Show p2 is Contained in <2> in Z[√-5]

  • Thread starter Firepanda
  • Start date
In summary, the conversation is about proving that p2 is contained in <2>. The participants discuss the form of p2 elements and how they can be contained in <2>. They also mention using generators and a basis to make the proof easier. One participant asks for a general rule on how to multiply certain elements, hoping it will help with the other proofs.
  • #1
Firepanda
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I'm doing part iii) here:

4uwjdz.png


So far for p2 = <2> I have:

Show p2 is containd in <2>

Since [we have already shown]

p={r + s√-5 | r = s mod 2}

Then p2 elements are of the form (r+s√-5)2

= (r2 -5s2) + 2rs√-5

Since r&s have the same sign, then (r2 -5s2) is always even, & 2rs is always even too

=> p2 elements are contained in <2>

Now, show <2> is contained in p2

=> show 2(a+b√-5) is an element of p2

so show 2a + 2b√-5 is an element of p2

From the above we know elements of p2 take the form of 2x + 2y√-5

So <2> is contained in p2

Am i done for this part?

Thanks
 
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  • #2
Firepanda said:
p={r + s√-5 | r = s mod 2}

Then p2 elements are of the form (r+s√-5)2
Actually, the elements of p2 are sums of elements of that form.


From the above we know elements of p2 take the form of 2x + 2y√-5
... but you haven't yet shown everything of that form is an element of p2.



As a general bit of advice, I find it a lot easier to work with generators rather than the entire ideal, or sometimes a basis for the ideal. To show p2 is contained in <2>, all you need to do is show that the generators of p2 are contained in <2>. Or to show that p2 = <2>, it's enough to show that some basis for p2 is also a basis for <2>.
 
  • #3
Hurkyl said:
Actually, the elements of p2 are sums of elements of that form.



... but you haven't yet shown everything of that form is an element of p2.



As a general bit of advice, I find it a lot easier to work with generators rather than the entire ideal, or sometimes a basis for the ideal. To show p2 is contained in <2>, all you need to do is show that the generators of p2 are contained in <2>. Or to show that p2 = <2>, it's enough to show that some basis for p2 is also a basis for <2>.

Ah ok, I'm not really sure how to work with the generators though.

I assume you mean something like

<2 , 1+ rt(-5)> = <4, 4(1+rt(-5)), 4> =<2>

Which I know its probably wrong but you get the idea. Can you give me a general rule on how to multiply together

<a, b + rt(c)><d, e + rt(c)>

please? It might help me do the others too.
 

Related to Show p2 is Contained in <2> in Z[√-5]

1. What is p2?

In this context, p2 refers to the prime number 2.

2. What does it mean for something to be contained in <2>?

In mathematical notation, <2> represents the ideal generated by the number 2. So, to say that p2 is contained in <2> means that the ideal generated by 2 contains the prime number 2.

3. What is Z[√-5]?

Z[√-5] is the set of all complex numbers of the form a + b√-5, where a and b are integers. It is also known as the ring of Gaussian integers.

4. How can we show that p2 is contained in <2> in Z[√-5]?

This can be shown using the definition of containment for ideals. We need to prove that for any element in p2, it can be expressed as a product of elements in <2>. In this case, it is enough to show that 2 is a multiple of p2, which is true since 2 is a prime number.

5. What implications does this have in the study of Z[√-5]?

This result is significant because it shows that p2 is a principal ideal in Z[√-5], meaning that it can be generated by a single element. This simplifies the study of this ring and its properties.

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