I am currently working on a problem from the book “Introductory Algebraic Number Theory” by Kenneth S. Williams and Saban Alaca, and I would like to verify my solution.
The problem is:
Factor $<6>$ into prime ideals in the ring $O_K$ of $K=Q(\sqrt{366})$.
Here’s my approach:
Square-Free Check:
I first noted that $366$ is square-free as it can be factored as $366=2⋅3⋅61$
Quadratic Field:
Given that $366$ is square-free, it follows that $Q(\sqrt{366})$ is a quadratic field
Ideal Factorization:
The ideal $<6>$ can be factored as $<6>=<2>.<3>$
Prime Ideal Factorization for $p=2$:
For $p=2$ and $m=366$
, we have $366 \equiv 2 \pmod{4}$
By applying Theorem $10.2.1$ from the book, we find that $<2> =P^2$, Therefore, $<2>=<2,\sqrt{366} >^2$
Prime Ideal Factorization for $p=3$:
For $p=3$
and $m=366$, since $p$ is greater than $2$ and $p$ divides $m$
, the same theorem gives $<3>=P^2$, Therefore, $<3>=<3,\sqrt{366}>^2$
Based on these calculations, it seems reasonable to conclude that $<6>=<2,\sqrt{366}>^2
. <3,\sqrt{366}>^2$
Is this correct?
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$\begingroup$ Please use LaTeX properly: your \sqrt366 comes out like $\sqrt{3}66$, which looks terrible. Put the entire number in braces, as \sqrt{366}. Then you get $\sqrt{366}$, which is what you want. $\endgroup$– KCdCommented Jun 6 at 3:20
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$\begingroup$ Your answer is correct, but since I don't have the book, I have no idea what Theorem 10.2.1 states. Reducing $X^2 - 366$ mod $2$ and $3$ will tell you the factorization of these ideals by Dedekind-Kummer (assuming you know that the ring of integers is $\mathbb{Z}[\sqrt{366}]$). $\endgroup$– Viktor VaughnCommented Jun 6 at 4:38
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$\begingroup$ @KCd got it , thanks $\endgroup$– Aseel .ACommented Jun 6 at 5:09
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$\begingroup$ @ViktorVaughn Thank you so much, I do... $\endgroup$– Aseel .ACommented Jun 6 at 5:10
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