Principal generators of prime ideals is $\mathbb{Q}[\sqrt{m}]$ for m=-1, -2, -3

algebraic-number-theoryidealsnumber theoryprime factorization

I'm reading Marcus "Number fields" book and at a certain point (page 52) in the chapter about prime decomposition he writes

We now consider in detail the way in which primes p $\in \mathbb{Z}$ split in quadratic fields.

Let $R=A \cap \mathbb{Q}[\sqrt{m}]$, m squarefree.

Recall that R has integral basis $\{1,
\sqrt{m}\}$ and
discriminant 4m when $m\equiv 2\; or\; 3\; (mod\; 4)$, and integral basis $\{1,\frac{1+\sqrt{m}}{2}\}$ and
discriminant m when $m\equiv 1\; (mod\; 4)$.

Let p be a prime in $\mathbb{Z}$. Theorem 21 shows that there are just three possibilities:
$$ pR=\begin{cases}
P^2&\Leftarrow f(P|p)=1\\
P&\Leftarrow f(P|p)=2\\
P_1P_2 &\Leftarrow f(P_1|p)=f(P_2|p)=1.
\end{cases}$$

Theorem 25 With notation as above, we have:

If p | m, then
$$ pR=(p,\sqrt
{m})^2.$$

If m is odd, then
$$ 2R= \begin{cases} (2,1+\sqrt
{m})^2&\text{if $m\equiv 3\pmod4$}\\
\left(2,\frac{1+\sqrt{m}}{2}\right)\left(2,\frac{1-\sqrt{m}}{2}\right) &
\text{if $m\equiv 1\pmod8$}\\
\text{prime if $m\equiv 5\pmod8$.}
\end{cases}$$

If p is odd, $p\not| m$ then
$$ pR=\begin{cases} (p,n+\sqrt{m})(p,n-\sqrt{m})\; \text{if $m\equiv n^2 \pmod p$}\\
\text{prime if $m$ is not a square mod $p$}
\end{cases}$$
where in all relevant cases the factors are distinct.

Proof. I'll skip this.

The prime ideals involved in these factorizations do not look like principal ideals,
but we know in certain cases they must be principal: for example when m = −1, -2
or −3 (exercises 7 and 14, chapter 1). Can you describe principal generators for the
various prime ideals in these two cases?

Now my problem is that I don't understand what it means in the last question, how do I find the principal ideals, and whose prime ideals is he referring to.

Any help in understanding both the question and how to solve it would be welcomed.

Best Answer

-1)

  • $p|m\Rightarrow p=\pm 1$ but then trivially $(\pm1,\sqrt{-1})=(1)=R$;
  • $-1$ is odd and $-1\equiv 3\; (mod\; 4)$ so we must check the second case and in it the first case $$ (2,1+\sqrt{-1})$$ here we suffice to notice that $$ (1+i)(1-i)=2$$ and so $$ (2,1+\sqrt{-1})=(1+\sqrt{-1})$$.
  • In this case any odd p works and both cases are possible since $4^2=16=17-1$ but for examples the squares mod 3 are 1 and 0 whislt $-1\equiv 2\; (mod\; 3)$ .\ Now we are working in an Euclidean domain so we have that, for this we suffice to be in a UFD since this is enoguh for the gcd to exist, $(p,n\pm i)\subset (gcd(p,n\pm i))$ but in a Euclidean domain we have the Euclidean alogrithm and so there is a combination of any two elements whose result is the gcd of them implying $$(p,n\pm i)= (gcd(p,n\pm i)).$$ The last case is trivial since $pR=(p).$

-2)

  • $p|m\Rightarrow p=\pm 1\wedge \pm 2$ but then $(\pm1,\sqrt{-1})=(1)=R$$$(\pm 2,\sqrt{-2})=(\sqrt{-2}) $$ an d actually this is true in all cases where m is a $\mathbb{Z}$ prime;

  • 2 is even so we can skip the second case;

  • In this case any odd p works and both cases are possible since for example $1^2+2\equiv0\; (mod\; 3)$ but we also have that squares mod 5 are 4,1 and 0 whilst $-2\equiv 3\; (mod\; 5)$ .\ Now we are working in an Euclidean domain so we have that, for this we suffice to be in a UFD since this is enoguh for the gcd to exist, $(p,n\pm \sqrt{-2})\subset (gcd(p,n\pm \sqrt{-2}))$ but in a Euclidean domain we have the Euclidean alogrithm and so there is a combination of any two elements whose result is the gcd of them implying $$(p,n\pm \sqrt{-2})= (gcd(p,n\pm \sqrt{-2})).$$ The last case is trivial since $pR=(p).$

-3)

  • For what we said in the previous case p=$\pm 1\Rightarrow (1,\sqrt{-3})=R$ and $p\neq \pm 1\Rightarrow (p,\sqrt{-3})=(\sqrt{-3})$ ;

  • -3 is odd and $-3\equiv 5\; (mod\; 8)$ so we must check the second case and in it the third case that is trivial since $2R=(2);$

  • In this case any odd p but 3 works and both cases are possible since $2^2=4\equiv-3\; (mod\; 7)$ but for examples the squares mod 5 are 4,1 and 0 whilst $-3\equiv 2\; (mod\; 5)$ .\ Now we are working in an Euclidean domain so we have that, for this we suffice to be in a UFD since this is enoguh for the gcd to exist, $(p,n\pm \sqrt{-3})\subset (gcd(p,n\pm \sqrt{-3}))$ but in a Euclidean domain we have the Euclidean alogrithm and so there is a combination of any two elements whose result is the gcd of them implying $$(p,n\pm \sqrt{-3})= (gcd(p,n\pm \sqrt{-3})).$$ The last case is trivial since $pR=(p).$

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