More norms of ideals

Dear Professor Kim,

I am unsure of how to calculate the norm of (2,2\sqrt{15}) in the ring Q(\sqrt{15}), which is on Sheet 5 Question 4a.

I can see that this ideal can be written as (2) so it will have norm 4. Also, in the ring Z[\sqrt{15}] a general element looks like n + m\sqrt{15}, where n,m belong to $ latex Z$. So if we calculate the norm using the principle

|(2)| = |Z[x]/(2)| = |Z[x]/(2(n + m\sqrt{15}))| = 4.

From what I understand this is the reasoning you give in ‘Some remarks on factorization’.

However, if we use the method which given further down that sheet I get:

Z[\sqrt{15}]/(2) = Z[x]/((x^2 - 15),2) = F_2[x]/(x^2 - 15)
= F_2[x]/(x^2 - 1) = F_2[x]/(x+1)(x-1) = F_2/(2)

|F_2/(2)| = 2

Please tell me where I’m going wrong.

Many Thanks!
—————————————————-

Reply:

first of all, I hope you can see that the line

|(2)| = |Z[x]/(2)| = |Z[x]/(2(n + m\sqrt{15}))| = 4

above doesn’t make too much sense. The second displayed equation is almost right, except an error occurs when computing

|F_2[x]/(x+1)(x-1)|.

Because the coefficients are in F_2, we have x-1=x+1. So

F_2[x]/(x+1)(x-1)=F_2[x]/(x-1)^2\simeq F_2[t]/(t^2),

from the isomorphism F_2[x]\simeq F_2[t] that takes x to t+1. It is easy to see that the F_2-vector space F_2[t]/(t^2) has dimension 2 with basis 1, t. Hence,

|F_2[x]/(x+1)(x-1)|=|F_2[t]/(t^2) | =4.

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