Can you find a domain where ax+by=1 has a solution for all a and b...

Josh Sizemore

Josh Sizemore

Answered

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?

Answer & Explanation

sirpsta3u

sirpsta3u

Expert

2022-01-03Added 42 answers

Step 1
Define a domain R0 as follows. Take a field K, adjoin an indeterminate x0, and localize at (x0) (that is, adjoin inverses to everything not a multiple of x0).
R0 has all its ideals principal and linearly ordered: (x0) contains (x02) contains (x03)
Now given Ri define Ri+1 inductively: Adjoin an indeterminate xi+1, so we have Ri[xi+1]. Quotient by (xi+12xi). Finally, localize at the prime ideal (xi+1)
This effectively just gives us one more principal ideal containing all the principal ideals from Ri:(xi+1) contains (xi+12=(xi) contains (xi2)
Now let R be the union of all the Ri, and its
Debbie Moore

Debbie Moore

Expert

2022-01-04Added 43 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 α and β such that αa+βb=d, where d is a gcd for a and b. However, the ideal
(2, 212, 214, 218, ,212n,)
is not principal, so the ring is not a PID.
karton

karton

Expert

2022-01-09Added 439 answers

Step 1
Given an integral domain R, we say that a, bR 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, yR 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 a, bR
WLOG, we can suppose that a0b
As R is a GCD domain, then d=gcd(a, b) exists.
By an elementary property of gcds we have that 1=gcd(ad, bd) and since R is Pre-Bézout then ad y bd are comaximal, which means that there are x, yR such that
adx+bdy=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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