An element lies in a prime ideal but not in its square
$begingroup$
Let $R$ be a UFD. I'm trying to understand whether the following statements are equivalent:
1) $rin R$ lies in some prime ideal $P$ but not in the ideal $P^2$.
2) $r$ is divisible by a prime element of $R$ but not by its square.
I think I can see why 2) implies 1). Suppose 2) holds, say $r$ is divisible by a prime element $p$ but not by $p^2$. Then $r$ lies in the ideal $P=(p)$, which is prime because $p$ is prime. Also $r$ does not lie in $P^2$, because otherwise it would be divisible by $p^2$. Is that right?
Suppose 1) holds. $P$ is an abstract prime ideal. Does it necessarily contain a prime element? If so, I would be able to conclude that $r$ is divisible by that prime element. (I think it must be finitely generated since $R$ is a UFD, do the generators have to be prime elements?) Does the condition $rnotin P^2$ imply that $r $ is not divisible by the square of that element?
abstract-algebra ideals maximal-and-prime-ideals unique-factorization-domains
asked Dec 1 '18 at 19:22
user437309user437309
630212
$endgroup$
add a comment |
$begingroup$
Let $R$ be a UFD. I'm trying to understand whether the following statements are equivalent:
1) $rin R$ lies in some prime ideal $P$ but not in the ideal $P^2$.
2) $r$ is divisible by a prime element of $R$ but not by its square.
I think I can see why 2) implies 1). Suppose 2) holds, say $r$ is divisible by a prime element $p$ but not by $p^2$. Then $r$ lies in the ideal $P=(p)$, which is prime because $p$ is prime. Also $r$ does not lie in $P^2$, because otherwise it would be divisible by $p^2$. Is that right?
Suppose 1) holds. $P$ is an abstract prime ideal. Does it necessarily contain a prime element? If so, I would be able to conclude that $r$ is divisible by that prime element. (I think it must be finitely generated since $R$ is a UFD, do the generators have to be prime elements?) Does the condition $rnotin P^2$ imply that $r $ is not divisible by the square of that element?
abstract-algebra ideals maximal-and-prime-ideals unique-factorization-domains
asked Dec 1 '18 at 19:22
user437309user437309
630212
$endgroup$
add a comment |
$begingroup$
Let $R$ be a UFD. I'm trying to understand whether the following statements are equivalent:
1) $rin R$ lies in some prime ideal $P$ but not in the ideal $P^2$.
2) $r$ is divisible by a prime element of $R$ but not by its square.
I think I can see why 2) implies 1). Suppose 2) holds, say $r$ is divisible by a prime element $p$ but not by $p^2$. Then $r$ lies in the ideal $P=(p)$, which is prime because $p$ is prime. Also $r$ does not lie in $P^2$, because otherwise it would be divisible by $p^2$. Is that right?
Suppose 1) holds. $P$ is an abstract prime ideal. Does it necessarily contain a prime element? If so, I would be able to conclude that $r$ is divisible by that prime element. (I think it must be finitely generated since $R$ is a UFD, do the generators have to be prime elements?) Does the condition $rnotin P^2$ imply that $r $ is not divisible by the square of that element?
abstract-algebra ideals maximal-and-prime-ideals unique-factorization-domains
asked Dec 1 '18 at 19:22
user437309user437309
630212
$endgroup$
Let $R$ be a UFD. I'm trying to understand whether the following statements are equivalent:
1) $rin R$ lies in some prime ideal $P$ but not in the ideal $P^2$.
2) $r$ is divisible by a prime element of $R$ but not by its square.
I think I can see why 2) implies 1). Suppose 2) holds, say $r$ is divisible by a prime element $p$ but not by $p^2$. Then $r$ lies in the ideal $P=(p)$, which is prime because $p$ is prime. Also $r$ does not lie in $P^2$, because otherwise it would be divisible by $p^2$. Is that right?
Suppose 1) holds. $P$ is an abstract prime ideal. Does it necessarily contain a prime element? If so, I would be able to conclude that $r$ is divisible by that prime element. (I think it must be finitely generated since $R$ is a UFD, do the generators have to be prime elements?) Does the condition $rnotin P^2$ imply that $r $ is not divisible by the square of that element?
abstract-algebra ideals maximal-and-prime-ideals unique-factorization-domains
abstract-algebra ideals maximal-and-prime-ideals unique-factorization-domains
asked Dec 1 '18 at 19:22
user437309user437309
630212
asked Dec 1 '18 at 19:22
user437309user437309
630212
edited Dec 1 '18 at 19:47
user437309
asked Dec 1 '18 at 19:22
user437309user437309
630212
asked Dec 1 '18 at 19:22
user437309user437309
630212
asked Dec 1 '18 at 19:22
user437309user437309
630212
630212
add a comment |
add a comment |
1 Answer
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Your reasoning that $(2) implies (1)$ is correct.
As for your questions in the end:
It is not true that $P$ must be finitely generated. If every prime ideal of a ring is finitely generated, then in fact every ideal is finitely generated, i.e. the ring is Noetherian (this is sometimes called Cohen's theorem). In general, UFDs are not Noetherian. A classic example is $K[x_1, x_2, x_3, ldots]$, the ring of polynomials in an infinite number of variables over a field.
However, this is not important for your problem.
In a UFD we can factor $r$ as $r = prod p_i^{n_i}$ with the $p_i$ distinct primes.
Show from the definition of prime ideal that $r in P$ prime implies $p_i in P$ for some $i$. Remember, $P$ prime means that $ab in P implies ain P text{ or } bin P$. (In general you've now shown that a prime ideal of a UFD always contains a prime element. In fact this behavior characterizes UFDs, and is sometimes called Kaplansky's theorem)
Finally, show from this and the assumption that $r notin P^2$ that $p_i mid r$ but $p_i^2 nmid r$
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
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add a comment |
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$begingroup$
Your reasoning that $(2) implies (1)$ is correct.
As for your questions in the end:
It is not true that $P$ must be finitely generated. If every prime ideal of a ring is finitely generated, then in fact every ideal is finitely generated, i.e. the ring is Noetherian (this is sometimes called Cohen's theorem). In general, UFDs are not Noetherian. A classic example is $K[x_1, x_2, x_3, ldots]$, the ring of polynomials in an infinite number of variables over a field.
However, this is not important for your problem.
In a UFD we can factor $r$ as $r = prod p_i^{n_i}$ with the $p_i$ distinct primes.
Show from the definition of prime ideal that $r in P$ prime implies $p_i in P$ for some $i$. Remember, $P$ prime means that $ab in P implies ain P text{ or } bin P$. (In general you've now shown that a prime ideal of a UFD always contains a prime element. In fact this behavior characterizes UFDs, and is sometimes called Kaplansky's theorem)
Finally, show from this and the assumption that $r notin P^2$ that $p_i mid r$ but $p_i^2 nmid r$
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
$endgroup$
add a comment |
$begingroup$
Your reasoning that $(2) implies (1)$ is correct.
As for your questions in the end:
It is not true that $P$ must be finitely generated. If every prime ideal of a ring is finitely generated, then in fact every ideal is finitely generated, i.e. the ring is Noetherian (this is sometimes called Cohen's theorem). In general, UFDs are not Noetherian. A classic example is $K[x_1, x_2, x_3, ldots]$, the ring of polynomials in an infinite number of variables over a field.
However, this is not important for your problem.
In a UFD we can factor $r$ as $r = prod p_i^{n_i}$ with the $p_i$ distinct primes.
Show from the definition of prime ideal that $r in P$ prime implies $p_i in P$ for some $i$. Remember, $P$ prime means that $ab in P implies ain P text{ or } bin P$. (In general you've now shown that a prime ideal of a UFD always contains a prime element. In fact this behavior characterizes UFDs, and is sometimes called Kaplansky's theorem)
Finally, show from this and the assumption that $r notin P^2$ that $p_i mid r$ but $p_i^2 nmid r$
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
$endgroup$
add a comment |
$begingroup$
Your reasoning that $(2) implies (1)$ is correct.
As for your questions in the end:
It is not true that $P$ must be finitely generated. If every prime ideal of a ring is finitely generated, then in fact every ideal is finitely generated, i.e. the ring is Noetherian (this is sometimes called Cohen's theorem). In general, UFDs are not Noetherian. A classic example is $K[x_1, x_2, x_3, ldots]$, the ring of polynomials in an infinite number of variables over a field.
However, this is not important for your problem.
In a UFD we can factor $r$ as $r = prod p_i^{n_i}$ with the $p_i$ distinct primes.
Show from the definition of prime ideal that $r in P$ prime implies $p_i in P$ for some $i$. Remember, $P$ prime means that $ab in P implies ain P text{ or } bin P$. (In general you've now shown that a prime ideal of a UFD always contains a prime element. In fact this behavior characterizes UFDs, and is sometimes called Kaplansky's theorem)
Finally, show from this and the assumption that $r notin P^2$ that $p_i mid r$ but $p_i^2 nmid r$
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
$endgroup$
Your reasoning that $(2) implies (1)$ is correct.
As for your questions in the end:
It is not true that $P$ must be finitely generated. If every prime ideal of a ring is finitely generated, then in fact every ideal is finitely generated, i.e. the ring is Noetherian (this is sometimes called Cohen's theorem). In general, UFDs are not Noetherian. A classic example is $K[x_1, x_2, x_3, ldots]$, the ring of polynomials in an infinite number of variables over a field.
However, this is not important for your problem.
In a UFD we can factor $r$ as $r = prod p_i^{n_i}$ with the $p_i$ distinct primes.
Show from the definition of prime ideal that $r in P$ prime implies $p_i in P$ for some $i$. Remember, $P$ prime means that $ab in P implies ain P text{ or } bin P$. (In general you've now shown that a prime ideal of a UFD always contains a prime element. In fact this behavior characterizes UFDs, and is sometimes called Kaplansky's theorem)
Finally, show from this and the assumption that $r notin P^2$ that $p_i mid r$ but $p_i^2 nmid r$
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
answered Dec 1 '18 at 22:46
Badam BaplanBadam Baplan
4,476722
4,476722
add a comment |
add a comment |
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