Does Z[root(-3)] Qualify as a GCD Domain?

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SUMMARY

Z[√(-3)] is confirmed as a GCD domain, exemplified by the elements 4 and -2+2√(-3), which do not share a greatest common divisor (gcd). The discussion clarifies that while 2 and 1+√(-3) have a gcd of 1, the proposed elements x and y demonstrate the absence of a gcd due to their factorization properties. The relationship between associates and gcd is critical in understanding the structure of this domain.

PREREQUISITES
  • Understanding of GCD domains and unique factorization domains (UFD)
  • Familiarity with algebraic integers in quadratic fields
  • Knowledge of factorization in the context of Z[√(-3)]
  • Basic concepts of divisibility and associates in ring theory
NEXT STEPS
  • Study the properties of GCD domains in algebraic number theory
  • Explore the implications of unique factorization in Z[√(-3)]
  • Investigate examples of non-UFDs and their characteristics
  • Learn about the role of associates in determining gcd in ring theory
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Mathematicians, algebraic number theorists, and students studying ring theory who are interested in the properties of GCD domains and unique factorization.

bw0young0math
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Hello. Today I study domains but I have some problems. Please help me.

I learned Z[root(-3)] is not UFD.
Then Z[root(-3)] is a example about gcd-domain( there exist some elts a,b in domain D such that it does not exist gcd(a,b).)

Let a=2, b-1+root(-3) .
a and b are not associates and they just have common divisor 1 and -1.
So gcd(a,b) does not exist.

Hum... I don't know why associate, 1/-1 have some relations with not existing gcd. Please......
 
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bw0young0math said:
Hello. Today I study domains but I have some problems. Please help me.

I learned Z[root(-3)] is not UFD.
Then Z[root(-3)] is a example about gcd-domain( there exist some elts a,b in domain D such that it does not exist gcd(a,b).)

Let a=2, b=1+root(-3) . I have changed a "-" to an "=" here. I hope that is what was intended.
a and b are not associates and they just have common divisor 1 and -1.
So gcd(a,b) does not exist.

Hum... I don't know why associate, 1/-1 have some relations with not existing gcd. Please......
There is something wrong here. The only common divisors of a and b are 1 and -1, and that means that a and b do have a gcd, namely 1.

If you want two elements of Z[√(-3)] without a gcd, then I suggest x = 4 and y = -2+2√(-3). Then a and b are both factors of x and of y, because
x = 2*2 = (1+√(-3))(1-√(-3)) and
y = 2(-1+√(-3)) = (1+√(-3))(1+√(-3)).
If x and y had a gcd then it would have to be a multiple of ab = 2+2√(-3). But ab does not divide x or y.
 

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