# Does the Bezout GCD equation hold in a UFD?

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 $$\langle a\rangle+\langle b\rangle = \langle d\rangle$$ I know that $$d = \gcd(a,b)$$. In this case, can I find $$x,y \in R$$ such that $$\gcd(a,b) = ax + by$$ ?

• 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. Commented Jul 13, 2017 at 15:30

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".

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.

There are simple examples of non-Bezout UFDs, e.g. any UFD of dimension $$> 1$$, e.g. $$\Bbb Z[x,y],\,$$ as follows by the below equivalent conditions for a UFD to be Bezout (or we can prove it 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.

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

Finally, yes: $$\,d\in (a)\!+\!(b) = aR + bR$$ $$\iff d = ar + br'$$ for some $$\,r,r'\in R$$