Factorization in prime elements of $mathbb{Z}left[sqrt{p}right]$ for a prime number $p$
I'm having troubles with the following problem:
Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.
At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.
Thanks in advance.
abstract-algebra ring-theory integers
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
add a comment |
I'm having troubles with the following problem:
Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.
At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.
Thanks in advance.
abstract-algebra ring-theory integers
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
add a comment |
I'm having troubles with the following problem:
Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.
At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.
Thanks in advance.
abstract-algebra ring-theory integers
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
I'm having troubles with the following problem:
Let $p$ be a prime number in $mathbb{Z}$, and $alphainmathbb{Z}left[sqrt{p}right]$ which is not a unit. Prove that $alpha$ have a factorization in irreducible elements of $mathbb{Z}left[sqrt{p}right]$.
At the beginning I though that that ring was a Euclidean Domain, but that fails for $mathbb{Z}left[sqrt{5}right]$. So, I don't know where to start now.
Thanks in advance.
abstract-algebra ring-theory integers
abstract-algebra ring-theory integers
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
edited Dec 10 '18 at 14:37
Eleven-Eleven
5,39072659
5,39072659
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
asked Dec 10 '18 at 14:31
José Alejandro Aburto Araneda
756110
756110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.
Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.
answered Dec 10 '18 at 14:53
lhf
163k10167386
I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
– Matt Samuel
Dec 10 '18 at 14:59
1
@MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
– Bill Dubuque
Dec 10 '18 at 21:43
@Bill Herstein and Artin undergraduate, Hungerford graduate.
– Matt Samuel
Dec 10 '18 at 23:29
@Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
– Bill Dubuque
Dec 10 '18 at 23:56
@Bill Then either we didn't cover them, or, heaven forbid...I forgot!
– Matt Samuel
Dec 10 '18 at 23:58
add a comment |
Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
$$alpha=a_1a_2$$
If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
$$alpha = a_3a_4a_2$$
In general there is an increasing chain of ideals
$$(a_1)subseteq (a_3)subseteq cdots$$
Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
$$alpha = a_1a_2$$
with $a_1$ irreducible. Thus you can iterate this to write
$$alpha = a_1cdots a_n$$
with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
$$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
This must also stabilize, until you end up with a factorization into irreducible elements.
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
At this level, one should justify the Noetherian claim.
– Bill Dubuque
Dec 10 '18 at 21:45
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.
Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.
answered Dec 10 '18 at 14:53
lhf
163k10167386
I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
– Matt Samuel
Dec 10 '18 at 14:59
1
@MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
– Bill Dubuque
Dec 10 '18 at 21:43
@Bill Herstein and Artin undergraduate, Hungerford graduate.
– Matt Samuel
Dec 10 '18 at 23:29
@Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
– Bill Dubuque
Dec 10 '18 at 23:56
@Bill Then either we didn't cover them, or, heaven forbid...I forgot!
– Matt Samuel
Dec 10 '18 at 23:58
add a comment |
Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.
Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.
answered Dec 10 '18 at 14:53
lhf
163k10167386
I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
– Matt Samuel
Dec 10 '18 at 14:59
1
@MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
– Bill Dubuque
Dec 10 '18 at 21:43
@Bill Herstein and Artin undergraduate, Hungerford graduate.
– Matt Samuel
Dec 10 '18 at 23:29
@Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
– Bill Dubuque
Dec 10 '18 at 23:56
@Bill Then either we didn't cover them, or, heaven forbid...I forgot!
– Matt Samuel
Dec 10 '18 at 23:58
add a comment |
Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.
Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.
answered Dec 10 '18 at 14:53
lhf
163k10167386
Hint: Consider the norm $N(a+bsqrt{p})=|a^2-b^2p|$. Prove that if $delta$ divides $alpha$, then $N(delta) le N(alpha)$. Then use induction on $N(alpha)$.
Or, more sophisticatedly, argue that $mathbb{Z}left[sqrt{p}right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.
answered Dec 10 '18 at 14:53
lhf
163k10167386
edited Dec 10 '18 at 15:52
answered Dec 10 '18 at 14:53
lhf
163k10167386
answered Dec 10 '18 at 14:53
lhf
163k10167386
answered Dec 10 '18 at 14:53
lhf
163k10167386
163k10167386
I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
– Matt Samuel
Dec 10 '18 at 14:59
1
@MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
– Bill Dubuque
Dec 10 '18 at 21:43
@Bill Herstein and Artin undergraduate, Hungerford graduate.
– Matt Samuel
Dec 10 '18 at 23:29
@Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
– Bill Dubuque
Dec 10 '18 at 23:56
@Bill Then either we didn't cover them, or, heaven forbid...I forgot!
– Matt Samuel
Dec 10 '18 at 23:58
add a comment |
I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
– Matt Samuel
Dec 10 '18 at 14:59
1
@MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
– Bill Dubuque
Dec 10 '18 at 21:43
@Bill Herstein and Artin undergraduate, Hungerford graduate.
– Matt Samuel
Dec 10 '18 at 23:29
@Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
– Bill Dubuque
Dec 10 '18 at 23:56
@Bill Then either we didn't cover them, or, heaven forbid...I forgot!
– Matt Samuel
Dec 10 '18 at 23:58
I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
– Matt Samuel
Dec 10 '18 at 14:59
I never learned about norms, ever, certainly not in an introductory algebra course, and I have a PhD focused mostly on algebra.
– Matt Samuel
Dec 10 '18 at 14:59
1
1
@MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
– Bill Dubuque
Dec 10 '18 at 21:43
@MattSamuel That's a rather serious gap. What algebra textbook(s) did you use?
– Bill Dubuque
Dec 10 '18 at 21:43
@Bill Herstein and Artin undergraduate, Hungerford graduate.
– Matt Samuel
Dec 10 '18 at 23:29
@Bill Herstein and Artin undergraduate, Hungerford graduate.
– Matt Samuel
Dec 10 '18 at 23:29
@Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
– Bill Dubuque
Dec 10 '18 at 23:56
@Matt Norms are in fact discussed in Artin (Ch. 13 on Quadratic Number Fields) and also in Hungerford (V.7 on Cyclic Extensions). It's not too surprising that it is missing from Herstein since that is much shorter so limited in the topics it can include.
– Bill Dubuque
Dec 10 '18 at 23:56
@Bill Then either we didn't cover them, or, heaven forbid...I forgot!
– Matt Samuel
Dec 10 '18 at 23:58
@Bill Then either we didn't cover them, or, heaven forbid...I forgot!
– Matt Samuel
Dec 10 '18 at 23:58
add a comment |
Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
$$alpha=a_1a_2$$
If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
$$alpha = a_3a_4a_2$$
In general there is an increasing chain of ideals
$$(a_1)subseteq (a_3)subseteq cdots$$
Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
$$alpha = a_1a_2$$
with $a_1$ irreducible. Thus you can iterate this to write
$$alpha = a_1cdots a_n$$
with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
$$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
This must also stabilize, until you end up with a factorization into irreducible elements.
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
At this level, one should justify the Noetherian claim.
– Bill Dubuque
Dec 10 '18 at 21:45
add a comment |
Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
$$alpha=a_1a_2$$
If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
$$alpha = a_3a_4a_2$$
In general there is an increasing chain of ideals
$$(a_1)subseteq (a_3)subseteq cdots$$
Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
$$alpha = a_1a_2$$
with $a_1$ irreducible. Thus you can iterate this to write
$$alpha = a_1cdots a_n$$
with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
$$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
This must also stabilize, until you end up with a factorization into irreducible elements.
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
At this level, one should justify the Noetherian claim.
– Bill Dubuque
Dec 10 '18 at 21:45
add a comment |
Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
$$alpha=a_1a_2$$
If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
$$alpha = a_3a_4a_2$$
In general there is an increasing chain of ideals
$$(a_1)subseteq (a_3)subseteq cdots$$
Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
$$alpha = a_1a_2$$
with $a_1$ irreducible. Thus you can iterate this to write
$$alpha = a_1cdots a_n$$
with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
$$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
This must also stabilize, until you end up with a factorization into irreducible elements.
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
Suppose $alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
$$alpha=a_1a_2$$
If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
$$alpha = a_3a_4a_2$$
In general there is an increasing chain of ideals
$$(a_1)subseteq (a_3)subseteq cdots$$
Since $mathbb{Z}[sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
$$alpha = a_1a_2$$
with $a_1$ irreducible. Thus you can iterate this to write
$$alpha = a_1cdots a_n$$
with $a_1,ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
$$(a_{n,1})subseteq (a_{n+1,2})subseteq cdots$$
This must also stabilize, until you end up with a factorization into irreducible elements.
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
answered Dec 10 '18 at 14:58
Matt Samuel
37.3k63465
37.3k63465
At this level, one should justify the Noetherian claim.
– Bill Dubuque
Dec 10 '18 at 21:45
add a comment |
At this level, one should justify the Noetherian claim.
– Bill Dubuque
Dec 10 '18 at 21:45
At this level, one should justify the Noetherian claim.
– Bill Dubuque
Dec 10 '18 at 21:45
At this level, one should justify the Noetherian claim.
– Bill Dubuque
Dec 10 '18 at 21:45
add a comment |
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