3. Assume that d € Z and √d & Q. Prove that if € Z[√d] has the property that |N₁(π)| is a prime number, then 7 is a prime of Z[√d]. (Hint: take inspiration from the example above.)
3. Assume that d € Z and √d & Q. Prove that if € Z[√d] has the property that |N₁(π)| is a prime number, then 7 is a prime of Z[√d]. (Hint: take inspiration from the example above.)
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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Question
![Let R be a ring and let p E R be an element which is nonzero and not a unit. We say
that p is prime if whenever p = xy with x, y € R, either x is a unit or y is a unit. This
definition does not quite reduce to the usual one for the integers: according to this general
definition, an integer n is prime if and only if it is of the form ±p, where p is a prime integer
in the usual sense (you should be able to prove this!). Note: The definition does not say that
PER is prime if it has a factorization p = xy with x or y a unit; any element r of R trivially
admits such a factorization since we can write r = 1Rr. The manner in which primality is
used in practice is as follows: if we know that pe R is prime, and we find ourselves with a
factorization p = xy, we can immediately conclude that x or y is a unit. It would not make
sense to start an argument with something like "because p is prime, there is a factorization
p=xy..." This is not consistent with what the definition actually says. On the other hand,
to prove that some p E R is prime, the definition suggests the following strategy: consider
a hypothetical factorization p = xy with x, y € R, and, somehow, show that x or y is a unit
in R. The "somehow" depends on the specific context, but this setup is always a good way
to approach a primality proof.
For an example, we work in the ring Z[i] of Gaussian integers, where we will prove that
π = 3 + 2i is prime. To this end, suppose that 7 = aß with a, ß Z[i].² We want to show
that either a or 3 is a unit. Our favorite approach for showing that a Gaussian integer (or
any quadratic integer) is a unit is to show that its norm is a unit in Z; in the case of a
Gaussian integer 6, because N-1(6) ≥ 0, we can say that 8 € Z[i]× if and only if N-1(8) = 1.
So, let's take norms of both sides of the equation π = aß:
N_1(T) = N_₁(aß) = N_1(a)N_1(3).
By definition, N-1(π) = 3² +2²= 13, and because a and 3 are necessarily nonzero (since
they are factors of the nonzero number 7), their norms are positive integers. Thus we have
a factorization 13 = N_1(a)N-1(3) in N. As 13 is a prime number in the usual sense, we
must have N-1(a) = 1 or N-1(3): 1; in the first case, a is a unit in Z[i], and in the second
case, is a unit in Z[i]. Having proved that the hypothetical factorization of in Z[i] is
trivial, we may conclude that is prime.
=
3. Assume that d € Z and √d ‡ Q. Prove that if 7 € Z[√d] has the property that |N₁(π)|
is a prime number, then 7 is a prime of Z[√d]. (Hint: take inspiration from the example
above.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febb913e1-4986-4d74-b6ce-ad576ddf43d3%2F79896e95-f061-4e4f-8d4d-dc53eddf3693%2Fptzhor_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Let R be a ring and let p E R be an element which is nonzero and not a unit. We say
that p is prime if whenever p = xy with x, y € R, either x is a unit or y is a unit. This
definition does not quite reduce to the usual one for the integers: according to this general
definition, an integer n is prime if and only if it is of the form ±p, where p is a prime integer
in the usual sense (you should be able to prove this!). Note: The definition does not say that
PER is prime if it has a factorization p = xy with x or y a unit; any element r of R trivially
admits such a factorization since we can write r = 1Rr. The manner in which primality is
used in practice is as follows: if we know that pe R is prime, and we find ourselves with a
factorization p = xy, we can immediately conclude that x or y is a unit. It would not make
sense to start an argument with something like "because p is prime, there is a factorization
p=xy..." This is not consistent with what the definition actually says. On the other hand,
to prove that some p E R is prime, the definition suggests the following strategy: consider
a hypothetical factorization p = xy with x, y € R, and, somehow, show that x or y is a unit
in R. The "somehow" depends on the specific context, but this setup is always a good way
to approach a primality proof.
For an example, we work in the ring Z[i] of Gaussian integers, where we will prove that
π = 3 + 2i is prime. To this end, suppose that 7 = aß with a, ß Z[i].² We want to show
that either a or 3 is a unit. Our favorite approach for showing that a Gaussian integer (or
any quadratic integer) is a unit is to show that its norm is a unit in Z; in the case of a
Gaussian integer 6, because N-1(6) ≥ 0, we can say that 8 € Z[i]× if and only if N-1(8) = 1.
So, let's take norms of both sides of the equation π = aß:
N_1(T) = N_₁(aß) = N_1(a)N_1(3).
By definition, N-1(π) = 3² +2²= 13, and because a and 3 are necessarily nonzero (since
they are factors of the nonzero number 7), their norms are positive integers. Thus we have
a factorization 13 = N_1(a)N-1(3) in N. As 13 is a prime number in the usual sense, we
must have N-1(a) = 1 or N-1(3): 1; in the first case, a is a unit in Z[i], and in the second
case, is a unit in Z[i]. Having proved that the hypothetical factorization of in Z[i] is
trivial, we may conclude that is prime.
=
3. Assume that d € Z and √d ‡ Q. Prove that if 7 € Z[√d] has the property that |N₁(π)|
is a prime number, then 7 is a prime of Z[√d]. (Hint: take inspiration from the example
above.)
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