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Based upon your examples, you may find it more natural to work with the slightly more general class of Prüfer domains (vs. Bezout domains) - i.e. finitely generated ideals$\:\ne 0$ are invertible (vs. principal). Prüfer domains are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations, e.g. CRT, or Gauss's Lemma for content ideals, or for ideals $\rm\ A\cap (B + C) = A\cap B + A\cap C\:,\:$ or $\rm\ (A + B)\ (A \cap B) = A\ B\$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my sci.math post for 30.

As a simple example I'll give the natural Prüfer domain proof of a generalization of your example, viz. the ideal-theoretic $\:$ Freshman's Dream $\rm\ \ (A + B)^n = A^n + B^n\:.\:$ This theorem identity is true for both arithmetic of $\:$ GCDs $\:$ and invertible ideals simply because, in both cases, multiplication is cancellative and addition is idempotent, i.e. $\rm\ A + A = A\$ for ideals and $\rm\ (A,A) = A\$ for GCDs. Combining this these properties with the associative, commutative, distributive laws of addition and multiplication we obtain an extremely elementary high-school-level proof of the Freshman's Dream . e.g. - which is best illustrated for $\rm\: n = 2$2\:,\:$viz.$\rm\quad\quad (A + B)^4 \ =\ A^4 + A^3 B + A^2 B^2 + AB^3 + B^4 \rm\quad\quad\phantom{(A + B)^4 }\ =\ A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2 \rm\quad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2 $So$\rm\ {(A + B)^2 }\ =\ \ A^2 + B^2\ $if$\rm\ A+B\ $is cancellative, e.g. if$\rm \rm\ A+B = 1$1\$ or if it's invertible.

The same proof generalizes for all $\rm\:n\:$ since, as above

$\rm\quad\quad (A + B)^{2n}\ =\ A^n\ (A^n + \cdots+ B^n) + (A^n +\cdots+ B^n)\ B^n$

$\rm\quad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n$

In the GCD case $\rm\ A+B\ := (A,B) = \gcd(A,B)\$ for $\rm\:A,B\:$ in a GCD-domain, i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$. Hence Here too the Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain. Note: (Note: one can unify the GCD and ideal cases by employing Divisor TheoryTheory).

In fact this yields yet another characterization: a domain is Prufer iff it satisfies the Freshman's Dream for all finitely generated ideals. See said sci.math post for further discussion.

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Based upon your examples, you may find it more natural to work with the slightly more general class of Prüfer domains (vs. Bezout domains) - i.e. finitely generated ideals$\:\ne 0$ are invertible (vs. principal). Prufer Prüfer domains are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations, e.g. CRT, or Gauss's Lemma for content ideals, or for ideals $\rm\ A\cap (B + C) = A\cap B + A\cap C\ C\:,\:$ or $\rm\ (A + B)\ (A \cap B) = A\ B\$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my sci.math post for 30.

As a simple example I'll give the natural Prufer Prüfer domain proof of a generalization of your example, viz. the ideal-theoretic $\:$ Freshman's Dream $\rm\ (A + B)^n = A^n + B^n\:.\:$ This theorem is true for both arithmetic of GCDs and invertible ideals simply because, in both cases, multiplication is cancellative and addition is idempotent, i.e. $\rm\ A + A = A\$ for ideals and $\rm\ (A,A) = A\$ for GCDs. Combining this with the associative, commutative, distributive laws of addition and multiplication we obtain an extremely elementary high-school-level proof of the Freshman's Dream. e.g. for $\rm\: n = 2$

$\rm\quad\quad (A + B)^4 \ =\ A^4 + A^3 B + A^2 B^2 + AB^3 + B^4$

$\rm\quad\quad\phantom{(A + B)^4 }\ =\ A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2$

$\rm\quad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2$

So $\rm\ {(A + B)^2 }\ =\ \ A^2 + B^2\$ if $\rm\ A+B\$ is cancellative, e.g. if $\rm A+B = 1$

The same proof generalizes for all $\rm\:n\:$ since, as above

$\rm\quad\quad (A + B)^{2n}\ =\ A^n\ (A^n + \cdots+ B^n) + (A^n +\cdots+ B^n)\ B^n$

$\rm\quad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n$

In the GCD case $\rm\ A+B\ := (A,B) = \gcd(A,B)\$ for $\rm\:A,B\:$ in a GCD-domain, i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$. Hence the Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain. Note: one can unify the GCD and ideal cases by employing Divisor Theory.

In fact this yields yet another characterization: a domain is Prufer iff it satisfies the Freshman's Dream for all finitely generated ideals. See said sci.math post for further discussion.

1

Based upon your examples, you may find it more natural to work with the slightly more general class of Prüfer domains (vs. Bezout domains) - i.e. finitely generated ideals$\:\ne 0$ are invertible (vs. principal). Prufer domains are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations, e.g. CRT, or Gauss's Lemma for content ideals, or for ideals $\rm\ A\cap (B + C) = A\cap B + A\cap C\$ or $\rm\ (A + B)\ (A \cap B) = A\ B\$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my sci.math post for 30.

As a simple example I'll give the natural Prufer domain proof of a generalization of your example, viz. the ideal-theoretic $\:$ Freshman's Dream $\rm\ (A + B)^n = A^n + B^n\:.\:$ This theorem is true for both arithmetic of GCDs and invertible ideals simply because, in both cases, multiplication is cancellative and addition is idempotent, i.e. $\rm\ A + A = A\$ for ideals and $\rm\ (A,A) = A\$ for GCDs. Combining this with the associative, commutative, distributive laws of addition and multiplication we obtain an extremely elementary high-school-level proof of the Freshman's Dream. e.g. for $\rm\: n = 2$

$\rm\quad\quad (A + B)^4 \ =\ A^4 + A^3 B + A^2 B^2 + AB^3 + B^4$

$\rm\quad\quad\phantom{(A + B)^4 }\ =\ A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2$

$\rm\quad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2$

So $\rm\ {(A + B)^2 }\ =\ \ A^2 + B^2\$ if $\rm\ A+B\$ is cancellative, e.g. if $\rm A+B = 1$

The same proof generalizes for all $\rm\:n\:$ since, as above

$\rm\quad\quad (A + B)^{2n}\ =\ A^n\ (A^n + \cdots+ B^n) + (A^n +\cdots+ B^n)\ B^n$

$\rm\quad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n$

In the GCD case $\rm\ A+B\ := (A,B) = \gcd(A,B)\$ for $\rm\:A,B\:$ in a GCD-domain, i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$. Hence the Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain. Note: one can unify the GCD and ideal cases by employing Divisor Theory.

In fact this yields yet another characterization: a domain is Prufer iff it satisfies the Freshman's Dream for all finitely generated ideals. See said sci.math post for further discussion.