(adsbygoogle = window.adsbygoogle || []).push({}); Prove (3, 2 + √−5)_{R}is not principal in Z[√−5].

Solution, where N = norm:

3 = m(a+b√−5) => N(a+b√−5) | N(3) = 9

and

2+√−5 = n(a+b√−5) => N(a+b√−5) | N(2+√−5) = 9

n,m in R

So,

N(a+b√−5) = a^{2}+5b^{2}= 1, 3 or 9

3 cannot be a solution clearly.

If N(a+b√−5) = 9 then => a=+-3 b=0 , or a=+-2 b=+-1

Take ideal I = (3)_{R}

=> 2+√−5 = 3r, r is in R

=> N(2+√−5) = 9.N(r)

=> 9 = 9.N(r)

so r is a unit, so all of r must be units in R => R is a field => only ideals in R are {0_{R}} and R which isn't (3)_{R}.

For a=2 b=1

=> (2+√−5)r = 3, r is in R, so by same arguement above this cannot be an ideal.

For N(a+b√−5) = 1

We have a=+-1 b = 0

Take I = (1)_{R}= R

=> 1 = 3m + (2+√−5)n

=> (2-√−5) = 3(2-√−5)m + 9n

=> 3 | (2-√−5)

=> 3k = 2-√−5

=> 9.N(k) = 9 => k unit and by above can be an ideal in R. But R is clearly not a field so this cannot be an ideal?

How do I show this is not principal?

Is the rest of this ok?

Wrong techniques?

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# Proving an ideal is not principal, have I gone wrong?

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