Contradictory results when computing the Ideal class group of $mathbb{Q}(sqrt{-7})$












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 0:44









André 3000

12.7k22243




12.7k22243










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


















  • $begingroup$
    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










1 Answer
1






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oldest

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3












$begingroup$

The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.






share|cite|improve this answer









$endgroup$









  • 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













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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.






share|cite|improve this answer









$endgroup$









  • 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


















3












$begingroup$

The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.






share|cite|improve this answer









$endgroup$









  • 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
















3












3








3





$begingroup$

The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.






share|cite|improve this answer









$endgroup$



The ring of integers of $mathbb{Q}(sqrt{-7})$ is not $mathbb{Z}[sqrt{-7}]$ but bigger, because $-7 equiv 1 pmod 4$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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




















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