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3. Let R be a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R. Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit.

3. Let R be a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R. Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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3. Let R be a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R. Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit.
Transcribed Image Text:3. Let R be a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R. Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit.
R is a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R. Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit. To prove that every nonzero element of norm m in a Euclidean Domain R is a unit, we need to show that it has a multiplicative inverse. Let's consider a nonzero element a in R with norm m. We want to find an element b in R such that ab = 1, where 1 is the multiplicative identity in R. Since R is a Euclidean Domain, it means that for any nonzero element b in (a), we can divide b by a and obtain a quotient q and a remainder r such that b = aq+r, and either r is zero or the norm of r is strictly less than the norm of a. Now, let's consider the element b = 1. By the above division algorithm, we can write 1 = aq+r, where r is either zero or has a norm strictly less than the norm of a. Since the norm of a is m, the norm of r must be less than m. If r is nonzero, it contradicts the minimality of m as the norm of a nonzero element in (a). Therefore, r must be zero, which implies that 1 = aq. This shows that a has a multiplicative inverse in R, and hence, a is a unit. Thus, we have proved that every nonzero element of norm m in a Euclidean Domain R is a unit. Now, let's consider a nonzero element à in R with norm zero. If such an element exists, it means that the norm of x is the smallest possible norm in R. Since x is nonzero, the norm is defined to be in Z+. Therefore, the only possibility is that the norm of x is zero. Hence, we can conclude that a nonzero element of norm zero, if it exists, is a unit. Therefore, in a Euclidean Domain, every nonzero element of norm m is a unit, and if a nonzero element of norm zero exists, it is also a unit.
Transcribed Image Text:R is a Euclidean Domain. Let m be the minimum integer in the set of norms of nonzero elements of R. Prove that every nonzero element of R of norm m is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit. To prove that every nonzero element of norm m in a Euclidean Domain R is a unit, we need to show that it has a multiplicative inverse. Let's consider a nonzero element a in R with norm m. We want to find an element b in R such that ab = 1, where 1 is the multiplicative identity in R. Since R is a Euclidean Domain, it means that for any nonzero element b in (a), we can divide b by a and obtain a quotient q and a remainder r such that b = aq+r, and either r is zero or the norm of r is strictly less than the norm of a. Now, let's consider the element b = 1. By the above division algorithm, we can write 1 = aq+r, where r is either zero or has a norm strictly less than the norm of a. Since the norm of a is m, the norm of r must be less than m. If r is nonzero, it contradicts the minimality of m as the norm of a nonzero element in (a). Therefore, r must be zero, which implies that 1 = aq. This shows that a has a multiplicative inverse in R, and hence, a is a unit. Thus, we have proved that every nonzero element of norm m in a Euclidean Domain R is a unit. Now, let's consider a nonzero element à in R with norm zero. If such an element exists, it means that the norm of x is the smallest possible norm in R. Since x is nonzero, the norm is defined to be in Z+. Therefore, the only possibility is that the norm of x is zero. Hence, we can conclude that a nonzero element of norm zero, if it exists, is a unit. Therefore, in a Euclidean Domain, every nonzero element of norm m is a unit, and if a nonzero element of norm zero exists, it is also a unit.
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