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

Expert

2022-01-03Added 42 answers

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 $({x}_{i+1}^{2}-{x}_{i})$ . 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 $({x}_{i+1}^{2}=\left({x}_{i}\right)$ contains $\left({x}_{i}^{2}\right)$

Now let R be the union of all the$R}_{i$ , and its

Define a domain

Now given

This effectively just gives us one more principal ideal containing all the principal ideals from

Now let R be the union of all the

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

$(2,\text{}{2}^{\frac{1}{2}},\text{}{2}^{\frac{1}{4}},\text{}{2}^{\frac{1}{8}},\text{}\cdots ,{2}^{\frac{1}{{2}^{n}}},\cdots )$

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

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

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

karton

Expert

2022-01-09Added 439 answers

Step 1

Given an integral domain R, we say that

On the other hand, we say that a and b are comaximal if there are

It's easy to see that comaximal

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)

Let

WLOG, we can suppose that

As R is a GCD domain, then

By an elementary property of gcds we have that

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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