1. Let d € Z with √d ¢ Q. (a) Let z € Q[√d] and let r and s be the unique rational numbers such that z = r+s√d. Prove that z € Z[√d] if and only if r, s € Z. (Hint: this is not as tautological as it might seem, but should not require more than a couple of sentences.) (b) Prove that Q[√d] is a field. (Hint: The ring Q[√d] is nontrivial because it contains the nontrivial ring Z[√d] a as a subring. Show that, if z € Q[√d] is nonzero, then N¿(z) is nonzero as well, so that z' = σa(z)/Na(z) is a well-defined complex number. Prove that z' is actually an element of Q[√d)] and that zz' = 1.)
1. Let d € Z with √d ¢ Q. (a) Let z € Q[√d] and let r and s be the unique rational numbers such that z = r+s√d. Prove that z € Z[√d] if and only if r, s € Z. (Hint: this is not as tautological as it might seem, but should not require more than a couple of sentences.) (b) Prove that Q[√d] is a field. (Hint: The ring Q[√d] is nontrivial because it contains the nontrivial ring Z[√d] a as a subring. Show that, if z € Q[√d] is nonzero, then N¿(z) is nonzero as well, so that z' = σa(z)/Na(z) is a well-defined complex number. Prove that z' is actually an element of Q[√d)] and that zz' = 1.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please help with 1a and 1b
![The field Q is the unique smallest field containing Z as a subring (it is an instance of a
general construction known as the field of fractions of an integral domain). If we replace Z
with a quadratic integer ring Z[√d] (where d € Z and √d & Q, as always), the role of Q is
played by the subset Q[√d] of the complex numbers defined by
z = r + s√d for some r, s € Q}.
Q[√d] = {z € C: z=r
The following facts can be verified using the same arguments employed for establishing the
analogous facts for Z[√d] (and you may take them for granted):
(i) Q[√d] is a subring of C containing Z[√d] as a subring;
(ii) if z € Q[√d], then there are unique rational numbers r and s such that z = r+s√d;
(iii) the conjugation function od on Z[√d] extends (uniquely) to a function on Q[√d] using
the same formula, i.e., σd(r + s√d) =r-s√d;
(iv) the conjugation function od on Q[√d] is multiplicative and additive, just as it is on
Z[√d];
(v) the norm function Na from Z[√d] to Z extends (uniquely) to a function from Q[√d]
to Q using the same formula, i.e., Na(z) = zoa(z), or, equivalently, if z = r + s√d,
then Na(z) = r² — s²d; and
(vi) the norm function Na : Q[√d] - → Q is multiplicative.
1. Let de Z with √d & Q.
(a) Let z € Q[√d] and let r and s be the unique rational numbers such that z =r+s√d.
Prove that z € Z[√d] if and only if r, s € Z. (Hint: this is not as tautological as it
might seem, but should not require more than a couple of sentences.)
(b) Prove that Q[√d] is a field. (Hint: The ring Q[√d] is nontrivial because it contains
the nontrivial ring Z[√d] as a subring. Show that, if z € Q[√d] is nonzero, then
Na(z) is nonzero as well, so that z' = oa(z)/Na(z) is a well-defined complex number.
Prove that z' is actually an element of Q[√d)] and that zz' = 1.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febb913e1-4986-4d74-b6ce-ad576ddf43d3%2F114ddae3-5cd2-43e9-8b2c-a75830c19f8b%2Fw2x53fn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The field Q is the unique smallest field containing Z as a subring (it is an instance of a
general construction known as the field of fractions of an integral domain). If we replace Z
with a quadratic integer ring Z[√d] (where d € Z and √d & Q, as always), the role of Q is
played by the subset Q[√d] of the complex numbers defined by
z = r + s√d for some r, s € Q}.
Q[√d] = {z € C: z=r
The following facts can be verified using the same arguments employed for establishing the
analogous facts for Z[√d] (and you may take them for granted):
(i) Q[√d] is a subring of C containing Z[√d] as a subring;
(ii) if z € Q[√d], then there are unique rational numbers r and s such that z = r+s√d;
(iii) the conjugation function od on Z[√d] extends (uniquely) to a function on Q[√d] using
the same formula, i.e., σd(r + s√d) =r-s√d;
(iv) the conjugation function od on Q[√d] is multiplicative and additive, just as it is on
Z[√d];
(v) the norm function Na from Z[√d] to Z extends (uniquely) to a function from Q[√d]
to Q using the same formula, i.e., Na(z) = zoa(z), or, equivalently, if z = r + s√d,
then Na(z) = r² — s²d; and
(vi) the norm function Na : Q[√d] - → Q is multiplicative.
1. Let de Z with √d & Q.
(a) Let z € Q[√d] and let r and s be the unique rational numbers such that z =r+s√d.
Prove that z € Z[√d] if and only if r, s € Z. (Hint: this is not as tautological as it
might seem, but should not require more than a couple of sentences.)
(b) Prove that Q[√d] is a field. (Hint: The ring Q[√d] is nontrivial because it contains
the nontrivial ring Z[√d] as a subring. Show that, if z € Q[√d] is nonzero, then
Na(z) is nonzero as well, so that z' = oa(z)/Na(z) is a well-defined complex number.
Prove that z' is actually an element of Q[√d)] and that zz' = 1.)
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