Quadratic integers

Question

Let consider the integer of $\mathbb{Q}(\sqrt{d})$, where $d$ is square free, i.e. if $d=4k+3$ or $d=4k+2$ it is $\mathcal{O}=\mathbb{Z}+\mathbb{Z}\sqrt{d}$. If $d=4k+1$, it is $\mathcal{O}=\mathbb{Z}+\mathbb{Z}\frac{1+\sqrt{d}}{2}$.

If $\mathcal{P}$ is a non-zero prime ideal of $\mathcal{O}$, then we have $\mathcal{P}\cap\mathbb{Z}=p\mathbb{Z}$ for some prime number $p$. What I want to prove is $\mathcal{O}/\mathcal{P}$ is a field, actually an at-most-two-degree extension of $\mathbb{F}_p$.

If we pick $[x]\neq[0]\in\mathcal{O}/\mathcal{P}$, we need to show $x\notin \mathcal{P}$, there exists $y\in \mathcal{O}$, such that $[yx]=1 \text{ mod } p$.

Further, if we consider the norm of $x=a+b\sqrt{d}$, how can we deduce $n(x)=\bar{x}x$ and $p$ are coprime? If we can deduce this property then $y:=c(a-b\sqrt{d})$ can be an inverse of $x$.

Answer 1

Here is a direct proof. As you noticed, for any non null prime ideal $\mathcal P$ of the ring of integers $\mathcal O$, we have $\mathcal O \cap \mathbf Z=p\mathbf Z $ for a certain rational prime $p$. Considering $\mathbf Z/p\mathbf Z$ as a subring of $\mathcal O/\mathcal P$ and using $\mathcal O=\mathbf Z[\alpha]$, we see that $\mathcal O/\mathcal P$ is an $\mathbf F_p$-vector space of dimension at most 2, hence is finite. But $\mathcal O/\mathcal P$ is a domain since $\mathcal P$ is prime, and a finite domain is necessarily a field.

Answer 2

Let $a$ be the algebraic integer generating $K=\Bbb Q(\sqrt d)$, so $a$ is either $\sqrt d$, or $(1+\sqrt d)/2$. And let $f$ be its minimal polynomial, seen as a polynomial of degree two in in $\Bbb Z[X]$. Then $$ \begin{aligned} K &= \Bbb Q(a)\ ,\\ \mathcal O_K &= \Bbb Z[a]=\Bbb Z[X]/f\ ,\\ \mathcal O_K/\mathcal P &= \Bbb Z/p=\Bbb F_p\ ,\\ &\text{ for a suitable unique prime $p$, for $\mathcal O_K/\mathcal P$ is a field, for $\mathcal P$ is prime,} \\ \mathcal O_K/p &= \Bbb Z[a]/p\\ &= (\Bbb Z[X]/f)/p\\ &= \Bbb Z[X]/(f,p)\\ &= (\Bbb Z[X]/p)/f\\ &= (\Bbb Z/p)[X]/f\\ &= \Bbb F_p[X]/f\ . \end{aligned} $$ Now i try to answer the posted question in a discussion. Since $\mathcal P$ is prime, the quotient ring is a field, and we can stop here. But the question wants to find the inverse of an element, say $ma+n$, $m,n\in \Bbb Z$, as explicit as possible.

Examples may make the situation transparent. We work in $\Bbb Q(\sqrt{ 2018})$, so $a= \sqrt {2018}$. Some primes in $\Bbb Z$ can be splitted further, some cannot be further splitted. For instance, using sage:

sage: K.<a> = QuadraticField(2018)
sage: ClK = K.class_group()
sage: ClK
Class group of order 2 with structure C2 of Number Field in a 
with defining polynomial x^2 - 2018
sage: K.class_number()
2
sage: for p in primes(40):
....:     print ( "Is p = %2s prime? %5s :: Is 2018 a square mod p? %s"
....:             % ( p, K(p).is_prime(), Zmod(p)(2018).is_square() ) )
....:     
Is p =  2 prime? False :: Is 2018 a square mod p? True
Is p =  3 prime?  True :: Is 2018 a square mod p? False
Is p =  5 prime?  True :: Is 2018 a square mod p? False
Is p =  7 prime? False :: Is 2018 a square mod p? True
Is p = 11 prime? False :: Is 2018 a square mod p? True
Is p = 13 prime? False :: Is 2018 a square mod p? True
Is p = 17 prime?  True :: Is 2018 a square mod p? False
Is p = 19 prime? False :: Is 2018 a square mod p? True
Is p = 23 prime?  True :: Is 2018 a square mod p? False
Is p = 29 prime?  True :: Is 2018 a square mod p? False
Is p = 31 prime?  True :: Is 2018 a square mod p? False
Is p = 37 prime?  True :: Is 2018 a square mod p? False

The pattern is transparent. (We eliminate $2$ and $1009$ from the next argument.) If $2018$ is a square mod $p$, then $p$ is not a prime in $\mathcal O_K$, and conversely.

• Consider for instance first $p=5$. Then with the above, it is a prime, since $\mathcal O_K/5\cong \Bbb F_5[X]/(X^2-2018)$, and the moded out polynomial has no roots in $\Bbb F_5$, so it is irreducible, so we get a field, so $5$ is a prime in $\Bbb O_K$. How to get the inverse of an element $\xi$ in this field? As the OP suggests, we build the conjugate $\bar\xi$ and have to show that we do not get the product zero in $\xi\,\bar\xi$. By other means, since this is already clear. So let us start with $m+na$, build $(m+na)(m-na)=m^2-n^2\cdot 2018$. If this is zero, then projecting the relation modulo $5$ we get that either $n=0$ mod five, with a quick end, or else $m/n$ is a square root of $2018$ mod five, contradiction.

• Consider further $p=7$. Then with the above sage computation, it is a not prime, humanly now, since $\mathcal O_K/7\cong \Bbb F_7[X]/(X^2-2018)=\Bbb F_7[X]/(X^2-9)\cong \Bbb F_7[X]/(X-3)\times \Bbb F_7[X]/(X+3)$, and we can also factor:

  sage: K.<a> = QuadraticField(2018)
  sage: K(7).factor()
  (-1) * (a - 45) * (a + 45)
  

so $7$ is not irreducible. The primes over $7$ are principal ideals, generated by $45\pm a$ each.

How to find an inverse in the case $\mathcal P=(45-a)$? (The other case is similar, by conjugation.) We choose a $\xi=m+na$ which is not zero mod $(45-a)$. Then modulo $\mathcal P=(45-a)$ we can replace the $a$ in the component $na$ by $45n$. And have to get the inverse of the integer number $m-45n$, if it is not zero. We do not need to pass to the conjugate.

• Consider finally $p=11$. Then with the above sage computation, it is an irreducible, but the ideal generated by $11$ is not prime. Using sage we can factor,

  sage: K.ideal(11).factor()
  (Fractional ideal (11, a + 4)) * (Fractional ideal (11, a + 7))
  

and both ideals over $11$ in $\mathcal O_K$ contain an element of the shape $a\pm 4$, again we replace $a$ in an element $m+na$ modulo $\mathcal P$ by the corresponding $\pm 4$, decide quickly if it is zero, if not, we simply build the inverse in the residue field.