Mixing
Is it true that: $a+bi$ is prime in $mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $mathbb{Z}$
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Is it true that $a+bi$ is prime in $mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $mathbb{Z}$?
How can I prove this? Can anybody help me please?
abstract-algebra
edited Jan 28 at 1:47
user549397
1,5761418
asked Apr 21 '13 at 4:32
habujihabuji
362
$endgroup$
|
show 1 more comment
$begingroup$
Is it true that $a+bi$ is prime in $mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $mathbb{Z}$?
How can I prove this? Can anybody help me please?
abstract-algebra
edited Jan 28 at 1:47
user549397
1,5761418
asked Apr 21 '13 at 4:32
habujihabuji
362
$endgroup$
$begingroup$
$a^2+b^2=(a+ib)(a-ib)$
$endgroup$
– Mathematician
Apr 21 '13 at 4:35
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this is obvious.but I could not understand how should I use this result
$endgroup$
– habuji
Apr 21 '13 at 4:40
13
$begingroup$
False. Three is prime in $mathbb{Z}[i]$, yet $3^2+0^2=9$ is not prime in $mathbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Apr 21 '13 at 4:43
1
$begingroup$
If the norm is prime, it can't be the product of Norms of two other numbers unless one is a unit. However the converse isn't always true.
$endgroup$
– Macavity
Apr 21 '13 at 4:57
2
$begingroup$
you would need restrictions on your statements, such as both a, b are nonzero, as @Jyrki has shown the hole there. Also, $a^2+b^2$ can not be of the form $4k+3$ (again, as is 3).
$endgroup$
– Eleven-Eleven
Apr 21 '13 at 5:01
|
show 1 more comment
$begingroup$
Is it true that $a+bi$ is prime in $mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $mathbb{Z}$?
How can I prove this? Can anybody help me please?
abstract-algebra
edited Jan 28 at 1:47
user549397
1,5761418
asked Apr 21 '13 at 4:32
habujihabuji
362
$endgroup$
Is it true that $a+bi$ is prime in $mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $mathbb{Z}$?
How can I prove this? Can anybody help me please?
abstract-algebra
abstract-algebra
edited Jan 28 at 1:47
user549397
1,5761418
asked Apr 21 '13 at 4:32
habujihabuji
362
edited Jan 28 at 1:47
user549397
1,5761418
asked Apr 21 '13 at 4:32
habujihabuji
362
edited Jan 28 at 1:47
user549397
1,5761418
edited Jan 28 at 1:47
user549397
1,5761418
edited Jan 28 at 1:47
user549397
1,5761418
1,5761418
asked Apr 21 '13 at 4:32
habujihabuji
362
asked Apr 21 '13 at 4:32
habujihabuji
362
asked Apr 21 '13 at 4:32
habujihabuji
362
362
$begingroup$
$a^2+b^2=(a+ib)(a-ib)$
$endgroup$
– Mathematician
Apr 21 '13 at 4:35
$begingroup$
this is obvious.but I could not understand how should I use this result
$endgroup$
– habuji
Apr 21 '13 at 4:40
13
$begingroup$
False. Three is prime in $mathbb{Z}[i]$, yet $3^2+0^2=9$ is not prime in $mathbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Apr 21 '13 at 4:43
1
$begingroup$
If the norm is prime, it can't be the product of Norms of two other numbers unless one is a unit. However the converse isn't always true.
$endgroup$
– Macavity
Apr 21 '13 at 4:57
2
$begingroup$
you would need restrictions on your statements, such as both a, b are nonzero, as @Jyrki has shown the hole there. Also, $a^2+b^2$ can not be of the form $4k+3$ (again, as is 3).
$endgroup$
– Eleven-Eleven
Apr 21 '13 at 5:01
|
show 1 more comment
$begingroup$
$a^2+b^2=(a+ib)(a-ib)$
$endgroup$
– Mathematician
Apr 21 '13 at 4:35
$begingroup$
this is obvious.but I could not understand how should I use this result
$endgroup$
– habuji
Apr 21 '13 at 4:40
13
$begingroup$
False. Three is prime in $mathbb{Z}[i]$, yet $3^2+0^2=9$ is not prime in $mathbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Apr 21 '13 at 4:43
1
$begingroup$
If the norm is prime, it can't be the product of Norms of two other numbers unless one is a unit. However the converse isn't always true.
$endgroup$
– Macavity
Apr 21 '13 at 4:57
2
$begingroup$
you would need restrictions on your statements, such as both a, b are nonzero, as @Jyrki has shown the hole there. Also, $a^2+b^2$ can not be of the form $4k+3$ (again, as is 3).
$endgroup$
– Eleven-Eleven
Apr 21 '13 at 5:01
$begingroup$
$a^2+b^2=(a+ib)(a-ib)$
$endgroup$
– Mathematician
Apr 21 '13 at 4:35
$begingroup$
$a^2+b^2=(a+ib)(a-ib)$
$endgroup$
– Mathematician
Apr 21 '13 at 4:35
$begingroup$
this is obvious.but I could not understand how should I use this result
$endgroup$
– habuji
Apr 21 '13 at 4:40
$begingroup$
this is obvious.but I could not understand how should I use this result
$endgroup$
– habuji
Apr 21 '13 at 4:40
13
13
$begingroup$
False. Three is prime in $mathbb{Z}[i]$, yet $3^2+0^2=9$ is not prime in $mathbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Apr 21 '13 at 4:43
$begingroup$
False. Three is prime in $mathbb{Z}[i]$, yet $3^2+0^2=9$ is not prime in $mathbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Apr 21 '13 at 4:43
1
1
$begingroup$
If the norm is prime, it can't be the product of Norms of two other numbers unless one is a unit. However the converse isn't always true.
$endgroup$
– Macavity
Apr 21 '13 at 4:57
$begingroup$
If the norm is prime, it can't be the product of Norms of two other numbers unless one is a unit. However the converse isn't always true.
$endgroup$
– Macavity
Apr 21 '13 at 4:57
2
2
$begingroup$
you would need restrictions on your statements, such as both a, b are nonzero, as @Jyrki has shown the hole there. Also, $a^2+b^2$ can not be of the form $4k+3$ (again, as is 3).
$endgroup$
– Eleven-Eleven
Apr 21 '13 at 5:01
$begingroup$
you would need restrictions on your statements, such as both a, b are nonzero, as @Jyrki has shown the hole there. Also, $a^2+b^2$ can not be of the form $4k+3$ (again, as is 3).
$endgroup$
– Eleven-Eleven
Apr 21 '13 at 5:01
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.
Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime. For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$. So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.
For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume
standard results that characterize the Gaussian primes. So we try not to use much machinery.
If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.
So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime. For any common divisor $delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime.
Now suppose that $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$.
Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,
Let $pi$ be a Gaussian prime that divides $p$. Then $pi$ divides one of $a+bi$ or $a-bi$. So $pi$ must be an associate of one of these, and the conjugate of $pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
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add a comment |
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Here is something that is true: $a+bi$ is prime in $defZ{Bbb Z}Z[i]$ implies that $(a+bi)Z[i]capZ=pZ$ for some (ordinary) prime $p$ of $Z$. This is because, (1) since $a+bi$ is irreducible and $Z[i]$ is a Euclidean and therefore unique factorisation domain, $(a+bi)Z[i]$ is a prime ideal of $Z[i]$, so (2) its intersection with $Z$ is a prime ideal of $Z$ (as is always true for the intersection of a prime ideal and a subring), and (3) the intersection is not reduced to$~{0}$ because $(a+bi)(a-bi)=a^2+b^2inZsetminus{0}$. (Here, like in the question, one does not have "if and only if": here the converse fails for $a+bi$ equal or associated to a prime number $pnotequiv3pmod4$, such as $p=2$ or $p=5$; then $(a+bi)Z[i]capZ=pZ$, but $p$ and therefore $a+bi$ are composite in $Z[i]$.)
The case $a+bi$ is prime in $Z[i]$ splits into two subcases. By the above there exists a prime number $p$ and $zinZ[i]$ with $p=(a+bi)z$; then $p^2=N(p)=N(a+bi)N(z)$, and either $z$ is non-invertible, in which case $N(a+bi)=p$ and $z=a-bi$, or $z$ is invertible, in which case $a+biin{p,ip,-p,-ip}$ and it can be shown that $pequiv3pmod4$. Indeed, the irreducibility of $p$ in the UFD $Z[i]$ means that $Z[i]/pZ[i]$ is an integral domain (it is a field), so the kernel $(X^2+1)$ of the ring morphism $(Z/pZ)[X]toZ[i]/pZ[i]$ sending $Xmapsto i$ is a prime ideal, so $X^2+1in(Z/pZ)[X]$ is irreducible, which excludes both $p=2$ (for which $X^2+1=(X+1)^2$) and $pequiv1pmod4$ (in which case $X^2+1=(X+a)(X-a)$ for some element $a$ of order $4$ in the cyclic group $(Z/pZ)^times$ of order $p-1$.
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
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The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring).
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– Math Gems
Apr 21 '13 at 16:20
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@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful.
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– Marc van Leeuwen
Apr 21 '13 at 16:45
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I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise.
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– Math Gems
Apr 21 '13 at 17:13
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@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better?
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– Marc van Leeuwen
Apr 21 '13 at 18:39
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Yes, that reads much more clearly. Thanks and +1.
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– Math Gems
Apr 21 '13 at 19:05
add a comment |
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Let me add another argument that I believe is simpler. First I need to prove a little proposition.
Let $p in mathbb{N}$ be a prime and $N$ be the norm function $N(alpha) = alpha overline{alpha}$.
If $alpha in mathbb{Z}[sqrt{n}]$ is prime then $N(alpha) =
p,-p,p^2,-p^2$
Furthermore, if $N(alpha) = p^2,-p^2$ then $alpha sim p$ where $sim$ is the associates relation on $mathbb{Z}[sqrt{n}]$.
Proof:
If $alpha$ is prime in $mathbb{Z}[sqrt{n}]$, then $N(alpha) neq 0,1,-1$ this is because $alpha$ is not a unit and the fact that $N(alpha) = alpha overline{alpha} neq 0$ comes because we are working in an integrity domain so that $alpha = 0 lor overline{alpha} = 0$ but since $n$ is square-free (by definition) it is necessary that if $alpha = a+bi$ then $a = b = 0$, therefore $alpha = 0$ and cannot be prime.
So let $N(alpha) = prod p_i$ be its prime factorization over the integers. Then $alpha|p_i$ since $alpha$ is prime. Therefore, $alpha beta = p implies N(alpha) N(beta) = p^2 implies N(alpha) in {p,-p,p^2,-p^2}$ as we wanted. Furthermore, if $N(alpha) = p,-p^2$ then $N(beta) = 1,-1$ which means that $beta$ is a unit and therefore $beta sim p$. $$tag*{$blacksquare$}$$
So now come to our problem over $mathbb{Z}[i]$:
$alpha = a+bi$ is prime in $mathbb{Z}[i] simeq p = N(a+bi) = a^2+b^2$ is prime in $mathbb{Z}$.
$impliedby)$ Since, $mathbb{Z}[i]$ is a UFD, $alpha$ is prime if and only if it is irreducible. But if, $alpha = beta gamma$ is a proper factorization of $alpha$ then $N(alpha) = N(beta) N(gamma)$ and you would have a proper factorization of $p$. Contradiction.
$implies)$ $alpha$ prime implies that $N(alpha) = p,-p,p^2,-p^2$. Since the norm is positive $N(alpha) = p,p^2$. If $N(alpha) = p$ then we are done. If $N(alpha) = p^2$ then $alpha sim p in mathbb{N}$ prime. Recalling the units of $mathbb{Z}[i]$ are $U(mathbb{Z}[i]) = {i,-i,1,-1}$ and multiplying you will see that necessarily $alpha in {p,-p,pi,-pi}$.
However, you cannot rule out this last case. For instance, take $3$ in $mathbb{Z}[i]$ which following a norm-argument can be shown to be prime, yet its norm is $p^2 = 9$ which is not prime in $mathbb{Z}$.
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
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Could you explain the last line where you claim that $ abneq 0 $? Take $ alpha=0+pi $, then $ a=0, b=p $, but $ ab=0 $ and we still have $ alphasim p $.
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– user549397
Jan 27 at 14:35
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@user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it
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– Javier
Jan 27 at 17:17
add a comment |
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3 Answers
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3 Answers
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$begingroup$
This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.
Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime. For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$. So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.
For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume
standard results that characterize the Gaussian primes. So we try not to use much machinery.
If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.
So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime. For any common divisor $delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime.
Now suppose that $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$.
Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,
Let $pi$ be a Gaussian prime that divides $p$. Then $pi$ divides one of $a+bi$ or $a-bi$. So $pi$ must be an associate of one of these, and the conjugate of $pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
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add a comment |
$begingroup$
This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.
Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime. For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$. So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.
For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume
standard results that characterize the Gaussian primes. So we try not to use much machinery.
If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.
So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime. For any common divisor $delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime.
Now suppose that $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$.
Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,
Let $pi$ be a Gaussian prime that divides $p$. Then $pi$ divides one of $a+bi$ or $a-bi$. So $pi$ must be an associate of one of these, and the conjugate of $pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
$endgroup$
add a comment |
$begingroup$
This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.
Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime. For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$. So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.
For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume
standard results that characterize the Gaussian primes. So we try not to use much machinery.
If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.
So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime. For any common divisor $delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime.
Now suppose that $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$.
Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,
Let $pi$ be a Gaussian prime that divides $p$. Then $pi$ divides one of $a+bi$ or $a-bi$. So $pi$ must be an associate of one of these, and the conjugate of $pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
$endgroup$
This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.
Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime. For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$. So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.
For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume
standard results that characterize the Gaussian primes. So we try not to use much machinery.
If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.
So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime. For any common divisor $delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime.
Now suppose that $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$.
Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,
Let $pi$ be a Gaussian prime that divides $p$. Then $pi$ divides one of $a+bi$ or $a-bi$. So $pi$ must be an associate of one of these, and the conjugate of $pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
edited Apr 21 '13 at 15:15
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
answered Apr 21 '13 at 5:16
André NicolasAndré Nicolas
454k36432819
454k36432819
add a comment |
add a comment |
$begingroup$
Here is something that is true: $a+bi$ is prime in $defZ{Bbb Z}Z[i]$ implies that $(a+bi)Z[i]capZ=pZ$ for some (ordinary) prime $p$ of $Z$. This is because, (1) since $a+bi$ is irreducible and $Z[i]$ is a Euclidean and therefore unique factorisation domain, $(a+bi)Z[i]$ is a prime ideal of $Z[i]$, so (2) its intersection with $Z$ is a prime ideal of $Z$ (as is always true for the intersection of a prime ideal and a subring), and (3) the intersection is not reduced to$~{0}$ because $(a+bi)(a-bi)=a^2+b^2inZsetminus{0}$. (Here, like in the question, one does not have "if and only if": here the converse fails for $a+bi$ equal or associated to a prime number $pnotequiv3pmod4$, such as $p=2$ or $p=5$; then $(a+bi)Z[i]capZ=pZ$, but $p$ and therefore $a+bi$ are composite in $Z[i]$.)
The case $a+bi$ is prime in $Z[i]$ splits into two subcases. By the above there exists a prime number $p$ and $zinZ[i]$ with $p=(a+bi)z$; then $p^2=N(p)=N(a+bi)N(z)$, and either $z$ is non-invertible, in which case $N(a+bi)=p$ and $z=a-bi$, or $z$ is invertible, in which case $a+biin{p,ip,-p,-ip}$ and it can be shown that $pequiv3pmod4$. Indeed, the irreducibility of $p$ in the UFD $Z[i]$ means that $Z[i]/pZ[i]$ is an integral domain (it is a field), so the kernel $(X^2+1)$ of the ring morphism $(Z/pZ)[X]toZ[i]/pZ[i]$ sending $Xmapsto i$ is a prime ideal, so $X^2+1in(Z/pZ)[X]$ is irreducible, which excludes both $p=2$ (for which $X^2+1=(X+1)^2$) and $pequiv1pmod4$ (in which case $X^2+1=(X+a)(X-a)$ for some element $a$ of order $4$ in the cyclic group $(Z/pZ)^times$ of order $p-1$.
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
$endgroup$
$begingroup$
The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring).
$endgroup$
– Math Gems
Apr 21 '13 at 16:20
$begingroup$
@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful.
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 16:45
$begingroup$
I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise.
$endgroup$
– Math Gems
Apr 21 '13 at 17:13
$begingroup$
@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better?
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 18:39
$begingroup$
Yes, that reads much more clearly. Thanks and +1.
$endgroup$
– Math Gems
Apr 21 '13 at 19:05
add a comment |
$begingroup$
Here is something that is true: $a+bi$ is prime in $defZ{Bbb Z}Z[i]$ implies that $(a+bi)Z[i]capZ=pZ$ for some (ordinary) prime $p$ of $Z$. This is because, (1) since $a+bi$ is irreducible and $Z[i]$ is a Euclidean and therefore unique factorisation domain, $(a+bi)Z[i]$ is a prime ideal of $Z[i]$, so (2) its intersection with $Z$ is a prime ideal of $Z$ (as is always true for the intersection of a prime ideal and a subring), and (3) the intersection is not reduced to$~{0}$ because $(a+bi)(a-bi)=a^2+b^2inZsetminus{0}$. (Here, like in the question, one does not have "if and only if": here the converse fails for $a+bi$ equal or associated to a prime number $pnotequiv3pmod4$, such as $p=2$ or $p=5$; then $(a+bi)Z[i]capZ=pZ$, but $p$ and therefore $a+bi$ are composite in $Z[i]$.)
The case $a+bi$ is prime in $Z[i]$ splits into two subcases. By the above there exists a prime number $p$ and $zinZ[i]$ with $p=(a+bi)z$; then $p^2=N(p)=N(a+bi)N(z)$, and either $z$ is non-invertible, in which case $N(a+bi)=p$ and $z=a-bi$, or $z$ is invertible, in which case $a+biin{p,ip,-p,-ip}$ and it can be shown that $pequiv3pmod4$. Indeed, the irreducibility of $p$ in the UFD $Z[i]$ means that $Z[i]/pZ[i]$ is an integral domain (it is a field), so the kernel $(X^2+1)$ of the ring morphism $(Z/pZ)[X]toZ[i]/pZ[i]$ sending $Xmapsto i$ is a prime ideal, so $X^2+1in(Z/pZ)[X]$ is irreducible, which excludes both $p=2$ (for which $X^2+1=(X+1)^2$) and $pequiv1pmod4$ (in which case $X^2+1=(X+a)(X-a)$ for some element $a$ of order $4$ in the cyclic group $(Z/pZ)^times$ of order $p-1$.
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
$endgroup$
$begingroup$
The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring).
$endgroup$
– Math Gems
Apr 21 '13 at 16:20
$begingroup$
@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful.
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 16:45
$begingroup$
I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise.
$endgroup$
– Math Gems
Apr 21 '13 at 17:13
$begingroup$
@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better?
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 18:39
$begingroup$
Yes, that reads much more clearly. Thanks and +1.
$endgroup$
– Math Gems
Apr 21 '13 at 19:05
add a comment |
$begingroup$
Here is something that is true: $a+bi$ is prime in $defZ{Bbb Z}Z[i]$ implies that $(a+bi)Z[i]capZ=pZ$ for some (ordinary) prime $p$ of $Z$. This is because, (1) since $a+bi$ is irreducible and $Z[i]$ is a Euclidean and therefore unique factorisation domain, $(a+bi)Z[i]$ is a prime ideal of $Z[i]$, so (2) its intersection with $Z$ is a prime ideal of $Z$ (as is always true for the intersection of a prime ideal and a subring), and (3) the intersection is not reduced to$~{0}$ because $(a+bi)(a-bi)=a^2+b^2inZsetminus{0}$. (Here, like in the question, one does not have "if and only if": here the converse fails for $a+bi$ equal or associated to a prime number $pnotequiv3pmod4$, such as $p=2$ or $p=5$; then $(a+bi)Z[i]capZ=pZ$, but $p$ and therefore $a+bi$ are composite in $Z[i]$.)
The case $a+bi$ is prime in $Z[i]$ splits into two subcases. By the above there exists a prime number $p$ and $zinZ[i]$ with $p=(a+bi)z$; then $p^2=N(p)=N(a+bi)N(z)$, and either $z$ is non-invertible, in which case $N(a+bi)=p$ and $z=a-bi$, or $z$ is invertible, in which case $a+biin{p,ip,-p,-ip}$ and it can be shown that $pequiv3pmod4$. Indeed, the irreducibility of $p$ in the UFD $Z[i]$ means that $Z[i]/pZ[i]$ is an integral domain (it is a field), so the kernel $(X^2+1)$ of the ring morphism $(Z/pZ)[X]toZ[i]/pZ[i]$ sending $Xmapsto i$ is a prime ideal, so $X^2+1in(Z/pZ)[X]$ is irreducible, which excludes both $p=2$ (for which $X^2+1=(X+1)^2$) and $pequiv1pmod4$ (in which case $X^2+1=(X+a)(X-a)$ for some element $a$ of order $4$ in the cyclic group $(Z/pZ)^times$ of order $p-1$.
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
$endgroup$
Here is something that is true: $a+bi$ is prime in $defZ{Bbb Z}Z[i]$ implies that $(a+bi)Z[i]capZ=pZ$ for some (ordinary) prime $p$ of $Z$. This is because, (1) since $a+bi$ is irreducible and $Z[i]$ is a Euclidean and therefore unique factorisation domain, $(a+bi)Z[i]$ is a prime ideal of $Z[i]$, so (2) its intersection with $Z$ is a prime ideal of $Z$ (as is always true for the intersection of a prime ideal and a subring), and (3) the intersection is not reduced to$~{0}$ because $(a+bi)(a-bi)=a^2+b^2inZsetminus{0}$. (Here, like in the question, one does not have "if and only if": here the converse fails for $a+bi$ equal or associated to a prime number $pnotequiv3pmod4$, such as $p=2$ or $p=5$; then $(a+bi)Z[i]capZ=pZ$, but $p$ and therefore $a+bi$ are composite in $Z[i]$.)
The case $a+bi$ is prime in $Z[i]$ splits into two subcases. By the above there exists a prime number $p$ and $zinZ[i]$ with $p=(a+bi)z$; then $p^2=N(p)=N(a+bi)N(z)$, and either $z$ is non-invertible, in which case $N(a+bi)=p$ and $z=a-bi$, or $z$ is invertible, in which case $a+biin{p,ip,-p,-ip}$ and it can be shown that $pequiv3pmod4$. Indeed, the irreducibility of $p$ in the UFD $Z[i]$ means that $Z[i]/pZ[i]$ is an integral domain (it is a field), so the kernel $(X^2+1)$ of the ring morphism $(Z/pZ)[X]toZ[i]/pZ[i]$ sending $Xmapsto i$ is a prime ideal, so $X^2+1in(Z/pZ)[X]$ is irreducible, which excludes both $p=2$ (for which $X^2+1=(X+1)^2$) and $pequiv1pmod4$ (in which case $X^2+1=(X+a)(X-a)$ for some element $a$ of order $4$ in the cyclic group $(Z/pZ)^times$ of order $p-1$.
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
edited Apr 21 '13 at 18:38
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
answered Apr 21 '13 at 11:37
Marc van LeeuwenMarc van Leeuwen
88.5k5111229
88.5k5111229
$begingroup$
The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring).
$endgroup$
– Math Gems
Apr 21 '13 at 16:20
$begingroup$
@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful.
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 16:45
$begingroup$
I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise.
$endgroup$
– Math Gems
Apr 21 '13 at 17:13
$begingroup$
@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better?
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 18:39
$begingroup$
Yes, that reads much more clearly. Thanks and +1.
$endgroup$
– Math Gems
Apr 21 '13 at 19:05
add a comment |
$begingroup$
The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring).
$endgroup$
– Math Gems
Apr 21 '13 at 16:20
$begingroup$
@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful.
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 16:45
$begingroup$
I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise.
$endgroup$
– Math Gems
Apr 21 '13 at 17:13
$begingroup$
@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better?
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 18:39
$begingroup$
Yes, that reads much more clearly. Thanks and +1.
$endgroup$
– Math Gems
Apr 21 '13 at 19:05
$begingroup$
The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring).
$endgroup$
– Math Gems
Apr 21 '13 at 16:20
$begingroup$
The first paragraph does not make clear precisely what is a consequence of said UFD property (that prime ideals are preserved under contraction is true in any ring).
$endgroup$
– Math Gems
Apr 21 '13 at 16:20
$begingroup$
@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful.
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 16:45
$begingroup$
@MathGems: The "since ... UFD" in the second sentence was awkwardly placed very early to indicate that that it is needed to justify the immediately following "is a prime ideal". And you are right, pulling back a prime ideal through a ring morphism always gives a prime ideal; maybe that is why they are so useful.
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 16:45
$begingroup$
I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise.
$endgroup$
– Math Gems
Apr 21 '13 at 17:13
$begingroup$
I surmised what you intended. But I fear that those beginning their studies might have more difficulty inferring the intended meaning. Whenever I encounter things like that I leave comments in the hope that the author might improve the exposition to eliminate ambiguities etc. Thankfully many folks do the same for me when I too do likewise.
$endgroup$
– Math Gems
Apr 21 '13 at 17:13
$begingroup$
@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better?
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 18:39
$begingroup$
@MathGems: Fine, thank you. Did I succeed in reducing the ambiguity, or if not what would be better?
$endgroup$
– Marc van Leeuwen
Apr 21 '13 at 18:39
$begingroup$
Yes, that reads much more clearly. Thanks and +1.
$endgroup$
– Math Gems
Apr 21 '13 at 19:05
$begingroup$
Yes, that reads much more clearly. Thanks and +1.
$endgroup$
– Math Gems
Apr 21 '13 at 19:05
add a comment |
$begingroup$
Let me add another argument that I believe is simpler. First I need to prove a little proposition.
Let $p in mathbb{N}$ be a prime and $N$ be the norm function $N(alpha) = alpha overline{alpha}$.
If $alpha in mathbb{Z}[sqrt{n}]$ is prime then $N(alpha) =
p,-p,p^2,-p^2$
Furthermore, if $N(alpha) = p^2,-p^2$ then $alpha sim p$ where $sim$ is the associates relation on $mathbb{Z}[sqrt{n}]$.
Proof:
If $alpha$ is prime in $mathbb{Z}[sqrt{n}]$, then $N(alpha) neq 0,1,-1$ this is because $alpha$ is not a unit and the fact that $N(alpha) = alpha overline{alpha} neq 0$ comes because we are working in an integrity domain so that $alpha = 0 lor overline{alpha} = 0$ but since $n$ is square-free (by definition) it is necessary that if $alpha = a+bi$ then $a = b = 0$, therefore $alpha = 0$ and cannot be prime.
So let $N(alpha) = prod p_i$ be its prime factorization over the integers. Then $alpha|p_i$ since $alpha$ is prime. Therefore, $alpha beta = p implies N(alpha) N(beta) = p^2 implies N(alpha) in {p,-p,p^2,-p^2}$ as we wanted. Furthermore, if $N(alpha) = p,-p^2$ then $N(beta) = 1,-1$ which means that $beta$ is a unit and therefore $beta sim p$. $$tag*{$blacksquare$}$$
So now come to our problem over $mathbb{Z}[i]$:
$alpha = a+bi$ is prime in $mathbb{Z}[i] simeq p = N(a+bi) = a^2+b^2$ is prime in $mathbb{Z}$.
$impliedby)$ Since, $mathbb{Z}[i]$ is a UFD, $alpha$ is prime if and only if it is irreducible. But if, $alpha = beta gamma$ is a proper factorization of $alpha$ then $N(alpha) = N(beta) N(gamma)$ and you would have a proper factorization of $p$. Contradiction.
$implies)$ $alpha$ prime implies that $N(alpha) = p,-p,p^2,-p^2$. Since the norm is positive $N(alpha) = p,p^2$. If $N(alpha) = p$ then we are done. If $N(alpha) = p^2$ then $alpha sim p in mathbb{N}$ prime. Recalling the units of $mathbb{Z}[i]$ are $U(mathbb{Z}[i]) = {i,-i,1,-1}$ and multiplying you will see that necessarily $alpha in {p,-p,pi,-pi}$.
However, you cannot rule out this last case. For instance, take $3$ in $mathbb{Z}[i]$ which following a norm-argument can be shown to be prime, yet its norm is $p^2 = 9$ which is not prime in $mathbb{Z}$.
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
$endgroup$
$begingroup$
Could you explain the last line where you claim that $ abneq 0 $? Take $ alpha=0+pi $, then $ a=0, b=p $, but $ ab=0 $ and we still have $ alphasim p $.
$endgroup$
– user549397
Jan 27 at 14:35
$begingroup$
@user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it
$endgroup$
– Javier
Jan 27 at 17:17
add a comment |
$begingroup$
Let me add another argument that I believe is simpler. First I need to prove a little proposition.
Let $p in mathbb{N}$ be a prime and $N$ be the norm function $N(alpha) = alpha overline{alpha}$.
If $alpha in mathbb{Z}[sqrt{n}]$ is prime then $N(alpha) =
p,-p,p^2,-p^2$
Furthermore, if $N(alpha) = p^2,-p^2$ then $alpha sim p$ where $sim$ is the associates relation on $mathbb{Z}[sqrt{n}]$.
Proof:
If $alpha$ is prime in $mathbb{Z}[sqrt{n}]$, then $N(alpha) neq 0,1,-1$ this is because $alpha$ is not a unit and the fact that $N(alpha) = alpha overline{alpha} neq 0$ comes because we are working in an integrity domain so that $alpha = 0 lor overline{alpha} = 0$ but since $n$ is square-free (by definition) it is necessary that if $alpha = a+bi$ then $a = b = 0$, therefore $alpha = 0$ and cannot be prime.
So let $N(alpha) = prod p_i$ be its prime factorization over the integers. Then $alpha|p_i$ since $alpha$ is prime. Therefore, $alpha beta = p implies N(alpha) N(beta) = p^2 implies N(alpha) in {p,-p,p^2,-p^2}$ as we wanted. Furthermore, if $N(alpha) = p,-p^2$ then $N(beta) = 1,-1$ which means that $beta$ is a unit and therefore $beta sim p$. $$tag*{$blacksquare$}$$
So now come to our problem over $mathbb{Z}[i]$:
$alpha = a+bi$ is prime in $mathbb{Z}[i] simeq p = N(a+bi) = a^2+b^2$ is prime in $mathbb{Z}$.
$impliedby)$ Since, $mathbb{Z}[i]$ is a UFD, $alpha$ is prime if and only if it is irreducible. But if, $alpha = beta gamma$ is a proper factorization of $alpha$ then $N(alpha) = N(beta) N(gamma)$ and you would have a proper factorization of $p$. Contradiction.
$implies)$ $alpha$ prime implies that $N(alpha) = p,-p,p^2,-p^2$. Since the norm is positive $N(alpha) = p,p^2$. If $N(alpha) = p$ then we are done. If $N(alpha) = p^2$ then $alpha sim p in mathbb{N}$ prime. Recalling the units of $mathbb{Z}[i]$ are $U(mathbb{Z}[i]) = {i,-i,1,-1}$ and multiplying you will see that necessarily $alpha in {p,-p,pi,-pi}$.
However, you cannot rule out this last case. For instance, take $3$ in $mathbb{Z}[i]$ which following a norm-argument can be shown to be prime, yet its norm is $p^2 = 9$ which is not prime in $mathbb{Z}$.
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
$endgroup$
$begingroup$
Could you explain the last line where you claim that $ abneq 0 $? Take $ alpha=0+pi $, then $ a=0, b=p $, but $ ab=0 $ and we still have $ alphasim p $.
$endgroup$
– user549397
Jan 27 at 14:35
$begingroup$
@user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it
$endgroup$
– Javier
Jan 27 at 17:17
add a comment |
$begingroup$
Let me add another argument that I believe is simpler. First I need to prove a little proposition.
Let $p in mathbb{N}$ be a prime and $N$ be the norm function $N(alpha) = alpha overline{alpha}$.
If $alpha in mathbb{Z}[sqrt{n}]$ is prime then $N(alpha) =
p,-p,p^2,-p^2$
Furthermore, if $N(alpha) = p^2,-p^2$ then $alpha sim p$ where $sim$ is the associates relation on $mathbb{Z}[sqrt{n}]$.
Proof:
If $alpha$ is prime in $mathbb{Z}[sqrt{n}]$, then $N(alpha) neq 0,1,-1$ this is because $alpha$ is not a unit and the fact that $N(alpha) = alpha overline{alpha} neq 0$ comes because we are working in an integrity domain so that $alpha = 0 lor overline{alpha} = 0$ but since $n$ is square-free (by definition) it is necessary that if $alpha = a+bi$ then $a = b = 0$, therefore $alpha = 0$ and cannot be prime.
So let $N(alpha) = prod p_i$ be its prime factorization over the integers. Then $alpha|p_i$ since $alpha$ is prime. Therefore, $alpha beta = p implies N(alpha) N(beta) = p^2 implies N(alpha) in {p,-p,p^2,-p^2}$ as we wanted. Furthermore, if $N(alpha) = p,-p^2$ then $N(beta) = 1,-1$ which means that $beta$ is a unit and therefore $beta sim p$. $$tag*{$blacksquare$}$$
So now come to our problem over $mathbb{Z}[i]$:
$alpha = a+bi$ is prime in $mathbb{Z}[i] simeq p = N(a+bi) = a^2+b^2$ is prime in $mathbb{Z}$.
$impliedby)$ Since, $mathbb{Z}[i]$ is a UFD, $alpha$ is prime if and only if it is irreducible. But if, $alpha = beta gamma$ is a proper factorization of $alpha$ then $N(alpha) = N(beta) N(gamma)$ and you would have a proper factorization of $p$. Contradiction.
$implies)$ $alpha$ prime implies that $N(alpha) = p,-p,p^2,-p^2$. Since the norm is positive $N(alpha) = p,p^2$. If $N(alpha) = p$ then we are done. If $N(alpha) = p^2$ then $alpha sim p in mathbb{N}$ prime. Recalling the units of $mathbb{Z}[i]$ are $U(mathbb{Z}[i]) = {i,-i,1,-1}$ and multiplying you will see that necessarily $alpha in {p,-p,pi,-pi}$.
However, you cannot rule out this last case. For instance, take $3$ in $mathbb{Z}[i]$ which following a norm-argument can be shown to be prime, yet its norm is $p^2 = 9$ which is not prime in $mathbb{Z}$.
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
$endgroup$
Let me add another argument that I believe is simpler. First I need to prove a little proposition.
Let $p in mathbb{N}$ be a prime and $N$ be the norm function $N(alpha) = alpha overline{alpha}$.
If $alpha in mathbb{Z}[sqrt{n}]$ is prime then $N(alpha) =
p,-p,p^2,-p^2$
Furthermore, if $N(alpha) = p^2,-p^2$ then $alpha sim p$ where $sim$ is the associates relation on $mathbb{Z}[sqrt{n}]$.
Proof:
If $alpha$ is prime in $mathbb{Z}[sqrt{n}]$, then $N(alpha) neq 0,1,-1$ this is because $alpha$ is not a unit and the fact that $N(alpha) = alpha overline{alpha} neq 0$ comes because we are working in an integrity domain so that $alpha = 0 lor overline{alpha} = 0$ but since $n$ is square-free (by definition) it is necessary that if $alpha = a+bi$ then $a = b = 0$, therefore $alpha = 0$ and cannot be prime.
So let $N(alpha) = prod p_i$ be its prime factorization over the integers. Then $alpha|p_i$ since $alpha$ is prime. Therefore, $alpha beta = p implies N(alpha) N(beta) = p^2 implies N(alpha) in {p,-p,p^2,-p^2}$ as we wanted. Furthermore, if $N(alpha) = p,-p^2$ then $N(beta) = 1,-1$ which means that $beta$ is a unit and therefore $beta sim p$. $$tag*{$blacksquare$}$$
So now come to our problem over $mathbb{Z}[i]$:
$alpha = a+bi$ is prime in $mathbb{Z}[i] simeq p = N(a+bi) = a^2+b^2$ is prime in $mathbb{Z}$.
$impliedby)$ Since, $mathbb{Z}[i]$ is a UFD, $alpha$ is prime if and only if it is irreducible. But if, $alpha = beta gamma$ is a proper factorization of $alpha$ then $N(alpha) = N(beta) N(gamma)$ and you would have a proper factorization of $p$. Contradiction.
$implies)$ $alpha$ prime implies that $N(alpha) = p,-p,p^2,-p^2$. Since the norm is positive $N(alpha) = p,p^2$. If $N(alpha) = p$ then we are done. If $N(alpha) = p^2$ then $alpha sim p in mathbb{N}$ prime. Recalling the units of $mathbb{Z}[i]$ are $U(mathbb{Z}[i]) = {i,-i,1,-1}$ and multiplying you will see that necessarily $alpha in {p,-p,pi,-pi}$.
However, you cannot rule out this last case. For instance, take $3$ in $mathbb{Z}[i]$ which following a norm-argument can be shown to be prime, yet its norm is $p^2 = 9$ which is not prime in $mathbb{Z}$.
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
edited Jan 27 at 17:21
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
answered Jan 18 '18 at 11:20
JavierJavier
2,08021234
2,08021234
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Could you explain the last line where you claim that $ abneq 0 $? Take $ alpha=0+pi $, then $ a=0, b=p $, but $ ab=0 $ and we still have $ alphasim p $.
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– user549397
Jan 27 at 14:35
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@user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it
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– Javier
Jan 27 at 17:17
add a comment |
$begingroup$
Could you explain the last line where you claim that $ abneq 0 $? Take $ alpha=0+pi $, then $ a=0, b=p $, but $ ab=0 $ and we still have $ alphasim p $.
$endgroup$
– user549397
Jan 27 at 14:35
$begingroup$
@user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it
$endgroup$
– Javier
Jan 27 at 17:17
$begingroup$
Could you explain the last line where you claim that $ abneq 0 $? Take $ alpha=0+pi $, then $ a=0, b=p $, but $ ab=0 $ and we still have $ alphasim p $.
$endgroup$
– user549397
Jan 27 at 14:35
$begingroup$
Could you explain the last line where you claim that $ abneq 0 $? Take $ alpha=0+pi $, then $ a=0, b=p $, but $ ab=0 $ and we still have $ alphasim p $.
$endgroup$
– user549397
Jan 27 at 14:35
$begingroup$
@user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it
$endgroup$
– Javier
Jan 27 at 17:17
$begingroup$
@user549397 you're right it seems i made a mistake, indeed above they are pointing the conjecture is wrong, so let me correct it
$endgroup$
– Javier
Jan 27 at 17:17
add a comment |
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$a^2+b^2=(a+ib)(a-ib)$
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– Mathematician
Apr 21 '13 at 4:35
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this is obvious.but I could not understand how should I use this result
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– habuji
Apr 21 '13 at 4:40
13
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False. Three is prime in $mathbb{Z}[i]$, yet $3^2+0^2=9$ is not prime in $mathbb{Z}$.
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– Jyrki Lahtonen
Apr 21 '13 at 4:43
1
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If the norm is prime, it can't be the product of Norms of two other numbers unless one is a unit. However the converse isn't always true.
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– Macavity
Apr 21 '13 at 4:57
2
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you would need restrictions on your statements, such as both a, b are nonzero, as @Jyrki has shown the hole there. Also, $a^2+b^2$ can not be of the form $4k+3$ (again, as is 3).
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– Eleven-Eleven
Apr 21 '13 at 5:01