Mixing
prove $(n)$ prime ideal of $mathbb{Z}$ iff $n$ is prime or zero
up vote
5
down vote
favorite
prove $(n)$ prime ideal of $mathbb{Z}$ iff $n$ is prime or zero
Defintions
Def of prime Ideal (n)
$$ abin (n) implies ain(n) vee bin(n) $$
Def 1] integer n is prime if $n neq 0,pm 1 $ and only divisors are $pm n,pm 1$
Def 2 of n is prime] If $nneq0,pm1$ only divisors of n are $pm1,pm n$
$$ n|ab implies n|a vee n|b $$
$Rightarrow $] (Prime Ideal $(n)$ of $mathbb{Z} $$Rightarrow$ $n$ prime or zero)
Now consider the case where $(n)neq (0)$. that is $nneq 0$
Using the def of prime Ideals $$begin{aligned} ab in (n) implies a in(n) vee b in (n) end{aligned} $$
Well, If an element $xin(n) iff x=q*n iff n|x$
$$n|ab implies n|a vee n|b$$
So, $n$ is prime ,nonzero.
In the case that $(n)=(0)$ clearly $n=0$ since there are no zero divisors in $mathbb{Z}$
$Leftarrow$] (n is prime or zero $Rightarrow $ $(n)$ is a prime ideal of $mathbb{Z}$)
Consider the case where $p is prime$ so $nneq 0, pm1$
$$ begin{aligned}
n|ab &implies n|a vee n|b \
ab in (n) &implies a in (n) vee bin(n) end{aligned}$$
in the case $n=0$, since $mathbb{Z}$ has no zero divisors
$$ab=0 implies a=0 vee b=0 $$
So (0) is a prime ideal.
Concern if this prove holds, also would be surprised if this question is not out there in this site. Did a search and clicked on similar questions and could not find it.And of course any other ways to prove it.
abstract-algebra prime-numbers ideals maximal-and-prime-ideals
asked Dec 8 '15 at 17:24
Tiger Blood
817724
add a comment |
up vote
5
down vote
favorite
prove $(n)$ prime ideal of $mathbb{Z}$ iff $n$ is prime or zero
Defintions
Def of prime Ideal (n)
$$ abin (n) implies ain(n) vee bin(n) $$
Def 1] integer n is prime if $n neq 0,pm 1 $ and only divisors are $pm n,pm 1$
Def 2 of n is prime] If $nneq0,pm1$ only divisors of n are $pm1,pm n$
$$ n|ab implies n|a vee n|b $$
$Rightarrow $] (Prime Ideal $(n)$ of $mathbb{Z} $$Rightarrow$ $n$ prime or zero)
Now consider the case where $(n)neq (0)$. that is $nneq 0$
Using the def of prime Ideals $$begin{aligned} ab in (n) implies a in(n) vee b in (n) end{aligned} $$
Well, If an element $xin(n) iff x=q*n iff n|x$
$$n|ab implies n|a vee n|b$$
So, $n$ is prime ,nonzero.
In the case that $(n)=(0)$ clearly $n=0$ since there are no zero divisors in $mathbb{Z}$
$Leftarrow$] (n is prime or zero $Rightarrow $ $(n)$ is a prime ideal of $mathbb{Z}$)
Consider the case where $p is prime$ so $nneq 0, pm1$
$$ begin{aligned}
n|ab &implies n|a vee n|b \
ab in (n) &implies a in (n) vee bin(n) end{aligned}$$
in the case $n=0$, since $mathbb{Z}$ has no zero divisors
$$ab=0 implies a=0 vee b=0 $$
So (0) is a prime ideal.
Concern if this prove holds, also would be surprised if this question is not out there in this site. Did a search and clicked on similar questions and could not find it.And of course any other ways to prove it.
abstract-algebra prime-numbers ideals maximal-and-prime-ideals
asked Dec 8 '15 at 17:24
Tiger Blood
817724
1
Can you use the fact that $mathbb{Z}/(n)$ is a field if $n$ is prime?
– Brandon Thomas Van Over
Dec 8 '15 at 21:56
$Z/(n)$ is a field iff $(n)$ is a maximal. So every maximal is a prime that is $(n)$ is prime. Thanks @SirJective :)
– Tiger Blood
Dec 8 '15 at 22:06
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
prove $(n)$ prime ideal of $mathbb{Z}$ iff $n$ is prime or zero
Defintions
Def of prime Ideal (n)
$$ abin (n) implies ain(n) vee bin(n) $$
Def 1] integer n is prime if $n neq 0,pm 1 $ and only divisors are $pm n,pm 1$
Def 2 of n is prime] If $nneq0,pm1$ only divisors of n are $pm1,pm n$
$$ n|ab implies n|a vee n|b $$
$Rightarrow $] (Prime Ideal $(n)$ of $mathbb{Z} $$Rightarrow$ $n$ prime or zero)
Now consider the case where $(n)neq (0)$. that is $nneq 0$
Using the def of prime Ideals $$begin{aligned} ab in (n) implies a in(n) vee b in (n) end{aligned} $$
Well, If an element $xin(n) iff x=q*n iff n|x$
$$n|ab implies n|a vee n|b$$
So, $n$ is prime ,nonzero.
In the case that $(n)=(0)$ clearly $n=0$ since there are no zero divisors in $mathbb{Z}$
$Leftarrow$] (n is prime or zero $Rightarrow $ $(n)$ is a prime ideal of $mathbb{Z}$)
Consider the case where $p is prime$ so $nneq 0, pm1$
$$ begin{aligned}
n|ab &implies n|a vee n|b \
ab in (n) &implies a in (n) vee bin(n) end{aligned}$$
in the case $n=0$, since $mathbb{Z}$ has no zero divisors
$$ab=0 implies a=0 vee b=0 $$
So (0) is a prime ideal.
Concern if this prove holds, also would be surprised if this question is not out there in this site. Did a search and clicked on similar questions and could not find it.And of course any other ways to prove it.
abstract-algebra prime-numbers ideals maximal-and-prime-ideals
asked Dec 8 '15 at 17:24
Tiger Blood
817724
prove $(n)$ prime ideal of $mathbb{Z}$ iff $n$ is prime or zero
Defintions
Def of prime Ideal (n)
$$ abin (n) implies ain(n) vee bin(n) $$
Def 1] integer n is prime if $n neq 0,pm 1 $ and only divisors are $pm n,pm 1$
Def 2 of n is prime] If $nneq0,pm1$ only divisors of n are $pm1,pm n$
$$ n|ab implies n|a vee n|b $$
$Rightarrow $] (Prime Ideal $(n)$ of $mathbb{Z} $$Rightarrow$ $n$ prime or zero)
Now consider the case where $(n)neq (0)$. that is $nneq 0$
Using the def of prime Ideals $$begin{aligned} ab in (n) implies a in(n) vee b in (n) end{aligned} $$
Well, If an element $xin(n) iff x=q*n iff n|x$
$$n|ab implies n|a vee n|b$$
So, $n$ is prime ,nonzero.
In the case that $(n)=(0)$ clearly $n=0$ since there are no zero divisors in $mathbb{Z}$
$Leftarrow$] (n is prime or zero $Rightarrow $ $(n)$ is a prime ideal of $mathbb{Z}$)
Consider the case where $p is prime$ so $nneq 0, pm1$
$$ begin{aligned}
n|ab &implies n|a vee n|b \
ab in (n) &implies a in (n) vee bin(n) end{aligned}$$
in the case $n=0$, since $mathbb{Z}$ has no zero divisors
$$ab=0 implies a=0 vee b=0 $$
So (0) is a prime ideal.
Concern if this prove holds, also would be surprised if this question is not out there in this site. Did a search and clicked on similar questions and could not find it.And of course any other ways to prove it.
abstract-algebra prime-numbers ideals maximal-and-prime-ideals
abstract-algebra prime-numbers ideals maximal-and-prime-ideals
asked Dec 8 '15 at 17:24
Tiger Blood
817724
asked Dec 8 '15 at 17:24
Tiger Blood
817724
asked Dec 8 '15 at 17:24
Tiger Blood
817724
asked Dec 8 '15 at 17:24
Tiger Blood
817724
asked Dec 8 '15 at 17:24
Tiger Blood
817724
817724
1
Can you use the fact that $mathbb{Z}/(n)$ is a field if $n$ is prime?
– Brandon Thomas Van Over
Dec 8 '15 at 21:56
$Z/(n)$ is a field iff $(n)$ is a maximal. So every maximal is a prime that is $(n)$ is prime. Thanks @SirJective :)
– Tiger Blood
Dec 8 '15 at 22:06
add a comment |
1
Can you use the fact that $mathbb{Z}/(n)$ is a field if $n$ is prime?
– Brandon Thomas Van Over
Dec 8 '15 at 21:56
$Z/(n)$ is a field iff $(n)$ is a maximal. So every maximal is a prime that is $(n)$ is prime. Thanks @SirJective :)
– Tiger Blood
Dec 8 '15 at 22:06
1
1
Can you use the fact that $mathbb{Z}/(n)$ is a field if $n$ is prime?
– Brandon Thomas Van Over
Dec 8 '15 at 21:56
Can you use the fact that $mathbb{Z}/(n)$ is a field if $n$ is prime?
– Brandon Thomas Van Over
Dec 8 '15 at 21:56
$Z/(n)$ is a field iff $(n)$ is a maximal. So every maximal is a prime that is $(n)$ is prime. Thanks @SirJective :)
– Tiger Blood
Dec 8 '15 at 22:06
$Z/(n)$ is a field iff $(n)$ is a maximal. So every maximal is a prime that is $(n)$ is prime. Thanks @SirJective :)
– Tiger Blood
Dec 8 '15 at 22:06
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Another way would be to show that $mathbb{Z}$ is a principal ideal domain or that it has unique factorization. Don't you also need the definition that a prime ideal has to be properly contained within the whole ring?
If $n = pm 1$, then $langle n rangle = mathbb{Z}$ and thus it can't be a prime ideal. If $n$ is composite and divisible by some prime $p$, then $langle n rangle$ is properly contained within $langle p rangle$ and thus $langle n rangle$ is not a prime ideal either.
And then you just proceed with what you have already demonstrated.
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
add a comment |
up vote
0
down vote
Yes, from $p mid a iff a in (p)$ we have more generally in a commutative ring with unity, $(p)$ is a prime ideal iff $p$ is prime (in the ring theory sense: $p mid ab implies p mid a vee p mid b$)
answered 5 hours ago
qwr
6,53842654
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Another way would be to show that $mathbb{Z}$ is a principal ideal domain or that it has unique factorization. Don't you also need the definition that a prime ideal has to be properly contained within the whole ring?
If $n = pm 1$, then $langle n rangle = mathbb{Z}$ and thus it can't be a prime ideal. If $n$ is composite and divisible by some prime $p$, then $langle n rangle$ is properly contained within $langle p rangle$ and thus $langle n rangle$ is not a prime ideal either.
And then you just proceed with what you have already demonstrated.
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
add a comment |
up vote
2
down vote
accepted
Another way would be to show that $mathbb{Z}$ is a principal ideal domain or that it has unique factorization. Don't you also need the definition that a prime ideal has to be properly contained within the whole ring?
If $n = pm 1$, then $langle n rangle = mathbb{Z}$ and thus it can't be a prime ideal. If $n$ is composite and divisible by some prime $p$, then $langle n rangle$ is properly contained within $langle p rangle$ and thus $langle n rangle$ is not a prime ideal either.
And then you just proceed with what you have already demonstrated.
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Another way would be to show that $mathbb{Z}$ is a principal ideal domain or that it has unique factorization. Don't you also need the definition that a prime ideal has to be properly contained within the whole ring?
If $n = pm 1$, then $langle n rangle = mathbb{Z}$ and thus it can't be a prime ideal. If $n$ is composite and divisible by some prime $p$, then $langle n rangle$ is properly contained within $langle p rangle$ and thus $langle n rangle$ is not a prime ideal either.
And then you just proceed with what you have already demonstrated.
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
Another way would be to show that $mathbb{Z}$ is a principal ideal domain or that it has unique factorization. Don't you also need the definition that a prime ideal has to be properly contained within the whole ring?
If $n = pm 1$, then $langle n rangle = mathbb{Z}$ and thus it can't be a prime ideal. If $n$ is composite and divisible by some prime $p$, then $langle n rangle$ is properly contained within $langle p rangle$ and thus $langle n rangle$ is not a prime ideal either.
And then you just proceed with what you have already demonstrated.
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
answered Dec 9 '15 at 3:25
Robert Soupe
10.6k21948
10.6k21948
add a comment |
add a comment |
up vote
0
down vote
Yes, from $p mid a iff a in (p)$ we have more generally in a commutative ring with unity, $(p)$ is a prime ideal iff $p$ is prime (in the ring theory sense: $p mid ab implies p mid a vee p mid b$)
answered 5 hours ago
qwr
6,53842654
add a comment |
up vote
0
down vote
Yes, from $p mid a iff a in (p)$ we have more generally in a commutative ring with unity, $(p)$ is a prime ideal iff $p$ is prime (in the ring theory sense: $p mid ab implies p mid a vee p mid b$)
answered 5 hours ago
qwr
6,53842654
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes, from $p mid a iff a in (p)$ we have more generally in a commutative ring with unity, $(p)$ is a prime ideal iff $p$ is prime (in the ring theory sense: $p mid ab implies p mid a vee p mid b$)
answered 5 hours ago
qwr
6,53842654
Yes, from $p mid a iff a in (p)$ we have more generally in a commutative ring with unity, $(p)$ is a prime ideal iff $p$ is prime (in the ring theory sense: $p mid ab implies p mid a vee p mid b$)
answered 5 hours ago
qwr
6,53842654
answered 5 hours ago
qwr
6,53842654
answered 5 hours ago
qwr
6,53842654
answered 5 hours ago
qwr
6,53842654
6,53842654
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1566111%2fprove-n-prime-ideal-of-mathbbz-iff-n-is-prime-or-zero%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Can you use the fact that $mathbb{Z}/(n)$ is a field if $n$ is prime?
– Brandon Thomas Van Over
Dec 8 '15 at 21:56
$Z/(n)$ is a field iff $(n)$ is a maximal. So every maximal is a prime that is $(n)$ is prime. Thanks @SirJective :)
– Tiger Blood
Dec 8 '15 at 22:06