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Showing elements in an extension field Z[sqrt(-5)] have gcd = 1

  • Thread starter tomelwood
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  • #1
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Homework Statement


Show that the elements 3 and 2+[tex]\sqrt{-5}[/tex] in [tex]\textbf{Z}[/tex][[tex]\sqrt{-5}[/tex]] have a greatest common divisor of 1, but the ideal I = (3, 2+[tex]\sqrt{-5}[/tex]) is not the total. Conclude that I is not principle.


Homework Equations





The Attempt at a Solution


I have got as far as attempting to show the first part. If I show that both 3 and 2+[tex]\sqrt{-5}[/tex] are both irreducible in [tex]\textbf{Z}[/tex][[tex]\sqrt{-5}[/tex]] then they are relatively prime and hence have gcd of 1?
Also, the norms of both the numbers are 9, so does this do anything at all?
The way I have argued that both 3 and 2+[tex]\sqrt{-5}[/tex] are irreducible is:
Let 3=ab with a and b in [tex]\textbf{Z}[/tex][[tex]\sqrt{-5}[/tex]].
Now |a|[tex]^{2}[/tex]*|b|[tex]^{2}[/tex]=9 which has only 1,3,9 as positive factors.
If |a|[tex]^{2}[/tex]=3 then a=x+y[tex]\sqrt{-5}[/tex] with x,y satisfying x[tex]^{2}[/tex]+5y[tex]^{2}[/tex]=3.
There are no such integers, so |a|[tex]^{2}[/tex]=/=3.
So either |a|=1 or |b| = 1. so either a or b is a unit, so 3 is irreducible.
Exactly the same for 2+[tex]\sqrt{-5}[/tex] since 4+5=9, then argue the same way.

But I don't really know what to do from here,
And how do I argue that the Ideal is not the total?

As for the last statement, is it true that if an ideal of two coprime numbers is not the whole ring, then the ideal is not principal?

Many thanks in advance.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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To prove that the ideal is not the total ring. Proceed by contradiction: assume that there exists [tex]a,b\in \mathbb{Z}{\sqrt{-5}}[/tex] such that

[tex]1=a.3+b.(2+\sqrt{-5})[/tex]

Now try to reach a contradiction...
 
  • #3
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OK.
So letting [tex]a=x+y\sqrt{-5}[/tex] and [tex]b=u+v\sqrt{-5}, x,y,u,v\in \textbf{Z}[/tex]
we can multiply out your expression to give
[tex]1=3x+3y\sqrt{-5}+2u+u\sqrt{-5}+2v\sqrt{-5}-5v[/tex]
Which can then be rearranged to give
[tex]x=\frac{(\sqrt{-5}(3y+u+2v)-1)}{3}[/tex]
which means that x is something multiplied by the square root of minus 5, so cannot be an integer, as assumed.
I presume similar arguments can be made for y,u,v, meaning that we have our required contradiction?
 
  • #4
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So the definition of an ideal being the total ring is that multiplying the elements of the ideal by arbitrary elements of the ring gives 1?
Why is this?
Also, does my argument show that the greatest common divisor is 1? As I have shown (have I?) that they are both irreducible and so are both prime. And two prime numbers have gcd 1.

And then to finish off, I can just say that since an ideal of two elements with gcd 1 is not the whole ring, then the ideal is not principal? Is this true? Or do I need to prove something else in the meantime? If so, what?

Many thanks.
 
  • #5
Tom! To prove that the gcd([tex]2+\sqrt{-5}[/tex])=1 you can use this:
divisors(3)=+-1,+-3. So you only have to try if those numbers divide [tex]2+\sqrt{-5}[/tex].
Of course, the don't. Therefore gcd=1.
 

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