== Definition 5.14. The function N: Z[i] \ {0} Z20. sti|s+ti|2 = s²+t² is called the norm on Z[i]. Proposition 5.15. The ring Z[i] is a Euclidean domain with valuation N, and hence also a PID. Proof. Let R = Z[i]. Note that N(a) ≥1 for all a € R\ {0}. So for a, b € R\ {0}, N(ab) = |ab|² = a²b² ≥ a2 N(a), and (ED1) is satisfied. Let a, b € R with b 0. Write a s+ti and b = u+vi with s, t, u, v € Z. We can form the quotient =1+mi Є C, where 1, m Є R. [To check (ED2), we want to use l+mi to define L+MI with L, M € Z as the element q in the expression a=qb+r we are looking for.] Let L, M Є Z be such that | L| ≤ ½ and m - M|≤ [such L, M can always be found]. Then =L+Mi + (1 L) + (m − M)i. Multiplying by b, we have Since a- done. a = (L+ Mi)b+ ((1 − L) + (m − M)i)b. - (L+ Mi)b Z[i], the term (1- L+ (m - M)i)b is in Z[i], too. If it is 0, we are So assume (L+(mM)i)b 0 (sol- L+ (mM)i 0). We have: N((I – L) + (m − M)i)b) = - |l-L+(m − M)i|³|b|ª S (+) ((글)2- + (-1)³) 16/2 = 16/2 N(b)
== Definition 5.14. The function N: Z[i] \ {0} Z20. sti|s+ti|2 = s²+t² is called the norm on Z[i]. Proposition 5.15. The ring Z[i] is a Euclidean domain with valuation N, and hence also a PID. Proof. Let R = Z[i]. Note that N(a) ≥1 for all a € R\ {0}. So for a, b € R\ {0}, N(ab) = |ab|² = a²b² ≥ a2 N(a), and (ED1) is satisfied. Let a, b € R with b 0. Write a s+ti and b = u+vi with s, t, u, v € Z. We can form the quotient =1+mi Є C, where 1, m Є R. [To check (ED2), we want to use l+mi to define L+MI with L, M € Z as the element q in the expression a=qb+r we are looking for.] Let L, M Є Z be such that | L| ≤ ½ and m - M|≤ [such L, M can always be found]. Then =L+Mi + (1 L) + (m − M)i. Multiplying by b, we have Since a- done. a = (L+ Mi)b+ ((1 − L) + (m − M)i)b. - (L+ Mi)b Z[i], the term (1- L+ (m - M)i)b is in Z[i], too. If it is 0, we are So assume (L+(mM)i)b 0 (sol- L+ (mM)i 0). We have: N((I – L) + (m − M)i)b) = - |l-L+(m − M)i|³|b|ª S (+) ((글)2- + (-1)³) 16/2 = 16/2 N(b)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.1: Inverse Functions
Problem 18E
Related questions
Question
Can you use the information provided to answer question 2 please.
![==
Definition 5.14. The function N: Z[i] \ {0} Z20. sti|s+ti|2 = s²+t² is called
the norm on Z[i].
Proposition 5.15. The ring Z[i] is a Euclidean domain with valuation N, and hence also a
PID.
Proof. Let R = Z[i].
Note that N(a) ≥1 for all a € R\ {0}. So for a, b € R\ {0}, N(ab) = |ab|² = a²b² ≥
a2 N(a), and (ED1) is satisfied.
Let a, b € R with b 0. Write a s+ti and b = u+vi with s, t, u, v € Z. We can form the
quotient =1+mi Є C, where 1, m Є R.
[To check (ED2), we want to use l+mi to define L+MI with L, M € Z as the
element q in the expression a=qb+r we are looking for.]
Let L, M Є Z be such that | L| ≤ ½ and m - M|≤
[such L, M can always be found].
Then
=L+Mi + (1 L) + (m − M)i. Multiplying by b, we have
Since a-
done.
a = (L+ Mi)b+ ((1 − L) + (m − M)i)b.
-
(L+ Mi)b Z[i], the term (1- L+ (m - M)i)b is in Z[i], too. If it is 0, we are
So assume (L+(mM)i)b 0 (sol- L+ (mM)i 0). We have:
N((I – L) + (m − M)i)b) =
-
|l-L+(m − M)i|³|b|ª
S
(+)
((글)2- + (-1)³) 16/2
=
16/2
N(b)<N(b).
We see that (ED2) holds, if we choose q = L+ Mi and r = ((1 L) + (mM)i)b.
Therefore Z[i] is a Euclidean domain, and hence also a PID by Theorem 5.8
2. Let a 6+3i, b = 1+2i be elements of Z[i]. Use the proof of Proposition 5.15
from lectures to find elements q,r Z[i] such that a qb+r and either (i) r = 0
or (ii) r 0 and N(r) < N(b), where N is the norm on Z[i] defined in lectures.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd0a2eb85-71bc-41a8-8f67-a7e3c7209cc3%2F187e2840-a933-4ca9-bab7-7b1dce06ab12%2F6c46rfl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:==
Definition 5.14. The function N: Z[i] \ {0} Z20. sti|s+ti|2 = s²+t² is called
the norm on Z[i].
Proposition 5.15. The ring Z[i] is a Euclidean domain with valuation N, and hence also a
PID.
Proof. Let R = Z[i].
Note that N(a) ≥1 for all a € R\ {0}. So for a, b € R\ {0}, N(ab) = |ab|² = a²b² ≥
a2 N(a), and (ED1) is satisfied.
Let a, b € R with b 0. Write a s+ti and b = u+vi with s, t, u, v € Z. We can form the
quotient =1+mi Є C, where 1, m Є R.
[To check (ED2), we want to use l+mi to define L+MI with L, M € Z as the
element q in the expression a=qb+r we are looking for.]
Let L, M Є Z be such that | L| ≤ ½ and m - M|≤
[such L, M can always be found].
Then
=L+Mi + (1 L) + (m − M)i. Multiplying by b, we have
Since a-
done.
a = (L+ Mi)b+ ((1 − L) + (m − M)i)b.
-
(L+ Mi)b Z[i], the term (1- L+ (m - M)i)b is in Z[i], too. If it is 0, we are
So assume (L+(mM)i)b 0 (sol- L+ (mM)i 0). We have:
N((I – L) + (m − M)i)b) =
-
|l-L+(m − M)i|³|b|ª
S
(+)
((글)2- + (-1)³) 16/2
=
16/2
N(b)<N(b).
We see that (ED2) holds, if we choose q = L+ Mi and r = ((1 L) + (mM)i)b.
Therefore Z[i] is a Euclidean domain, and hence also a PID by Theorem 5.8
2. Let a 6+3i, b = 1+2i be elements of Z[i]. Use the proof of Proposition 5.15
from lectures to find elements q,r Z[i] such that a qb+r and either (i) r = 0
or (ii) r 0 and N(r) < N(b), where N is the norm on Z[i] defined in lectures.
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