Show that $q$ must divide one of the prime integer factors of $N(q)$.
$begingroup$
The following is from Aluffi's Algebra
Why this is true? :
.. since q is prime in Z[i] ⊇ Z, q must divide one of
the prime integer factors of N(q).
abstract-algebra number-theory algebraic-number-theory gaussian-integral
asked Dec 3 '18 at 13:07
72D72D
512116
$endgroup$
add a comment |
$begingroup$
The following is from Aluffi's Algebra
Why this is true? :
.. since q is prime in Z[i] ⊇ Z, q must divide one of
the prime integer factors of N(q).
abstract-algebra number-theory algebraic-number-theory gaussian-integral
asked Dec 3 '18 at 13:07
72D72D
512116
$endgroup$
1
$begingroup$
Write $N(q) = p_1 times cdots times p_n$ a product of $mathbb Z$-primes. Now, in $mathbb Z[i]$, $q mid p_1 times cdots times p_n$. By definition of primality, there is an $m$ such that $q mid p_m$.
$endgroup$
– Mees de Vries
Dec 3 '18 at 13:12
$begingroup$
@MeesdeVries, wow, that simple! thanks :)
$endgroup$
– 72D
Dec 3 '18 at 13:17
add a comment |
$begingroup$
The following is from Aluffi's Algebra
Why this is true? :
.. since q is prime in Z[i] ⊇ Z, q must divide one of
the prime integer factors of N(q).
abstract-algebra number-theory algebraic-number-theory gaussian-integral
asked Dec 3 '18 at 13:07
72D72D
512116
$endgroup$
The following is from Aluffi's Algebra
Why this is true? :
.. since q is prime in Z[i] ⊇ Z, q must divide one of
the prime integer factors of N(q).
abstract-algebra number-theory algebraic-number-theory gaussian-integral
abstract-algebra number-theory algebraic-number-theory gaussian-integral
asked Dec 3 '18 at 13:07
72D72D
512116
asked Dec 3 '18 at 13:07
72D72D
512116
asked Dec 3 '18 at 13:07
72D72D
512116
asked Dec 3 '18 at 13:07
72D72D
512116
asked Dec 3 '18 at 13:07
72D72D
512116
512116
1
$begingroup$
Write $N(q) = p_1 times cdots times p_n$ a product of $mathbb Z$-primes. Now, in $mathbb Z[i]$, $q mid p_1 times cdots times p_n$. By definition of primality, there is an $m$ such that $q mid p_m$.
$endgroup$
– Mees de Vries
Dec 3 '18 at 13:12
$begingroup$
@MeesdeVries, wow, that simple! thanks :)
$endgroup$
– 72D
Dec 3 '18 at 13:17
add a comment |
1
$begingroup$
Write $N(q) = p_1 times cdots times p_n$ a product of $mathbb Z$-primes. Now, in $mathbb Z[i]$, $q mid p_1 times cdots times p_n$. By definition of primality, there is an $m$ such that $q mid p_m$.
$endgroup$
– Mees de Vries
Dec 3 '18 at 13:12
$begingroup$
@MeesdeVries, wow, that simple! thanks :)
$endgroup$
– 72D
Dec 3 '18 at 13:17
1
1
$begingroup$
Write $N(q) = p_1 times cdots times p_n$ a product of $mathbb Z$-primes. Now, in $mathbb Z[i]$, $q mid p_1 times cdots times p_n$. By definition of primality, there is an $m$ such that $q mid p_m$.
$endgroup$
– Mees de Vries
Dec 3 '18 at 13:12
$begingroup$
Write $N(q) = p_1 times cdots times p_n$ a product of $mathbb Z$-primes. Now, in $mathbb Z[i]$, $q mid p_1 times cdots times p_n$. By definition of primality, there is an $m$ such that $q mid p_m$.
$endgroup$
– Mees de Vries
Dec 3 '18 at 13:12
$begingroup$
@MeesdeVries, wow, that simple! thanks :)
$endgroup$
– 72D
Dec 3 '18 at 13:17
$begingroup$
@MeesdeVries, wow, that simple! thanks :)
$endgroup$
– 72D
Dec 3 '18 at 13:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This might make a lot more sense to you if you look at specific numbers. I suggest $2$ and $3$.
Although $2$ is prime in $mathbb{Z}$, it's not prime in $mathbb{Z}[i]$, since $(1 - i)(1 + i) = 2$. So $N(2) = 4$, no surprise there. We also see that $N(1 - i) = 2$, and $N(1 + i) = 2$ as well. These facts may not seem obvious but they are easily verified.
And these facts mesh perfectly well with the multiplicative property of the norm function. If $ab = 2$, then $N(a)$ can only be $1$, $2$ or $4$, and the choice of $b$ follows immediately.
Compare $3$, which is indeed prime in both $mathbb{Z}$ and $mathbb{Z}[i]$. Its norm in the latter is still its square, and since it neither ramifies nor splits in this domain, it likewise follows that $N(z) = 3$ is impossible for $z in mathbb{Z}[i]$.
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
This might make a lot more sense to you if you look at specific numbers. I suggest $2$ and $3$.
Although $2$ is prime in $mathbb{Z}$, it's not prime in $mathbb{Z}[i]$, since $(1 - i)(1 + i) = 2$. So $N(2) = 4$, no surprise there. We also see that $N(1 - i) = 2$, and $N(1 + i) = 2$ as well. These facts may not seem obvious but they are easily verified.
And these facts mesh perfectly well with the multiplicative property of the norm function. If $ab = 2$, then $N(a)$ can only be $1$, $2$ or $4$, and the choice of $b$ follows immediately.
Compare $3$, which is indeed prime in both $mathbb{Z}$ and $mathbb{Z}[i]$. Its norm in the latter is still its square, and since it neither ramifies nor splits in this domain, it likewise follows that $N(z) = 3$ is impossible for $z in mathbb{Z}[i]$.
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
$endgroup$
add a comment |
$begingroup$
This might make a lot more sense to you if you look at specific numbers. I suggest $2$ and $3$.
Although $2$ is prime in $mathbb{Z}$, it's not prime in $mathbb{Z}[i]$, since $(1 - i)(1 + i) = 2$. So $N(2) = 4$, no surprise there. We also see that $N(1 - i) = 2$, and $N(1 + i) = 2$ as well. These facts may not seem obvious but they are easily verified.
And these facts mesh perfectly well with the multiplicative property of the norm function. If $ab = 2$, then $N(a)$ can only be $1$, $2$ or $4$, and the choice of $b$ follows immediately.
Compare $3$, which is indeed prime in both $mathbb{Z}$ and $mathbb{Z}[i]$. Its norm in the latter is still its square, and since it neither ramifies nor splits in this domain, it likewise follows that $N(z) = 3$ is impossible for $z in mathbb{Z}[i]$.
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
$endgroup$
add a comment |
$begingroup$
This might make a lot more sense to you if you look at specific numbers. I suggest $2$ and $3$.
Although $2$ is prime in $mathbb{Z}$, it's not prime in $mathbb{Z}[i]$, since $(1 - i)(1 + i) = 2$. So $N(2) = 4$, no surprise there. We also see that $N(1 - i) = 2$, and $N(1 + i) = 2$ as well. These facts may not seem obvious but they are easily verified.
And these facts mesh perfectly well with the multiplicative property of the norm function. If $ab = 2$, then $N(a)$ can only be $1$, $2$ or $4$, and the choice of $b$ follows immediately.
Compare $3$, which is indeed prime in both $mathbb{Z}$ and $mathbb{Z}[i]$. Its norm in the latter is still its square, and since it neither ramifies nor splits in this domain, it likewise follows that $N(z) = 3$ is impossible for $z in mathbb{Z}[i]$.
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
$endgroup$
This might make a lot more sense to you if you look at specific numbers. I suggest $2$ and $3$.
Although $2$ is prime in $mathbb{Z}$, it's not prime in $mathbb{Z}[i]$, since $(1 - i)(1 + i) = 2$. So $N(2) = 4$, no surprise there. We also see that $N(1 - i) = 2$, and $N(1 + i) = 2$ as well. These facts may not seem obvious but they are easily verified.
And these facts mesh perfectly well with the multiplicative property of the norm function. If $ab = 2$, then $N(a)$ can only be $1$, $2$ or $4$, and the choice of $b$ follows immediately.
Compare $3$, which is indeed prime in both $mathbb{Z}$ and $mathbb{Z}[i]$. Its norm in the latter is still its square, and since it neither ramifies nor splits in this domain, it likewise follows that $N(z) = 3$ is impossible for $z in mathbb{Z}[i]$.
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
answered Dec 7 '18 at 0:10
Mr. BrooksMr. Brooks
16911237
16911237
add a comment |
add a comment |
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$begingroup$
Write $N(q) = p_1 times cdots times p_n$ a product of $mathbb Z$-primes. Now, in $mathbb Z[i]$, $q mid p_1 times cdots times p_n$. By definition of primality, there is an $m$ such that $q mid p_m$.
$endgroup$
– Mees de Vries
Dec 3 '18 at 13:12
$begingroup$
@MeesdeVries, wow, that simple! thanks :)
$endgroup$
– 72D
Dec 3 '18 at 13:17