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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 $i,j$?

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  • $\begingroup$ 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. $\endgroup$ – Arthur Mar 7 '18 at 9:32
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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 $gcd(a,b)=1$.

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  • $\begingroup$ 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? $\endgroup$ – ant Mar 8 '18 at 1:20
  • $\begingroup$ Yes, indeed. For example, the gcd in $R=K[x]$ the ring of polynomials over a field $K$. $\endgroup$ – Dietrich Burde Mar 8 '18 at 8:58
  • $\begingroup$ 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. $\langle x , y \rangle \neq C[x,y]$? $\endgroup$ – Sean Haight Oct 10 '19 at 22:40
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Oct 11 '19 at 7:54

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