Mixing
Find a maximal ideal in $R = mathbb{Z}[sqrt{−5}]$ containing the principal ideal (3)
$begingroup$
I think I need to choose an element to pair it with 3, i.e. (3,x). But I don't see how to find such an element.
abstract-algebra algebraic-number-theory
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
asked Jan 16 at 18:49
davidhdavidh
1745
$endgroup$
|
show 3 more comments
$begingroup$
I think I need to choose an element to pair it with 3, i.e. (3,x). But I don't see how to find such an element.
abstract-algebra algebraic-number-theory
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
asked Jan 16 at 18:49
davidhdavidh
1745
$endgroup$
$begingroup$
An element with norm that is a multiple of 3 might help. The norm is the mapping $N(a+bsqrt{-5})=a^2+5b^2$. Hint: the missing element appears in the "famous" non-unique factorization in this ring.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:02
$begingroup$
@JyrkiLahtonen Okay, so I know 3 is irreducible but not prime in this ring. And 3 factors out to be $(2+sqrt{-5})(2-sqrt(-5))$, but the norm is 9.
$endgroup$
– davidh
Jan 16 at 19:09
1
$begingroup$
What if you use $3$ and one of those factors as generators? Correct, their product is nine, so they aren't really factors of $3$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:11
$begingroup$
@JyrkiLahtonen $(2+sqrt{-5},3)$ seems to be maximal, since there's no 1 in there.
$endgroup$
– davidh
Jan 16 at 19:16
$begingroup$
Correct, David. It is the same ideal as $(-1+sqrt{-5},3)$. The first generator has norm six only. There is also another maximal ideal containing $3$, but if I got it right you only needed one. But, you need to be a bit more careful in proving maxiimality. Can you show that your ideal has index three as a subgroup? That would prove maxiimality (because three is a prime).
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:19
|
show 3 more comments
$begingroup$
I think I need to choose an element to pair it with 3, i.e. (3,x). But I don't see how to find such an element.
abstract-algebra algebraic-number-theory
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
asked Jan 16 at 18:49
davidhdavidh
1745
$endgroup$
I think I need to choose an element to pair it with 3, i.e. (3,x). But I don't see how to find such an element.
abstract-algebra algebraic-number-theory
abstract-algebra algebraic-number-theory
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
asked Jan 16 at 18:49
davidhdavidh
1745
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
asked Jan 16 at 18:49
davidhdavidh
1745
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
edited Jan 17 at 3:16
J. W. Tanner
2,3751117
2,3751117
asked Jan 16 at 18:49
davidhdavidh
1745
asked Jan 16 at 18:49
davidhdavidh
1745
asked Jan 16 at 18:49
davidhdavidh
1745
1745
$begingroup$
An element with norm that is a multiple of 3 might help. The norm is the mapping $N(a+bsqrt{-5})=a^2+5b^2$. Hint: the missing element appears in the "famous" non-unique factorization in this ring.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:02
$begingroup$
@JyrkiLahtonen Okay, so I know 3 is irreducible but not prime in this ring. And 3 factors out to be $(2+sqrt{-5})(2-sqrt(-5))$, but the norm is 9.
$endgroup$
– davidh
Jan 16 at 19:09
1
$begingroup$
What if you use $3$ and one of those factors as generators? Correct, their product is nine, so they aren't really factors of $3$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:11
$begingroup$
@JyrkiLahtonen $(2+sqrt{-5},3)$ seems to be maximal, since there's no 1 in there.
$endgroup$
– davidh
Jan 16 at 19:16
$begingroup$
Correct, David. It is the same ideal as $(-1+sqrt{-5},3)$. The first generator has norm six only. There is also another maximal ideal containing $3$, but if I got it right you only needed one. But, you need to be a bit more careful in proving maxiimality. Can you show that your ideal has index three as a subgroup? That would prove maxiimality (because three is a prime).
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:19
|
show 3 more comments
$begingroup$
An element with norm that is a multiple of 3 might help. The norm is the mapping $N(a+bsqrt{-5})=a^2+5b^2$. Hint: the missing element appears in the "famous" non-unique factorization in this ring.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:02
$begingroup$
@JyrkiLahtonen Okay, so I know 3 is irreducible but not prime in this ring. And 3 factors out to be $(2+sqrt{-5})(2-sqrt(-5))$, but the norm is 9.
$endgroup$
– davidh
Jan 16 at 19:09
1
$begingroup$
What if you use $3$ and one of those factors as generators? Correct, their product is nine, so they aren't really factors of $3$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:11
$begingroup$
@JyrkiLahtonen $(2+sqrt{-5},3)$ seems to be maximal, since there's no 1 in there.
$endgroup$
– davidh
Jan 16 at 19:16
$begingroup$
Correct, David. It is the same ideal as $(-1+sqrt{-5},3)$. The first generator has norm six only. There is also another maximal ideal containing $3$, but if I got it right you only needed one. But, you need to be a bit more careful in proving maxiimality. Can you show that your ideal has index three as a subgroup? That would prove maxiimality (because three is a prime).
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:19
$begingroup$
An element with norm that is a multiple of 3 might help. The norm is the mapping $N(a+bsqrt{-5})=a^2+5b^2$. Hint: the missing element appears in the "famous" non-unique factorization in this ring.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:02
$begingroup$
An element with norm that is a multiple of 3 might help. The norm is the mapping $N(a+bsqrt{-5})=a^2+5b^2$. Hint: the missing element appears in the "famous" non-unique factorization in this ring.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:02
$begingroup$
@JyrkiLahtonen Okay, so I know 3 is irreducible but not prime in this ring. And 3 factors out to be $(2+sqrt{-5})(2-sqrt(-5))$, but the norm is 9.
$endgroup$
– davidh
Jan 16 at 19:09
$begingroup$
@JyrkiLahtonen Okay, so I know 3 is irreducible but not prime in this ring. And 3 factors out to be $(2+sqrt{-5})(2-sqrt(-5))$, but the norm is 9.
$endgroup$
– davidh
Jan 16 at 19:09
1
1
$begingroup$
What if you use $3$ and one of those factors as generators? Correct, their product is nine, so they aren't really factors of $3$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:11
$begingroup$
What if you use $3$ and one of those factors as generators? Correct, their product is nine, so they aren't really factors of $3$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:11
$begingroup$
@JyrkiLahtonen $(2+sqrt{-5},3)$ seems to be maximal, since there's no 1 in there.
$endgroup$
– davidh
Jan 16 at 19:16
$begingroup$
@JyrkiLahtonen $(2+sqrt{-5},3)$ seems to be maximal, since there's no 1 in there.
$endgroup$
– davidh
Jan 16 at 19:16
$begingroup$
Correct, David. It is the same ideal as $(-1+sqrt{-5},3)$. The first generator has norm six only. There is also another maximal ideal containing $3$, but if I got it right you only needed one. But, you need to be a bit more careful in proving maxiimality. Can you show that your ideal has index three as a subgroup? That would prove maxiimality (because three is a prime).
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:19
$begingroup$
Correct, David. It is the same ideal as $(-1+sqrt{-5},3)$. The first generator has norm six only. There is also another maximal ideal containing $3$, but if I got it right you only needed one. But, you need to be a bit more careful in proving maxiimality. Can you show that your ideal has index three as a subgroup? That would prove maxiimality (because three is a prime).
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:19
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
In general, given a positive number $p$ that is prime in $mathbb Z$, and a squarefree number $d$ coprime to $p$ such that $$left(frac{d}{p}right) = 1,$$ find the smallest positive integer $n$ such that $n^2 equiv d pmod p$, then $langle p rangle = langle p, n - sqrt d rangle langle p, n + sqrt d rangle$.
In this instance $p$ is 3 and $d = -5$. Indeed $$left(frac{-5}{3}right) = 1,$$ $-5 equiv 1 pmod 3$, and obviously $1^2 equiv 1 pmod 3$. Then $1 - sqrt{-5}$ and $1 + sqrt{-5}$ are both numbers with norm divisible by 3.
Furthermore, any number in $mathbb Z[sqrt{-5}]$ with norm divisible by 3 belongs in either one of these two ideals: $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$. For example, $4 + sqrt{-5} = 3 + (1 + sqrt{-5})$ and hence $(4 + sqrt{-5}) in langle 3, 1 + sqrt{-5} rangle$. Likewise $(4 - sqrt{-5}) in langle 3, 1 - sqrt{-5} rangle$.
Neither of these two ideals can contain numbers with norm divisible by 2 but not by 3. For example, prove that $(3 + sqrt{-5}) notin langle 3, 1 + sqrt{-5} rangle$.
Since norms are multiplicative, it's a given that all nonzero numbers in $langle 3 rangle$ have norms divisible by 9 and all nonzero numbers in $langle 1 - sqrt{-5} rangle$ have norms divisible by 6. Same goes for $langle 1 + sqrt{-5} rangle$.
Then you just need to prove that all numbers of the form $3x + y(1 pm sqrt{-5})$, with both $x$ and $y$ being any numbers from $mathbb Z[sqrt{-5}]$, have norms divisible by 3. And lastly that any ideal properly containing $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$ is the whole ring.
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
$endgroup$
add a comment |
$begingroup$
The ideals $Isubset R$ containing the principal ideal $(3)subset R$ correspond bijectively with the ideals of $R/(3)$. This quotient is isomorphic to $Bbb{F}_3timesBbb{F}_3$, which has four ideals. Which of these correspond to maximal ideals of $R$?
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
$endgroup$
$begingroup$
I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)simeqBbb{F}_3timesBbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already..
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:23
$begingroup$
For example, $(11)$ is a maximal ideal of $R$, and $R/(11)simeq Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:43
$begingroup$
@JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$Bbb{Z}[sqrt{-5}]/(3)cong(Bbb{Z}[X]/(X^2+5))/(3)cong(Bbb{Z}[X]/(3))(X^2+5)congBbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,sqrt{-5}-1)$ and $(3,sqrt{-5}+1)$.
$endgroup$
– Servaes
Jan 16 at 20:58
$begingroup$
Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker?
$endgroup$
– Jyrki Lahtonen
Jan 16 at 21:02
$begingroup$
Sure, I figured the isomorphism $Bbb{F}_3[X]/(X^2-1)congBbb{F}_3timesBbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader.
$endgroup$
– Servaes
Jan 16 at 21:06
add a comment |
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2 Answers
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2 Answers
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$begingroup$
In general, given a positive number $p$ that is prime in $mathbb Z$, and a squarefree number $d$ coprime to $p$ such that $$left(frac{d}{p}right) = 1,$$ find the smallest positive integer $n$ such that $n^2 equiv d pmod p$, then $langle p rangle = langle p, n - sqrt d rangle langle p, n + sqrt d rangle$.
In this instance $p$ is 3 and $d = -5$. Indeed $$left(frac{-5}{3}right) = 1,$$ $-5 equiv 1 pmod 3$, and obviously $1^2 equiv 1 pmod 3$. Then $1 - sqrt{-5}$ and $1 + sqrt{-5}$ are both numbers with norm divisible by 3.
Furthermore, any number in $mathbb Z[sqrt{-5}]$ with norm divisible by 3 belongs in either one of these two ideals: $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$. For example, $4 + sqrt{-5} = 3 + (1 + sqrt{-5})$ and hence $(4 + sqrt{-5}) in langle 3, 1 + sqrt{-5} rangle$. Likewise $(4 - sqrt{-5}) in langle 3, 1 - sqrt{-5} rangle$.
Neither of these two ideals can contain numbers with norm divisible by 2 but not by 3. For example, prove that $(3 + sqrt{-5}) notin langle 3, 1 + sqrt{-5} rangle$.
Since norms are multiplicative, it's a given that all nonzero numbers in $langle 3 rangle$ have norms divisible by 9 and all nonzero numbers in $langle 1 - sqrt{-5} rangle$ have norms divisible by 6. Same goes for $langle 1 + sqrt{-5} rangle$.
Then you just need to prove that all numbers of the form $3x + y(1 pm sqrt{-5})$, with both $x$ and $y$ being any numbers from $mathbb Z[sqrt{-5}]$, have norms divisible by 3. And lastly that any ideal properly containing $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$ is the whole ring.
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
$endgroup$
add a comment |
$begingroup$
In general, given a positive number $p$ that is prime in $mathbb Z$, and a squarefree number $d$ coprime to $p$ such that $$left(frac{d}{p}right) = 1,$$ find the smallest positive integer $n$ such that $n^2 equiv d pmod p$, then $langle p rangle = langle p, n - sqrt d rangle langle p, n + sqrt d rangle$.
In this instance $p$ is 3 and $d = -5$. Indeed $$left(frac{-5}{3}right) = 1,$$ $-5 equiv 1 pmod 3$, and obviously $1^2 equiv 1 pmod 3$. Then $1 - sqrt{-5}$ and $1 + sqrt{-5}$ are both numbers with norm divisible by 3.
Furthermore, any number in $mathbb Z[sqrt{-5}]$ with norm divisible by 3 belongs in either one of these two ideals: $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$. For example, $4 + sqrt{-5} = 3 + (1 + sqrt{-5})$ and hence $(4 + sqrt{-5}) in langle 3, 1 + sqrt{-5} rangle$. Likewise $(4 - sqrt{-5}) in langle 3, 1 - sqrt{-5} rangle$.
Neither of these two ideals can contain numbers with norm divisible by 2 but not by 3. For example, prove that $(3 + sqrt{-5}) notin langle 3, 1 + sqrt{-5} rangle$.
Since norms are multiplicative, it's a given that all nonzero numbers in $langle 3 rangle$ have norms divisible by 9 and all nonzero numbers in $langle 1 - sqrt{-5} rangle$ have norms divisible by 6. Same goes for $langle 1 + sqrt{-5} rangle$.
Then you just need to prove that all numbers of the form $3x + y(1 pm sqrt{-5})$, with both $x$ and $y$ being any numbers from $mathbb Z[sqrt{-5}]$, have norms divisible by 3. And lastly that any ideal properly containing $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$ is the whole ring.
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
$endgroup$
add a comment |
$begingroup$
In general, given a positive number $p$ that is prime in $mathbb Z$, and a squarefree number $d$ coprime to $p$ such that $$left(frac{d}{p}right) = 1,$$ find the smallest positive integer $n$ such that $n^2 equiv d pmod p$, then $langle p rangle = langle p, n - sqrt d rangle langle p, n + sqrt d rangle$.
In this instance $p$ is 3 and $d = -5$. Indeed $$left(frac{-5}{3}right) = 1,$$ $-5 equiv 1 pmod 3$, and obviously $1^2 equiv 1 pmod 3$. Then $1 - sqrt{-5}$ and $1 + sqrt{-5}$ are both numbers with norm divisible by 3.
Furthermore, any number in $mathbb Z[sqrt{-5}]$ with norm divisible by 3 belongs in either one of these two ideals: $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$. For example, $4 + sqrt{-5} = 3 + (1 + sqrt{-5})$ and hence $(4 + sqrt{-5}) in langle 3, 1 + sqrt{-5} rangle$. Likewise $(4 - sqrt{-5}) in langle 3, 1 - sqrt{-5} rangle$.
Neither of these two ideals can contain numbers with norm divisible by 2 but not by 3. For example, prove that $(3 + sqrt{-5}) notin langle 3, 1 + sqrt{-5} rangle$.
Since norms are multiplicative, it's a given that all nonzero numbers in $langle 3 rangle$ have norms divisible by 9 and all nonzero numbers in $langle 1 - sqrt{-5} rangle$ have norms divisible by 6. Same goes for $langle 1 + sqrt{-5} rangle$.
Then you just need to prove that all numbers of the form $3x + y(1 pm sqrt{-5})$, with both $x$ and $y$ being any numbers from $mathbb Z[sqrt{-5}]$, have norms divisible by 3. And lastly that any ideal properly containing $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$ is the whole ring.
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
$endgroup$
In general, given a positive number $p$ that is prime in $mathbb Z$, and a squarefree number $d$ coprime to $p$ such that $$left(frac{d}{p}right) = 1,$$ find the smallest positive integer $n$ such that $n^2 equiv d pmod p$, then $langle p rangle = langle p, n - sqrt d rangle langle p, n + sqrt d rangle$.
In this instance $p$ is 3 and $d = -5$. Indeed $$left(frac{-5}{3}right) = 1,$$ $-5 equiv 1 pmod 3$, and obviously $1^2 equiv 1 pmod 3$. Then $1 - sqrt{-5}$ and $1 + sqrt{-5}$ are both numbers with norm divisible by 3.
Furthermore, any number in $mathbb Z[sqrt{-5}]$ with norm divisible by 3 belongs in either one of these two ideals: $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$. For example, $4 + sqrt{-5} = 3 + (1 + sqrt{-5})$ and hence $(4 + sqrt{-5}) in langle 3, 1 + sqrt{-5} rangle$. Likewise $(4 - sqrt{-5}) in langle 3, 1 - sqrt{-5} rangle$.
Neither of these two ideals can contain numbers with norm divisible by 2 but not by 3. For example, prove that $(3 + sqrt{-5}) notin langle 3, 1 + sqrt{-5} rangle$.
Since norms are multiplicative, it's a given that all nonzero numbers in $langle 3 rangle$ have norms divisible by 9 and all nonzero numbers in $langle 1 - sqrt{-5} rangle$ have norms divisible by 6. Same goes for $langle 1 + sqrt{-5} rangle$.
Then you just need to prove that all numbers of the form $3x + y(1 pm sqrt{-5})$, with both $x$ and $y$ being any numbers from $mathbb Z[sqrt{-5}]$, have norms divisible by 3. And lastly that any ideal properly containing $langle 3, 1 - sqrt{-5} rangle$ or $langle 3, 1 + sqrt{-5} rangle$ is the whole ring.
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
answered Jan 19 at 6:55
Robert SoupeRobert Soupe
11.2k21950
11.2k21950
add a comment |
add a comment |
$begingroup$
The ideals $Isubset R$ containing the principal ideal $(3)subset R$ correspond bijectively with the ideals of $R/(3)$. This quotient is isomorphic to $Bbb{F}_3timesBbb{F}_3$, which has four ideals. Which of these correspond to maximal ideals of $R$?
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
$endgroup$
$begingroup$
I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)simeqBbb{F}_3timesBbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already..
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:23
$begingroup$
For example, $(11)$ is a maximal ideal of $R$, and $R/(11)simeq Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:43
$begingroup$
@JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$Bbb{Z}[sqrt{-5}]/(3)cong(Bbb{Z}[X]/(X^2+5))/(3)cong(Bbb{Z}[X]/(3))(X^2+5)congBbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,sqrt{-5}-1)$ and $(3,sqrt{-5}+1)$.
$endgroup$
– Servaes
Jan 16 at 20:58
$begingroup$
Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker?
$endgroup$
– Jyrki Lahtonen
Jan 16 at 21:02
$begingroup$
Sure, I figured the isomorphism $Bbb{F}_3[X]/(X^2-1)congBbb{F}_3timesBbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader.
$endgroup$
– Servaes
Jan 16 at 21:06
add a comment |
$begingroup$
The ideals $Isubset R$ containing the principal ideal $(3)subset R$ correspond bijectively with the ideals of $R/(3)$. This quotient is isomorphic to $Bbb{F}_3timesBbb{F}_3$, which has four ideals. Which of these correspond to maximal ideals of $R$?
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
$endgroup$
$begingroup$
I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)simeqBbb{F}_3timesBbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already..
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:23
$begingroup$
For example, $(11)$ is a maximal ideal of $R$, and $R/(11)simeq Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:43
$begingroup$
@JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$Bbb{Z}[sqrt{-5}]/(3)cong(Bbb{Z}[X]/(X^2+5))/(3)cong(Bbb{Z}[X]/(3))(X^2+5)congBbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,sqrt{-5}-1)$ and $(3,sqrt{-5}+1)$.
$endgroup$
– Servaes
Jan 16 at 20:58
$begingroup$
Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker?
$endgroup$
– Jyrki Lahtonen
Jan 16 at 21:02
$begingroup$
Sure, I figured the isomorphism $Bbb{F}_3[X]/(X^2-1)congBbb{F}_3timesBbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader.
$endgroup$
– Servaes
Jan 16 at 21:06
add a comment |
$begingroup$
The ideals $Isubset R$ containing the principal ideal $(3)subset R$ correspond bijectively with the ideals of $R/(3)$. This quotient is isomorphic to $Bbb{F}_3timesBbb{F}_3$, which has four ideals. Which of these correspond to maximal ideals of $R$?
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
$endgroup$
The ideals $Isubset R$ containing the principal ideal $(3)subset R$ correspond bijectively with the ideals of $R/(3)$. This quotient is isomorphic to $Bbb{F}_3timesBbb{F}_3$, which has four ideals. Which of these correspond to maximal ideals of $R$?
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
answered Jan 16 at 19:11
ServaesServaes
25.8k33996
25.8k33996
$begingroup$
I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)simeqBbb{F}_3timesBbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already..
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– Jyrki Lahtonen
Jan 16 at 19:23
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For example, $(11)$ is a maximal ideal of $R$, and $R/(11)simeq Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:43
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@JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$Bbb{Z}[sqrt{-5}]/(3)cong(Bbb{Z}[X]/(X^2+5))/(3)cong(Bbb{Z}[X]/(3))(X^2+5)congBbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,sqrt{-5}-1)$ and $(3,sqrt{-5}+1)$.
$endgroup$
– Servaes
Jan 16 at 20:58
$begingroup$
Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker?
$endgroup$
– Jyrki Lahtonen
Jan 16 at 21:02
$begingroup$
Sure, I figured the isomorphism $Bbb{F}_3[X]/(X^2-1)congBbb{F}_3timesBbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader.
$endgroup$
– Servaes
Jan 16 at 21:06
add a comment |
$begingroup$
I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)simeqBbb{F}_3timesBbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already..
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:23
$begingroup$
For example, $(11)$ is a maximal ideal of $R$, and $R/(11)simeq Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:43
$begingroup$
@JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$Bbb{Z}[sqrt{-5}]/(3)cong(Bbb{Z}[X]/(X^2+5))/(3)cong(Bbb{Z}[X]/(3))(X^2+5)congBbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,sqrt{-5}-1)$ and $(3,sqrt{-5}+1)$.
$endgroup$
– Servaes
Jan 16 at 20:58
$begingroup$
Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker?
$endgroup$
– Jyrki Lahtonen
Jan 16 at 21:02
$begingroup$
Sure, I figured the isomorphism $Bbb{F}_3[X]/(X^2-1)congBbb{F}_3timesBbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader.
$endgroup$
– Servaes
Jan 16 at 21:06
$begingroup$
I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)simeqBbb{F}_3timesBbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already..
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:23
$begingroup$
I'm not sure I understand what you are trying to do here. I don't see an easy way of proving that $R/(3)simeqBbb{F}_3timesBbb{F}_3$. I mean, a proof that wouldn't rely on having the key two maximal ideals at hand already..
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:23
$begingroup$
For example, $(11)$ is a maximal ideal of $R$, and $R/(11)simeq Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:43
$begingroup$
For example, $(11)$ is a maximal ideal of $R$, and $R/(11)simeq Bbb{F}_{11^2}$. So the argument works with prime $3$ but does not work with the prime $11$. A different kind of a problem appears with $R/(2)$ which has non-zero nilpotent elements.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:43
$begingroup$
@JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$Bbb{Z}[sqrt{-5}]/(3)cong(Bbb{Z}[X]/(X^2+5))/(3)cong(Bbb{Z}[X]/(3))(X^2+5)congBbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,sqrt{-5}-1)$ and $(3,sqrt{-5}+1)$.
$endgroup$
– Servaes
Jan 16 at 20:58
$begingroup$
@JyrkiLahtonen The argument is very standard, though not the most convenient for this particular problem; the isomorphisms $$Bbb{Z}[sqrt{-5}]/(3)cong(Bbb{Z}[X]/(X^2+5))/(3)cong(Bbb{Z}[X]/(3))(X^2+5)congBbb{F}_3[X]/(X^2-1).$$ tell you that the maximal ideals are $(3,sqrt{-5}-1)$ and $(3,sqrt{-5}+1)$.
$endgroup$
– Servaes
Jan 16 at 20:58
$begingroup$
Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker?
$endgroup$
– Jyrki Lahtonen
Jan 16 at 21:02
$begingroup$
Oh yeah! Thanks! How did I miss/forget that :-) You still need Chinese remainder theorem at the end, but hat may be ok for the asker?
$endgroup$
– Jyrki Lahtonen
Jan 16 at 21:02
$begingroup$
Sure, I figured the isomorphism $Bbb{F}_3[X]/(X^2-1)congBbb{F}_3timesBbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader.
$endgroup$
– Servaes
Jan 16 at 21:06
$begingroup$
Sure, I figured the isomorphism $Bbb{F}_3[X]/(X^2-1)congBbb{F}_3timesBbb{F}_3$ was okay to assume (or leave as a small exercise), though that may depend on the level of the reader.
$endgroup$
– Servaes
Jan 16 at 21:06
add a comment |
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$begingroup$
An element with norm that is a multiple of 3 might help. The norm is the mapping $N(a+bsqrt{-5})=a^2+5b^2$. Hint: the missing element appears in the "famous" non-unique factorization in this ring.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:02
$begingroup$
@JyrkiLahtonen Okay, so I know 3 is irreducible but not prime in this ring. And 3 factors out to be $(2+sqrt{-5})(2-sqrt(-5))$, but the norm is 9.
$endgroup$
– davidh
Jan 16 at 19:09
1
$begingroup$
What if you use $3$ and one of those factors as generators? Correct, their product is nine, so they aren't really factors of $3$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:11
$begingroup$
@JyrkiLahtonen $(2+sqrt{-5},3)$ seems to be maximal, since there's no 1 in there.
$endgroup$
– davidh
Jan 16 at 19:16
$begingroup$
Correct, David. It is the same ideal as $(-1+sqrt{-5},3)$. The first generator has norm six only. There is also another maximal ideal containing $3$, but if I got it right you only needed one. But, you need to be a bit more careful in proving maxiimality. Can you show that your ideal has index three as a subgroup? That would prove maxiimality (because three is a prime).
$endgroup$
– Jyrki Lahtonen
Jan 16 at 19:19