Josh Sizemore

2022-01-02

Can you find a domain where $ax+by=1$ has a solution for all a and b relatively prime, but which is not a PID?

sirpsta3u

Expert

Step 1
Define a domain ${R}_{0}$ as follows. Take a field K, adjoin an indeterminate ${x}_{0}$, and localize at $\left({x}_{0}\right)$ (that is, adjoin inverses to everything not a multiple of ${x}_{0}$).
${R}_{0}$ has all its ideals principal and linearly ordered: $\left({x}_{0}\right)$ contains $\left({x}_{0}^{2}\right)$ contains $\left({x}_{0}^{3}\right)$
Now given ${R}_{i}$ define ${R}_{i+1}$ inductively: Adjoin an indeterminate ${x}_{i+1}$, so we have ${R}_{i}\left[{x}_{i+1}\right]$. Quotient by $\left({x}_{i+1}^{2}-{x}_{i}\right)$. Finally, localize at the prime ideal $\left({x}_{i+1}\right)$
This effectively just gives us one more principal ideal containing all the principal ideals from ${R}_{i}:\left({x}_{i+1}\right)$ contains $\left({x}_{i+1}^{2}=\left({x}_{i}\right)$ contains $\left({x}_{i}^{2}\right)$
Now let R be the union of all the ${R}_{i}$, and its

Debbie Moore

Expert

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 $\alpha$ and $\beta$ such that $\alpha a+\beta b=d$, where d is a gcd for a and b. However, the ideal

is not principal, so the ring is not a PID.

karton

Expert

Step 1
Given an integral domain R, we say that are coprime if gcd(a,b) exists and .
On the other hand, we say that a and b are comaximal if there are such that $ax+by=1$.
It's easy to see that comaximal $⇒$ coprime, but the other implication isn't necessarily true. Domains where coprime $⇒$ 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) $⇒$ ii) It's immediate.
ii) $⇒$ i)
Let
WLOG, we can suppose that $a\ne 0\ne b$
As R is a GCD domain, then exists.
By an elementary property of gcds we have that and since R is Pre-Bézout then $\frac{a}{d}$ y $\frac{b}{d}$ are comaximal, which means that there are 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.

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