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We know that in every PID, for every two elements, there is one GCD (up to being associate). Now, since the set of real numbers is a field and consequently PID, this theorem holds for them.

Then, what's the GCD of a rational (for instance $2$) and an irrational number (for instance $\sqrt2)$)?

MohammadJavad Vaez
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  • @Peter But we are looking for greatest common divisor, not the smallest one. Aren't we? – MohammadJavad Vaez May 20 '21 at 18:13
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    You are right, but since every non-zero real number is a common dicisor there is neither a greatest common divisior. – Peter May 20 '21 at 18:14
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    Gcd is up to units. But in a field all nonzero elements are units. – Dietrich Burde May 20 '21 at 18:17
  • @DietrichBurde so, is it common to say all of the nonzero elements are gcd of two elements, or there is no greatest common divisor? – MohammadJavad Vaez May 20 '21 at 18:19
  • We need to define "greatest" by divisibility, too. Think of the PID $\Bbb Z[i]$. See here for the definition. – Dietrich Burde May 20 '21 at 18:21
  • @DietrichBurde so, I should define the GCD of two elements in the question. It must have two properties: the first one is that it must divide both elements. The second condition is that if another element divides them, then it must divide the GCD. $a|b$ iff there is an element $k$ belonging to the field s.t $b=ak$ – MohammadJavad Vaez May 20 '21 at 18:26
  • Yes, you are right. For fields, however, this notion of a gcd is not very useful, see the duplicate. – Dietrich Burde May 20 '21 at 18:52

1 Answers1

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Suppose that $D$ is an integral domain (so associates are unit multiples).

If $\,\forall i\!: a\mid a_i\,$ then $\,(a,a_1,a_2,\ldots) = (a)\ $ so $\,\gcd(a,a_1,a_2,\ldots) \approx a,\,$ where $\approx$ means associate.

In particular a unit $a$ divides all elements $\in D$ so any ideal containing it $=(a)=(1)$ and any gcd containing it is $\approx a\approx 1\,$ (usually unit ideals and gcds are unit normalized to $\,1)$

In particular the above applies to every $\,a\neq 0\,$ in a field, since then $a$ is a unit.

Bill Dubuque
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