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While working with rings, groups and other structures, I find the following

$ gcd(a,b) = 1 \iff ai + bj = 1$ for some $ i , j\in Z$

often arrises. I feel like I encountered one side of this implication or the opposite many times especially when dealing with rings, PID, UFD and other such structures, but I am never allowed to assume this. Is there a proof for this statement or a counter example, I am not able to find any elegant proof that comes to mind immediately, neither a counter example.

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gangsignwall
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    A proof is in the opening chapters of every introductory Number Theory textbook ever written – a proof for ${\bf Z}$, that is. For some rings it holds, for some, there isn't even such a thing as a gcd. – Gerry Myerson Nov 19 '19 at 05:59
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    It's true in a UFD $ D \iff \dim D \le 1$ (non0 prime ideals are max) $\iff D$ is Bezout (every ideal $(a,b)$ is principal) $\iff D$ is a PID. See here for proofs. – Bill Dubuque Nov 19 '19 at 06:21

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Let $R$ be a commutative ring with identity. Let's say that two elements $a, b \in R$ have "GCD equal to $1$" if there is no irreducible element $p$ of $R$ that divides both $a$ and $b$. You're asking when the following property holds:

Property: If $a, b \in R$, then $\operatorname{GCD}(a,b) = 1$ if and only if the ideal generated by $a$ and $b$ is all of $R$.

The implication $\Leftarrow$ is obviously always true, since if $xa + yb = 1$ for some $x, y \in R$, then any irreducible element of $R$ which divides both $a$ and $b$ would have to divide $1$, which is impossible, because irreducible elements are by definition nonunits.

The converse is true for principal ideal domains, but false for general unique factorization domains. If $k$ is a field, the ring $R = k[X,Y]$ is a UFD, the elements $X$ and $Y$ are relatively prime, but you cannot find polynomials $f, g \in R$ such that $X f(X,Y) + Y g(X,Y) = 1$.

This is because the ideal $(X,Y)$ generated by $X$ and $Y$ is properly contained in $R$, being the kernel of the surjective ring homomorphism $R \rightarrow k, h(X,Y) \mapsto h(0,0)$. So $\Rightarrow$ is false for the ring $k[X,Y]$.

Proof that the property holds for principal ideal domains:

Let $R$ be a PID, and let $a, b \in R$ be elements with GCD equal to one. Let $J$ be the ideal generated by $a$ and $b$. We want to show that $J = R$. Since $R$ is a PID, we have $J = Rc$ for some $c \in R$. If $c$ is a unit, we are done. Otherwise, $c$ must be divisible by some irreducible element $p$. The inclusions

$$(a), (b) \subset (c) \subset (p)$$ tell us that $p$ divides $a$ and $b$. Therefore $a$ and $b$ are not relatively prime, contradiction.

D_S
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  • The correct definition of $\gcd(a,b) = 1,$ is $,c\mid a,b\iff c\mid 1.,$ i.e. every common divisor of $,a,b,$ is a unit (invertible). Your definition doesn't work in domains with nonunits having no irreducible factor. It's true in a UFD iff the UFD is a PID - see my comment on the question. – Bill Dubuque Nov 19 '19 at 06:34
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You can prove a more general statement.

In order that there exist integers $i$ and $j$ satisfying the equation $$ ai+bj = c, $$ it is both necessary and sufficient that $\gcd(a,b)$ divides $c.$

Proof. Let $a = nd$ and $b = md.$ Then, if it is true that $ai+bj = c,$ we have $$ ndi+mdj = d(ni+mj) = c. $$ Thus, $d$ divides $c.$ The same argument works in the reverse direction.

Letting $c = 1,$ you recover your desired proof.

N.N.
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    This only shows that if $ai+bj = c$, then $\operatorname{gcd}(a,b)$ divides $c$. – D_S Nov 19 '19 at 05:43
  • @D_S You're right. However, I believe the same argument works in the reverse direction; I should have mentioned this. I will edit the post. – N.N. Nov 19 '19 at 05:46
  • Simpler: if $,d = \gcd(a,b),$ then $\ (a,b)\ni c \iff \overbrace{(a,b)}^{\textstyle (d)} \supseteq (c) \iff d\mid c\ $ by contains = divides for principal ideals. – Bill Dubuque Nov 19 '19 at 06:09