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We're in an integral domain with unity 1 $\neq$ 0. Suppose that the highest common factor between x,y is 1 and the highest common factor for x,z is 1.

Show that $x \mid yz$ implies that $x$ is a unit, or provide a counterexample.

I'm stuck. I don't have that we are in a unique factorization domain, I don't have that this ring is Noetherian.

  • Thanks, I wrote the problem wrong :). Just fixed it. –  Apr 14 '12 at 21:40
  • Did you mean, the highest common factor between $x$ and $y$, and between $x$ and $z$ is $1$? If so, look at $\mathbb{Z}[\sqrt{-5}]$, $x=2$, $y=1+\sqrt{-5}$, and $z=1-\sqrt{-5}$. – Arturo Magidin Apr 14 '12 at 21:40
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    What does a "highest" common factor mean in a general integral domain (that is, without any assumption that it is Euclidean)? – hmakholm left over Monica Apr 14 '12 at 21:46
  • (It is probably reasonable to interpret it as "all common factors of $x$ and $y$ are units", or equivalently "the ideal generated by $x$ and $y$ is the entire ring" -- but would be nice to have that confirmed explicitly). – hmakholm left over Monica Apr 14 '12 at 21:52
  • it means it is a multiple of any other common factor. –  Apr 14 '12 at 21:53
  • @Henning: It means that it divides both, and anything that divides both will divide it. In the instant case, it just means that the only common factors between them are units. – Arturo Magidin Apr 14 '12 at 21:53
  • @Henning In any ring, $\rm:gcd(a,b) = 1:$ means $\rm:c:|:a,b:\Rightarrow:c:|:1.:$ This follows from the universal definition of the gcd. – Bill Dubuque Apr 14 '12 at 21:57

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Hint $\ $ For the special case $\rm\:x\:$ is irreducible, a counterexample would be an irreducible element that is not prime. These are easy to find in non-UFD number rings. See also this post.

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Bill Dubuque
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  • This is interesting. So then I am clearly stuck on a problem. I am asked to find that if $a$ and $a$ have a highest common factor, then they have a least common multiple. So here is my strategy: I let $(a,b)$ be the hcf, write $(a,b)a'=a$ and $(a,b)b'=b$ and claim that my candidate is $(a,b)a'b'$ is my lcm. I will repost this as a new question. –  Apr 14 '12 at 21:55