Contradictory results when computing the Ideal class group of $mathbb{Q}(sqrt{-7})$
$begingroup$
I seem to have arrived at some contradictory results in computing this group, would you mind helping me resolve this?
By Sage + internet I find that it should be true that this class group is trivial.
We know that Minkowski's bound for this group is:
$$
M_k = frac{1}{2}frac{pi}{4}sqrt{7*4} approx 2.08.
$$
Therefore the only prime we need to check is $2 mathcal{O}_k$. Notice that $x^2 + 7 equiv_2 x^2 + 1 equiv_2 (x+1)^2$, which implies that it is totally ramified. So then we have to consider $mathfrak{p}_2 = (2, sqrt{-7}+1)$. We know that $mathfrak{p}_2^2 = (2)$, which implies that we know that the order of the ideal class group is less than or equal to 2. So then we consider $(2, sqrt{-7}+1)$ and want to show that this is principle (otherwise sage/internet is wrong).
Suppose we have some element $z$ such that $(z) = (2, 1 + sqrt{-7})$. Then it must be true that $N(z) mid N(2)$ and $N(z) mid N(1+sqrt{-7}) implies N(z) mid 4, 8 implies N(z)$ is one of $1, 2, 4$. However, since $N(a+bsqrt{-7}) = a^2 + 7b^2$, we see that it is impossible for it to take on the values $2$ and $4$, therefore if such $z$ exists it must be $1$.
However, by this answer on stack exchange, such a $z$ is not possible!
Therefore I must conclude that the class group is isomorphic to $mathbb{Z}/2mathbb{Z}$, but this is clearly not true.
Could you point out my error?
abstract-algebra algebraic-number-theory ideal-class-group
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
$endgroup$
add a comment |
$begingroup$
I seem to have arrived at some contradictory results in computing this group, would you mind helping me resolve this?
By Sage + internet I find that it should be true that this class group is trivial.
We know that Minkowski's bound for this group is:
$$
M_k = frac{1}{2}frac{pi}{4}sqrt{7*4} approx 2.08.
$$
Therefore the only prime we need to check is $2 mathcal{O}_k$. Notice that $x^2 + 7 equiv_2 x^2 + 1 equiv_2 (x+1)^2$, which implies that it is totally ramified. So then we have to consider $mathfrak{p}_2 = (2, sqrt{-7}+1)$. We know that $mathfrak{p}_2^2 = (2)$, which implies that we know that the order of the ideal class group is less than or equal to 2. So then we consider $(2, sqrt{-7}+1)$ and want to show that this is principle (otherwise sage/internet is wrong).
Suppose we have some element $z$ such that $(z) = (2, 1 + sqrt{-7})$. Then it must be true that $N(z) mid N(2)$ and $N(z) mid N(1+sqrt{-7}) implies N(z) mid 4, 8 implies N(z)$ is one of $1, 2, 4$. However, since $N(a+bsqrt{-7}) = a^2 + 7b^2$, we see that it is impossible for it to take on the values $2$ and $4$, therefore if such $z$ exists it must be $1$.
However, by this answer on stack exchange, such a $z$ is not possible!
Therefore I must conclude that the class group is isomorphic to $mathbb{Z}/2mathbb{Z}$, but this is clearly not true.
Could you point out my error?
abstract-algebra algebraic-number-theory ideal-class-group
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
$endgroup$
$begingroup$
Minkowski bound say we have to check ideals of norm $le 2$, that is $(1)$ and the possible prime ideals above $(2)$. Indeed $(2) =(frac{1+sqrt{-7}}{2})^2$ is a product of principal ideals so the ideal class group is trivial.
$endgroup$
– reuns
Dec 13 '18 at 0:12
$begingroup$
I see now, I was trying to factor in $Z[sqrt{-7}]$ but the ring of integers is larger than that by @RicardoBuring.
$endgroup$
– TrostAft
Dec 13 '18 at 0:14
add a comment |
$begingroup$
I seem to have arrived at some contradictory results in computing this group, would you mind helping me resolve this?
By Sage + internet I find that it should be true that this class group is trivial.
We know that Minkowski's bound for this group is:
$$
M_k = frac{1}{2}frac{pi}{4}sqrt{7*4} approx 2.08.
$$
Therefore the only prime we need to check is $2 mathcal{O}_k$. Notice that $x^2 + 7 equiv_2 x^2 + 1 equiv_2 (x+1)^2$, which implies that it is totally ramified. So then we have to consider $mathfrak{p}_2 = (2, sqrt{-7}+1)$. We know that $mathfrak{p}_2^2 = (2)$, which implies that we know that the order of the ideal class group is less than or equal to 2. So then we consider $(2, sqrt{-7}+1)$ and want to show that this is principle (otherwise sage/internet is wrong).
Suppose we have some element $z$ such that $(z) = (2, 1 + sqrt{-7})$. Then it must be true that $N(z) mid N(2)$ and $N(z) mid N(1+sqrt{-7}) implies N(z) mid 4, 8 implies N(z)$ is one of $1, 2, 4$. However, since $N(a+bsqrt{-7}) = a^2 + 7b^2$, we see that it is impossible for it to take on the values $2$ and $4$, therefore if such $z$ exists it must be $1$.
However, by this answer on stack exchange, such a $z$ is not possible!
Therefore I must conclude that the class group is isomorphic to $mathbb{Z}/2mathbb{Z}$, but this is clearly not true.
Could you point out my error?
abstract-algebra algebraic-number-theory ideal-class-group
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
$endgroup$
I seem to have arrived at some contradictory results in computing this group, would you mind helping me resolve this?
By Sage + internet I find that it should be true that this class group is trivial.
We know that Minkowski's bound for this group is:
$$
M_k = frac{1}{2}frac{pi}{4}sqrt{7*4} approx 2.08.
$$
Therefore the only prime we need to check is $2 mathcal{O}_k$. Notice that $x^2 + 7 equiv_2 x^2 + 1 equiv_2 (x+1)^2$, which implies that it is totally ramified. So then we have to consider $mathfrak{p}_2 = (2, sqrt{-7}+1)$. We know that $mathfrak{p}_2^2 = (2)$, which implies that we know that the order of the ideal class group is less than or equal to 2. So then we consider $(2, sqrt{-7}+1)$ and want to show that this is principle (otherwise sage/internet is wrong).
Suppose we have some element $z$ such that $(z) = (2, 1 + sqrt{-7})$. Then it must be true that $N(z) mid N(2)$ and $N(z) mid N(1+sqrt{-7}) implies N(z) mid 4, 8 implies N(z)$ is one of $1, 2, 4$. However, since $N(a+bsqrt{-7}) = a^2 + 7b^2$, we see that it is impossible for it to take on the values $2$ and $4$, therefore if such $z$ exists it must be $1$.
However, by this answer on stack exchange, such a $z$ is not possible!
Therefore I must conclude that the class group is isomorphic to $mathbb{Z}/2mathbb{Z}$, but this is clearly not true.
Could you point out my error?
abstract-algebra algebraic-number-theory ideal-class-group
abstract-algebra algebraic-number-theory ideal-class-group
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
edited Dec 13 '18 at 0:44
André 3000
12.7k22243
12.7k22243
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
asked Dec 12 '18 at 23:50
TrostAftTrostAft
423412
423412
$begingroup$
Minkowski bound say we have to check ideals of norm $le 2$, that is $(1)$ and the possible prime ideals above $(2)$. Indeed $(2) =(frac{1+sqrt{-7}}{2})^2$ is a product of principal ideals so the ideal class group is trivial.
$endgroup$
– reuns
Dec 13 '18 at 0:12
$begingroup$
I see now, I was trying to factor in $Z[sqrt{-7}]$ but the ring of integers is larger than that by @RicardoBuring.
$endgroup$
– TrostAft
Dec 13 '18 at 0:14
add a comment |
$begingroup$
Minkowski bound say we have to check ideals of norm $le 2$, that is $(1)$ and the possible prime ideals above $(2)$. Indeed $(2) =(frac{1+sqrt{-7}}{2})^2$ is a product of principal ideals so the ideal class group is trivial.
$endgroup$
– reuns
Dec 13 '18 at 0:12
$begingroup$
I see now, I was trying to factor in $Z[sqrt{-7}]$ but the ring of integers is larger than that by @RicardoBuring.
$endgroup$
– TrostAft
Dec 13 '18 at 0:14
$begingroup$
Minkowski bound say we have to check ideals of norm $le 2$, that is $(1)$ and the possible prime ideals above $(2)$. Indeed $(2) =(frac{1+sqrt{-7}}{2})^2$ is a product of principal ideals so the ideal class group is trivial.
$endgroup$
– reuns
Dec 13 '18 at 0:12
$begingroup$
Minkowski bound say we have to check ideals of norm $le 2$, that is $(1)$ and the possible prime ideals above $(2)$. Indeed $(2) =(frac{1+sqrt{-7}}{2})^2$ is a product of principal ideals so the ideal class group is trivial.
$endgroup$
– reuns
Dec 13 '18 at 0:12
$begingroup$
I see now, I was trying to factor in $Z[sqrt{-7}]$ but the ring of integers is larger than that by @RicardoBuring.
$endgroup$
– TrostAft
Dec 13 '18 at 0:14
$begingroup$
I see now, I was trying to factor in $Z[sqrt{-7}]$ but the ring of integers is larger than that by @RicardoBuring.
$endgroup$
– TrostAft
Dec 13 '18 at 0:14
add a comment |
1 Answer
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The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
2
$begingroup$
Oh of course, it's actually $mathbb{Z}[ (1 + sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks.
$endgroup$
– TrostAft
Dec 13 '18 at 0:12
add a comment |
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1 Answer
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$begingroup$
The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
2
$begingroup$
Oh of course, it's actually $mathbb{Z}[ (1 + sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks.
$endgroup$
– TrostAft
Dec 13 '18 at 0:12
add a comment |
$begingroup$
The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
2
$begingroup$
Oh of course, it's actually $mathbb{Z}[ (1 + sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks.
$endgroup$
– TrostAft
Dec 13 '18 at 0:12
add a comment |
$begingroup$
The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
answered Dec 13 '18 at 0:09
Ricardo BuringRicardo Buring
1,4711334
1,4711334
2
$begingroup$
Oh of course, it's actually $mathbb{Z}[ (1 + sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks.
$endgroup$
– TrostAft
Dec 13 '18 at 0:12
add a comment |
2
$begingroup$
Oh of course, it's actually $mathbb{Z}[ (1 + sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks.
$endgroup$
– TrostAft
Dec 13 '18 at 0:12
2
2
$begingroup$
Oh of course, it's actually $mathbb{Z}[ (1 + sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks.
$endgroup$
– TrostAft
Dec 13 '18 at 0:12
$begingroup$
Oh of course, it's actually $mathbb{Z}[ (1 + sqrt{-7})/2 ]$, and then we immediately see it. That was crazy of me, thanks.
$endgroup$
– TrostAft
Dec 13 '18 at 0:12
add a comment |
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$begingroup$
Minkowski bound say we have to check ideals of norm $le 2$, that is $(1)$ and the possible prime ideals above $(2)$. Indeed $(2) =(frac{1+sqrt{-7}}{2})^2$ is a product of principal ideals so the ideal class group is trivial.
$endgroup$
– reuns
Dec 13 '18 at 0:12
$begingroup$
I see now, I was trying to factor in $Z[sqrt{-7}]$ but the ring of integers is larger than that by @RicardoBuring.
$endgroup$
– TrostAft
Dec 13 '18 at 0:14