# Ideals in Z[√-5]

I'm doing part iii) here: So far for p2 = <2> I have:

Show p2 is containd in <2>

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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Hurkyl
Staff Emeritus
Gold Member
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>.

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.