1. Classify the Primes in
Ingredient: the norm map
√
Z[ 2]
√
N (a + b 2) = a2 − 2b2
and maybe some frobenius
reciprocity which I'll explain later.
•
Suppose that
√
x ∈Z[ 2]
were a prime element. That is to say, when-
x|ab it implies that
√ either x|a or x|b
√.
√
• Then N (x) = N (x1 + 2x2 ) = ±(x1 + 2x2 )(x1 − 2x2 ) = p1 p2 ...pn where
the pi form the unique factorization of N (x) in Z. (They aren't necessarily
ever
distinct)
•
x1 +
√
p1 p2 ...p
√ n , it must
z ∈ Z[ 2]
• Therefore, taking the norm of that relationship N (x · z) = N (pi ) we have
2
2
that N (x)N (z) = pi . And therefore p1 p2 ...pn N (z) = pi in Z.
• By unique factorization in Z we have that N (x) is either pi or p2i where
pi is a prime in Z
√
is a necessary condition for x to be a prime element in Z[ 2]. Now we want
Since
2x2
divide one of the
This
is prime, and it divides the product
pi .
In other words
x · z = pi
for some
to nd sucient conditions
•
Suppose
√
Z[ 2]
N (a +
√
2b) = p
where
p
is a prime in
Z
then
a+
√
2b
is prime in
most of you showed this.
√
2b) = p2 . This breaks into two cases.
√
Case 1. Suppose p is a prime in Z[ 2] and in Z.
√
√
• Then since p2√= ±(a + 2b)(a − 2b), since we're in a unique factorization
domain, a +
2b is an associate to p.
√
Case 2. Suppose that p isn't prime in Z[ 2]
√
• Suppose
√ (by2 contradiction)
√ that x = a +√ 2b is a prime element. Then,
(a + 2b)|p =⇒ (a + 2b)|p =⇒ (a + √
2b) · z1 = p.
√
• We could do a similar thing with x̄ = a − 2b so that (a − 2b) · z2 = p.
(you have to use the easy fact that x is prime ⇐⇒ x̄ is prime )
√
√
2
• Combining the above two points we have
that p = ±z1 z2 (a− 2b)(a+ 2b)
√
√
2
but, on the other hand, p = ±(a −
2b)(a + 2b) so z1 and z2 are units.
√
• Therefore x is associate to p which we supposed was not a prime in Z[ 2]
so x is not prime.
2
Therefore, a sucient condition for x to be prime is either N (x) = p or N (x) = p
√
where p is a prime in Z and Z[ 2].
√
In summary, x is a prime in Z[ 2] if and only if one of the following is true
• N (x) = p where p is a prime in Z
√
• N (x) = p2 where p is a prime in Z that remains prime in Z[ 2].
√
All that remains is to nd the primes p which are primes in both Z and Z[ 2].
√
In other words, which primes in Z remain prime when we extend to Z[ 2]. You
can do this with quadratic reciprocity. The basic theorem tells you that p is NOT
prime if and only if p ≡ ±1 mod 8.
•
Now suppose that
N (a +
2. Finding non-trivial solutions to
Lemma.
a2 − db2 = 1
Let d be a square free integer, then for innitely many pairs of integerss
PN , QN
1
2
√
d − PN < 1
QN
Q2N
Most of you showed this easily.
(yay!).
Now, take the relationship
PN ).
√
√ P d− N <
QN
1
and multiply both sides by
Q2N
√
QN ( dQN +
We get that
Where the second
d−
The proof involves the pigeonhole principal
√
√
2
dQN − PN2 < d + PN < 2 d + 1
QN
√
< follows from that fact that d −
PN
QN .
Now, there are only nitely many integers less than
(QN, PN )
such that
√
2
dQ − P 2 < 2 d + 1.
N
N
PN
QN
√
2 d + 1,
< 1 =⇒ −1 <
but innitely pairs
So in order to t an innite number
of pairs into a nite number of boxes, an innite number of pairs have to go to the
same thing, say
n,
which, remember is less than
now, lets restrict
So
dQ2 − P 2 = n.
N
N
√
2 d + 1.
ourselves to the innite set of pairs
(QN, PN )
such that
Of this innite set of pairs, there must be an innite subset such that
mod n.
but only a nite number of congruences
mod
n.
Okay, so now restrict to the innite set of pairs
congruent to eachother
mod
n.
Now we have an innite set of pairs
(2)
(3)
2
dQ − P 2 = n
N
N
Qi ≡ Qj mod n
Pi ≡ Pj mod n.
where all the
QN
are
Pi ≡ Pj (mod
(QN, PN ) satisfying
n).
√
√
A = Pi +√Qi d and B = Pj + Qj d. Now theres just
A
A
that
B ∈ Z[ d and that N ( B ) = 1. I'll leave that to you.
For any two these pairs, let
a bit of algebra to show
(QN, PN )
We can do the same thing again, and of these
pairs nd an innite subset where, in addition
(1)
Qi ≡ Qj
QN
This is again the pigeonhole principal. There are an innite number of