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 $\gcd(a,b) = ax + by$ ?

  • 1
    $\begingroup$ 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

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.


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".


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 $\,r,r'\in R$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.