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I'm wondering when Bezout's Theorem is valid.

I know when the ring $R$ is a Euclidean Domain or a Principal Ideal Domain there is always $\gcd(a,b)$ and I can find $x,y \in R$ such that $\gcd(a,b) = ax + by.$

I know when the ring $R$ is a Unique Factorization Domain, there is always $\gcd(a,b)$. My question is, in this case can I find $x,y \in R$ such that $\gcd(a,b) = ax + by.$ ?

Now if $R$ is just a commutative ring, if there is $d \in R$ such that $<a>+<b> = <d> $ I know that $d = \gcd(a,b)$. In this case, can I find $x,y \in R$ such that $\gcd(a,b) = ax + by$ ?

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    $\begingroup$ In $\mathbb Z[x]$, a UFD, $2$ and $x$ have GCD $1$, but there is no solution to $2f(x)+xg(x)=1$. More generally, a UFD in which Bezout is true is always a principal ideal domain. I think. $\endgroup$ – Thomas Andrews Jul 13 '17 at 15:30
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It is not necessarily true in a UFD. For instance, if $k$ is a field, then the polynomial ring $k[s,t]$ is a UFD. But $\gcd(s,t)=1$, and there do not exist $x$ and $y$ such that $sx+ty=1$.

For your second question, the answer is yes by definition. Indeed, by definition, $\langle a\rangle+\langle b\rangle$ is the set of elements of the form $ax+by$. So if $d\in \langle a\rangle+\langle b\rangle$, then there exist $x,y\in R$ such that $d=ax+by$.

The converse holds as well: if $d=\gcd(a,b)$ can be written in the form $ax+by$, then that means $d\in \langle a\rangle+\langle b\rangle$, and it follows that $\langle a\rangle+\langle b\rangle=\langle d\rangle$.

Thus a commutative ring has the property that the GCD of any two elements $a$ and $b$ can be written in the form $ax+by$ iff the sum of any two principal ideals is principal. By induction on the number of generators, this is equivalent to any finitely generated ideal being principal. Such a ring is known as a Bezout ring. A Noetherian Bezout ring is the same thing as a principal ideal ring, but there exist non-Noetherian examples as well. For instance, the ring of holomorphic functions on a connected open subset of $\mathbb{C}$ is a Bezout domain but has ideals which are not finitely generated.

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The rings where it is valid Bézout's lemma are known as Bézout rings. In the case of integral domains rings we have the domains known as Bézout domains.

In general, a UFD doesn't have to be a Bézout domain. The classical example is $\Bbb{Z}[x]$, as Thomas Andrews pointed out in his comment. Theoretically there is a reason for which the above is true. We have the following:

Theorem 1: If $D$ is both a UFD and a Bézout domain, then $D$ is a PID.

Proof: See exercise 11 of section 8.3 of the book "Abstract Algebra" by Dummit and Foote.

By this theorem, if every UFD were also a Bézout domain, we would conclude that every UFD is a PID, and this is false (look at $\Bbb{Z}[x]$ again), as you surely know.

More generally, for Bézout domains we have the following result:

Theorem 2: Let $D$ be a Bézout domain. TFAE:

i) $D$ is a PID.

ii) $D$ is Noetherian.

iii) $D$ is a UFD.

iv) $D$ satisfies the ACCP (ascending chain condition on principal ideals).

v) $D$ is an atomic domain.

Proof: This is theorem 46 given in Pete L. Clark's notes on factorization in integral domains

Your second question has been also answered, but let me include a general version of what you wanted to prove.

Theorem 3: Let $a_1,a_2,\ldots a_n$ be nonzero elements of a commutative ring $R$. Then $a_1,a_2,\ldots a_n$ have a greatest common divisor $d$, expressible in the form $$d=r_1a_1+r_2a_2+\cdots +r_na_n$$ if only if the ideal $(a_1,a_2,\ldots a_n)$ is principal.

Note that $(a_1,a_2,\ldots a_n)$ is the same as $(a_1)+(a_2)+\cdots +(a_n)$.

Proof: This is theorem 6-3 in Burton's book "First Course in Rings and Ideals".

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There are simple examples of non-Bezout UFDs, e.g. any UFD of dimension $> 1$, e.g. $\Bbb Z[x,y],\,$ by the below equivalent conditions for a UFD to be Bezout (or directly: $\,\gcd(x,y)=1\,$ but $\,xf+y\,g = 1\Rightarrow 0 = 1\,$ via eval at $\,x=0=y)$.

Theorem $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$

$(1)\ \ $ prime ideals are maximal if nonzero, $ $ i.e. $\rm\ dim\,\ D \le 1$
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1, $ i.e. $ $ coprime $\Rightarrow$ comaximal
$(5)\ \ $ $\rm D$ is Bezout, i.e. all ideals $\,\rm (a,b)\,$ are principal.
$(6)\ \ $ $\rm D$ is a $\rm PID$

Proof $\ $ (sketch of $\,1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\ $ where $\rm\,p_i,\,P\,$ denote primes $\neq 0$

$(1\Rightarrow 2)$ $\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$ some $\rm\,p_j\in P\,$ so $\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$ by dim $\le1$
$(2\Rightarrow 3)$ $ \ $ max ideals are prime, so principal by $(2)$
$(3\Rightarrow 4)$ $\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p) $ for all max $\rm\,(p),\,$ so $\rm\ (a,b) = 1$
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$
$(5\Rightarrow 6)$ $\ $ Ideals $\neq 0\,$ in Bezout UFDs are generated by an elt with least #prime factors
$(6\Rightarrow 1)$ $\ \ \rm (d) \supsetneq (p)$ properly $\rm\Rightarrow\,d\mid p\,$ properly $\rm\,\Rightarrow\,d\,$ unit $\,\rm\Rightarrow\,(d)=(1),\,$ so $\rm\,(p)\,$ is max


As for the gcd as the ideal sum in a PID, that follows using "contains = divides", viz.

$$ c\mid a,b\iff (c)\supseteq (a),(b)\iff (c)\supseteq (a)\!+\!(b)\!=\!(d)\iff c\mid d$$

And, yes, we have $\,d\in (a)+(b) = aR + bR\, $ $\Rightarrow\, d = ar + br'$ for some $\,r,r'\in R$

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