Problem: Want to determine whether 6-sqrt(14) is irreducible over
Z[sqrt(14)], the ring of integer joined by square root of 14.
Attempt: I tried to set up the equation (a+b*sqrt(14)) * (c
+d*sqrt(14)) = 6-sqrt(14). And try to solve for a,b,c,d in Z. However
this method gets too complicated and there are more than 10 cases to
discuss. Is there a simpler way of doing this? Thanks in advance.
Hint: Consider norms.
Let
z = 6 - sqrt(14)
x = a + b*sqrt(14)
y = c + d*sqrt(14)
Then x*y = z => N(x)*N(y) = N(z)
quasi
Thanks!
it's not irreducible sine $ (5+\sqrt{14})(4-\sqrt{14})=6-\sqrt{14} $
and neither $ 5+\sqrt{14} $ nor $ 4-\sqrt{14} $ is a unit.
Right -- I saw that too (but wanted to let the OP play).
Finding that factorization is easy by trial and error.
But to bound the search in advance is a little tricky.
Of course, if the absolute value of the norm was prime, irreducibility
would be automatic, but in this case, the norm is 22.
Assuming reducibility, one of the factors must have norm 2 or -2.
A norm of -2 is impossible since the equation a^2 - 14b^2 = -2 has no
solutions mod 7.
A norm of 2 _can_ be achieved (and the search for it can be bounded in
advance).
The other factor would then have to have a norm of 11, and that norm
can also be achieved (and once again, the search can be bounded in
advance).
Once examples of both are found, their product will of course have
norm 22.
Now if the ring R=Z[sqrt(14)] is a UFD (I haven't checked), then
unique factorization of 22 in R forces 6 - sqrt(14) to be divisible by
both of the factors found above (the ones with norms 2 and 11,
respectively). To avoid having the product miss by a unit factor,
simply divide 6 - sqrt(14) by either one of those factors to get a
valid cofactor.
Thus, what I am trying to get at is this ...
In the case where the norm is composite, searching for the factors can
be problematic, especially if it turns out that the given number is
actually irreducible.
Here are two followup questions ...
Let d be a squarefree positive integer, with d > 1, and let R be the
ring of integers of the real quadratic number field Q(sqrt(d)).
Let z in R be such that N(z) is composite.
Question (1): Must z be reducible in R?
Question (2): If there exist nonunits x,y in R such that N(x)N(y) =
N(z), must z be reducible in R?
Remarks:
Obviously, if the answer to question (1) is yes, then the answer to
question (2) is also yes.
Also, as discussed above, if R is a UFD, then for that R, the answer
to question (2) is yes.
Thus, an example, if any, showing that the answer to question (2) is
no, will require that R is a non-UFD.
quasi
It would not be surprising if the answer to Question (1) were always
no. Certainly it is sometimes no. For example, look at 5 +
sqrt(15). The norm is 10, but x^2 - 15y^2 cannot be equal to 2 or -2.
No, e.g. inert primes p have N(p) = p^2
A number ring is a PID iff its irreducibles have prime power norms,
i.e. exactly one prime p occurs in norms of an irreducible element
N(q) = p^n, p,q atoms in resp. rings. The proof is very easy.
> Question (2): If there exist nonunits x,y in R such that
> N(x) N(y) = N(z), must z be reducible in R?
No, e.g. x = y = 3, z = 5 + 2 sqrt(-14).
Bumby and Dade [2] classified those quadratic number fields K where
reducibility depends only on the norm. Denoting the class group by H,
they proved that K satisfies this property iff
(a) H has exponent 2 or
(b) H is odd or
(c) K is real with positive fundamental unit and the 2-Sylow subgroup
of the narrow class group is cyclic.
Note the above example Q(/-14) has class group K = C(4) with exponent 4.
See Coykendall's paper [1] for a recent discussion of related results
and generalizations; see also the math reviews below.
--Bill Dubuque
[1] Jim Coykendall. Properties of the normset relating to the class group.
http://www.ams.org/proc/1996-124-12/S0002-9939-96-03387-4
http://www.math.ndsu.nodak.edu/faculty/coykenda/paper3.pdf
------------------------------------------------------------------------------
[2] 35 #4186 10.65 (12.00)
Bumby, R. T. Irreducible integers in Galois extensions.
Pacific J. Math. 22 1967 221--229.
------------------------------------------------------------------------------
This paper continues a study, initiated in a joint paper by the author and
E. C. Dade [same J. 22 (1967), 15--18; MR 35#2857], of the following property.
An extension K/k of number fields is said to satisfy Property N if whenever
x and y are two integers of K with the same relative norm and x is
irreducible so is y . This condition is analyzed by homological and
combinatorial techniques. It is shown that this property implies very severe
restrictions on the structure of the ideal class group of K as G-module, but
a complete classification is given only for absolutely quadratic number fields.
Reviewed by A. Brumer
------------------------------------------------------------------------------
35 #2857 10.66
Bumby, R. T.; Dade, E. C. Remark on a problem of Niven and Zuckerman.
Pacific J. Math. 22 1967 15-18.
------------------------------------------------------------------------------
Let H be the ideal class group of a quadratic field K . Two nonzero
fractional ideals I,J are said to be strongly equivalent if I J^{-1} = (b)
is a principal ideal generated by an element \g of positive norm. The
strong equivalence classes form a group H' called the extended class group.
Say that K has property N if for every pair of integers of equal norm in K the
irreducibility (i.e., no proper divisors) of one implies the irreducibility of
the other. It is proved that K has property N if and only if
(i) H has exponent 2, or
(ii) H has odd order, or
(iii) K is real, the norm of every unit in K is +1,
and the 2-Sylow subgroup of H' is cyclic.
Reviewed by I. Niven
------------------------------------------------------------------------------
89k:11103 11R27
Lettl, G.(A-GRAZ)
Characterization of irreducible algebraic integers by their norms.
Colloq. Math. 54 (1987), no. 2, 325--332.
------------------------------------------------------------------------------
The partial characterization of "Property N" given by the reviewer
[Pacific J. Math. 22 (1967), 221--229; MR 35 #4186] is extended to a
complete characterization of a related "Property N^*". Thus, it is now
known that the Galois extensions L/K of algebraic number fields for which
elements a and b with N(a) = N(b) are of only three types, determined
by the action of the Galois group, G , on the ideal class group, H , of L .
The three types are: (a) H ~ C_2 (+) C_2; (b) G acting trivially on any H;
(c) every element of G acting as +-1 on an H of odd order.
Reviewed by Richard T. Bumby
Zbl 0651.12002
A finite extension L of an algebraic number field K is said to have property
(N^*), if for two algebraic integers a,b in L with N(a) associated with N(b)
in K, a and b are either both irreducible or both not. The main result of
the paper under review is the following: A normal extension L|K with
Galois group G has property (N^*) if and only if one of the following holds:
(a) The class group of L written additively is isomorphic to Z/2Z (+) Z/2Z;
(b) G acts trivially on the class group of L;
(c) The class number of L is odd and there exists an algebraic number field
L_0, K <= L_0 <= L such that [L_0:K] = 2.
The author also shows that the property (N^*) depends on the G-module structure
of the class group of L written additively.
------------------------------------------------------------------------------
97b:11130 11R29 (11R11)
Coykendall, Jim(1-CRNL)
Properties of the normset relating to the class group.
Proc. Amer. Math. Soc. 124 (1996), no. 12, 3587--3593.
------------------------------------------------------------------------------
Let F/K be a Galois extension of number fields with rings of integers T
and R , and suppose that T is a principal ideal domain. The set of norms
S, from R to T, is said to be strongly saturated if whenever a,b in R
with N(a)/N(b) in T, there's a c dividing a in R for which N(c) = N(b).
When norms are identified up to multiplication by units, the resulting similar
property is called saturation, and strong saturation implies saturation. The
author studies the relation between saturation and the ideal class group of F .
He proves that if [F:K] = p_1^{e_1}...p_r^{e_r}, and S is saturated, then the
ideal class group of F is the product of its p_i Sylow subgroups, 1 <= i <= r,
each of which has exponents not exceeding p_i^{e_i} . He then proves that if
F is quadratic over K, then S is saturated if and only if the ideal class
group of S is elementary abelian of 2-power order, obtaining a converse in the
quadratic case. He also proves that S is saturated if and only if Gal(K/F)
acts trivially on the ideal class group. He then shows that saturation implies
a property studied by R. T. Bumby and E. C. Dade [Pacific J. Math. 22 (1967),
15-18; MR 35#2857] whereby irreducible integers are completely determined by
their norms.
Reviewed by Alan Candiotti
------------------------------------------------------------------------------
2001m:11191 11R27 (13B05)
Coykendall, Jim(1-NDS); Teymuroglu, Ayse A.
Divisor properties inherited by normsets of rings of integers.
Advances in commutative ring theory (Fez, 1997), 227--239,
Lecture Notes in Pure and Appl. Math., 205, Dekker, New York, 1999.
------------------------------------------------------------------------------
Let K <= F be algebraic number fields with respective rings of integers
T <= R, and let N be the norm map from R to T. This paper proves that
if F/K is Galois, then R is a unique factorization domain if and only if
the monoid of norms of principal ideals of R has unique factorization. As a
corollary, it further proves that R is a unique factorization domain if and
only if the set of norms from R to T has unique factorization.
The authors then study three saturation conditions on the set S of norms
from R to T . If a and b are elements of R and N(b) divides N(a)
in T, then S is strongly saturated if there must exist c in R for which
c divides a and N(c) = N(b); S is medium saturated if N(b) must divide
N(a) in S; and S is saturated if there must exist c in R such that
N(a) = N(b) N(c) u, where u is a unit in T. They prove that the first two
conditions are equivalent, but give an example of a quadratic number field
whose normset is saturated but not strictly saturated. Finally, they prove
that if the normset of a quadratic number field is saturated, then it is
strictly saturated if and only if either -1 is a norm or -n^{-2} is not
a norm for any n in Z.
Reviewed by Alan Candiotti
Thanks, Bill.
Those are some cool results.
I have lots to learn.
quasi
It's not a UFD.
http://journals.smc.math.ca/cgi-bin/vault/view/harper2205
showing that it is Euclidean. As a consequence it is a UFD.
i agree.