# Definition of coprime in a ring

If $R$ is a unique factorisation domain, what is meant by two elements $x,y\in R$ being coprime? We decompose them as $x=up_1\dots p_n$ and $y=u'q_1\dots q_r$, such taht these $p_i,q_j$ are prime elements. Does $x,y$ being coprime mean that $p_i\ne q_j$ for each $i,j$?

• I would think it means that if $a\in R$ such that $a\mid x$ and $a\mid y$, then $a$ is a unit. But I don't have a book with a definition with me at the moment, so I can't confirm. – Arthur Mar 7 '18 at 9:32

## 1 Answer

One definition for a PID is that $$a$$ and $$b$$ are called coprime if $$Ra+Rb=R$$, i.e., they generate coprime ideals. How this is related is explained here:

If coprime elements generate coprime ideals, does it imply for any $a,b\in R$ that $\langle a\rangle+\langle b\rangle=\langle \gcd (a,b)\rangle$?

Since $$R$$ is a UFD, it is a gcd-domain, so there exists a gcd for all non-zero elements. Then $$a$$ and $$b$$ are coprime, iff $$gcd(a,b)=1$$.

• When we say $gcd(a,b)=1$ we mean that it is unique up to associates, and $1$ is a representative for this class of associates? – ant Mar 8 '18 at 1:20
• Yes, indeed. For example, the gcd in $R=K[x]$ the ring of polynomials over a field $K$. – Dietrich Burde Mar 8 '18 at 8:58
• By your definition, is it true that elements are co-prime if and only if $gcd(a,b) = 1$? In the UFD $\mathbb{C}[x,y]$ the elements $x$ and $y$ have gcd 1 but are not coprime, i.e. $\langle x , y \rangle \neq C[x,y]$? – Sean Haight Oct 10 '19 at 22:40
• You are right, $Ra+Rb=R$ is for PID. For UFD's it may not be true, see your example. Then only no prime element is a common divisor. – Dietrich Burde Oct 11 '19 at 7:54