 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?

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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? Debbie Moore
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. karton

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.