## 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!
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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,

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