\ b)\) exists.
By an elementary property of gcds we have that \(1=gcd(\frac{a}{d},\ \frac{b}{d})\) and since R is Pre-Bézout then \(\frac{a}{d}\) y \(\frac{b}{d}\) are comaximal, which means that there are \(x,\ y\in R\) such that
\(\frac{a}{d}x+\frac{b}{d}y=1\)
Finally, if we multiply by d the above equality we get
ax+by=d
Thus d is a R-linear combination of a and b. Hence, R is Bézout domain.
In conclusion, according to Cohn, the class of domains you are looking for are known as Pre-Bézout domains, and these aren't necessarily Bézout domains, let alone PIDs.

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-01-15
How can one show (hopefully in an elementary manner) that there exist irreducible polynomials of arbitrary degree and number of variables over arbitrary field?
Does \(\displaystyle\forall{n},{d}\in{\mathbb{{{N}}}}\forall\) field \(\displaystyle{\mathbb{{{F}}}}\) exist an irreducible \(\displaystyle{f}\in{\mathbb{{{F}}}}{\left[{x}_{{{1}}},\cdots,{x}_{{{n}}}\right]}\) of degree d?
asked 2022-01-15
If S is a set of functions from X to Y then I can consider the action of a group G on S via its action on X and Y by the formula \(\displaystyle{\left({g}\cdot{f}\right)}{\left({x}\right)}={g}\cdot{f{{\left({g}^{{-{1}}}\cdot{x}\right)}}}\),
So we are considering left actions both on X and Y.
Then, the definition of equivariant map pops out but I don't really understand how it is related to previous statement.
An function \(\displaystyle{f}:{X}\rightarrow{Y}\) is equivariant if it satisfies
\(\displaystyle{f{{\left({g}\cdot{x}\right)}}}={g}\cdot{f{{\left({x}\right)}}}\forall{g}\in{G}\).
What's happening here? Are we assuming that the group G acts trivially on Y? How to get this definition from the previous statement?
asked 2022-01-13
What reduced quadratic equation has solutions \(\displaystyle\sqrt{{{2}}}\) and \(\displaystyle\sqrt{{{3}}}\)?
asked 2022-01-14
Proof attempt: Since f is irreducible and has a as root then K(a) is isomorphic to \(\displaystyle{\frac{{{K}{\left[{x}\right]}}}{{{\left({f}\right)}}}}\) with the isomorphism \(\displaystyle{X}+{f}\rightarrow{a}\). And we can do the same for b and we get \(\displaystyle{K}{\left({b}\right)}\) is isomorphic to \(\displaystyle{\frac{{{K}{\left[{x}\right]}}}{{{\left({f}\right)}}}}\). Therefore \(\displaystyle{K}{\left({a}\right)}\) is isomorphic with \(\displaystyle{K}{\left({b}\right)}\).
a) f is irreducible and has a as root then \(\displaystyle{K}{\left({a}\right)}\) is isomorphic to \(\displaystyle{\frac{{{K}{\left[{x}\right]}}}{{{\left({f}\right)}}}}\) we took it for granted in the lecture but why does this hold?
asked 2022-01-12
An automorphism of a group G is an isomorphism from G to itself. Denote by Aut G the set of all automorphisms of G.
a) Prove that Aut G is a group with respect to the operation of composition
b) Give an example of an abelian G such that Aut G is not abelian
asked 2020-11-09

Show that if \(\displaystyle{K}\subseteq{F}\) is an extension of degree k, every element \(\displaystyle\alpha\in{F}\) has a minimal polynomial over K of degree <=k. Find theminimal polymials of complex numers over \(\displaystyle\mathbb{R}\) explicitly to verify this fact

asked 2022-01-14
Find a polynomial \(\displaystyle{P}_{{{n}}}\in{C}{\left[{X}\right]}\) such as \(\displaystyle{S}_{{{n}}}={k}{e}{r}{P}_{{{n}}}{\left({D}\right)}\)
With \(\displaystyle{S}_{{{n}}}{\left({n}\geq{1}\right)}\) all of the functions such as \(\displaystyle{y}^{{{\left({n}\right)}}}={y}\) (where \(\displaystyle{y}^{{{\left({n}\right)}}}\) is the nth derivative of y)
D the endomorphism sending the functions \(\displaystyle{C}^{{\infty}}\) on their derivatives \(\displaystyle{D}:{y}\rightarrow{y}'\)
I already proved that D is an endomorphism and \(\displaystyle{S}_{{{n}}}\) a vector space containing the functions \(\displaystyle{k}{e}^{{{x}}}\).
But my probleme is that i dont really understand the meaning of \(\displaystyle{P}_{{{n}}}{\left({D}\right)}\), and so dont see what \(\displaystyle{S}_{{{n}}}={k}{e}{r}\ {P}_{{{n}}}{\left({D}\right)}\) could be. It probably have to do with eigenvalues, eigenvectors...But really I dont see anything.

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question
...