How do I show that Z[sqrt(5)] is an integral domain
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How do I show that Z[sqrt(5)] is an integral domain

[From: ] [author: ] [Date: 11-04-24] [Hit: ]
(Being an integral domain is hereditary.As for the second question, this is a much more interesting question.If a + b√5 is a unit in Z[√5],(a + b√5)(c + d√5) = 1.(a - b√5)(c - d√5) = 1.......
Given a + b√5, c + d√5 in Z[√5] (for some integers a, b, c, d):

Suppose that (a + b√5)(c + d√5) = 0.
(We want to show that a + b√5 = 0 or c + d√5 = 0.)

The painless way to prove this is to simply observe that Z[√5] is a subring of R, which is an integral domain itself. (Being an integral domain is hereditary.)
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As for the second question, this is a much more interesting question.

If a + b√5 is a unit in Z[√5], then there exists c + d√5 in Z[√5] such that
(a + b√5)(c + d√5) = 1.

Taking conjugates of both sides:
(a - b√5)(c - d√5) = 1.

Now, multiply the last two equations together:
(a^2 - 5b^2)(c^2 - 5d^2) = 1.

Since this is an equation in Z, this has integer solutions iff
a^2 - 5b^2 = ±1 = c^2 - 5d^2.

In particular, we want integer solutions to a^2 - 5b^2 = ±1.

This is an instance of Pell's Equation, which is known to have infinitely many integer solutions.
Besides (a,b) = (±1, 0), which corresponds to ±1, here are a few nontrivial solutions
(±2, 1) and (±2, -1) <==> ± 2 + √5 and ± 2 - √5
(±9, 4) and (±9, -4) <==> ± 9 + 4√5 and ± 9 - 4√5.

Link:
http://en.wikipedia.org/wiki/Pell%27s_eq…

I hope this helps!
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