History of Math/Abstract Algebra: Why is Z[i] is a UFD but Z[Sqrt(-5)] is not
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History of Math/Abstract Algebra: Why is Z[i] is a UFD but Z[Sqrt(-5)] is not

[From: ] [author: ] [Date: 12-02-03] [Hit: ]
I find out that 5 = (2 + i)(2 - i) = (1 + 2i)(1 - 2i), however its been stated almost everywhere that Z[i] is a UFD. How on earth could this be? These are clearly different factorizations.-A unit is defined as a number that has a multiplicative inverse.In Z[i],......
I'm trying to gain some conceptual understanding here. I am enrolled in a class called "History of Math" and the instructor is asking us to write a paper related to either Abstract Algebra or Analysis.

(1 + sqrt(-5))(1 - sqrt(-5)) = 6 = 2*3 so 2 and 3 are both primes because if that were not the case when ab = 2 and cq = 3. Taking the norms, Norm(a)Norm(b) = 4, but there exists no integer solutions to the equation x^2 + 5y^2 = 2. The same goes for 3. My understanding is that the early mathematicians all knew that complex numbers had to be used to justify whether a ring was a UFD or not.

But in Z[i], I find out that 5 = (2 + i)(2 - i) = (1 + 2i)(1 - 2i), however its been stated almost everywhere that Z[i] is a UFD. How on earth could this be? These are clearly different factorizations.

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A unit is defined as a number that has a multiplicative inverse.

In Z[i], the units are 1, -1, i, and -i.

Two elements in a ring are called associates if one is equal to the other times a unit.

abc = def are considered to be the same factorization if a and d, b and e, and c and f are all associates of each other. You should be able to figure out how this statements generalizes to factorizations with other than three factors.

In your example, 2 + i, and 1 - 2i are associates since 2 + i = i(1 - 2i)
and 2 - i and 1 + 2i are associates since 2 - i = -i(1 + 2i).
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