189-235A: Basic Algebra I Assignment 7

Due: Monday, November 4.

Section 6.1: 34, 40, 43, 44.

Section 6.2: 29, 31, 33.

8. Let Q(\sqrt{−5}) = \{a+b\sqrt{−5}, a, b ∈ Q\}, and Z[\sqrt{−5}] = \{a+b\sqrt{−5}, a, b ∈ Z\}.

a) Show that Q(\sqrt{−5}) is a field, and that Z[\sqrt{−5}] is a subring. It is called the ring of integers of Q(\sqrt{−5}) and plays the role of the usual integers in the arithmetic of Q(\sqrt{−5}).

b) Show that the invertible elements in Z[\sqrt{−5}] are exactly 1 and −1.

c) Show that the elements 2, 3, 1 + \sqrt{−5} and 1 − \sqrt{−5} are irreducible. (I.e., they cannot be written in the form ab where a, b ≠ ±1.)

d) Using c), show that the ring Z[\sqrt{−5}] is not a unique factorization ring. (I.e., the “integers” in Z[\sqrt{−5}] cannot be written uniquely as a product of irreducible elements.)

e) Show that the ideals (2, 1 + \sqrt{−5}), (3, 1 + \sqrt{−5}), and (3, 1 − \sqrt{−5}) are not principal, and that they are irreducible, i.e.,they cannot be factored further into products of non-trivial ideals.

f) If I and J are ideals, define the product IJ to be the ideal generated by the elements of the form ij with i ∈ I and j ∈ J. Show that (2, 1 + \sqrt{5})^{2} = (2), (3, 1+\sqrt{−5})(3, 1−\sqrt{−5}) = (3), and conclude that the ideal (6) factorizes as a product of 4 (non-principal) ideals: (6) = (2, 1+\sqrt{−5})^{2}(3, 1+\sqrt{−5})(3, 1− \sqrt{−5}). Remark: It can be shown that this factorization of the principal ideal (6) into a product of irreducible ideals is unique, up to the order of the factors. This is a general phenomenon: although the ring Z[\sqrt{−5}] fails to satisfy unique factorization, its ideals can be expressed uniquely as products of irreducible ideals. The introduction of ideals in the late 19-th century by Dedekind was an attempt to salvage unique factorization in such rings, by showing it was true on the level of ideals which were viewed as a kind of “ideal number”. This is where the terminology comes from...