Quotients of a ring of integersWhat does the d-slice of a weighted polynomial algebra look like?Big image theorems for products of modular forms?Basic arithmetic behind ramification in quadratic number fieldsReference request for a basic result on relative differents & discriminantsClass groups of ordersWhat is the decomposition group at $p$ in the Galois group unramified outside $ell?$Can a product of conjugates be a Pisot number again?Distinct projections along factors of splitting primeComputational complexity of finding the class numberProper ideals are invertible
Quotients of a ring of integers
What does the d-slice of a weighted polynomial algebra look like?Big image theorems for products of modular forms?Basic arithmetic behind ramification in quadratic number fieldsReference request for a basic result on relative differents & discriminantsClass groups of ordersWhat is the decomposition group at $p$ in the Galois group unramified outside $ell?$Can a product of conjugates be a Pisot number again?Distinct projections along factors of splitting primeComputational complexity of finding the class numberProper ideals are invertible
$begingroup$
Let $L$ be a number field, let $p$ be a prime number, and let $I$ be a ideal of $mathcalO_L$ containing $p$. I am not assuming that $mathcalO_L$ or that $I$ is prime. The quotient ring $mathcalO_L/I$ has a natural structure of $mathbbF_p$-algebra.
Question. Do we have an isomorphism of $mathbbF_p$-algebras $$mathcalO_L/Isimeq mathbbF_p[X]/(f)$$ for some nonzero monic polynomial $f$ ?
I know that the answer if YES in several cases.
1) For example, it is true if $mathcalO_L=mathbbZ[alpha]$ for some $alpha$.
Indeed, since $I$ contains $p$, evaluation at $alpha$ induces a morphism of $mathbbF_p$-algebras $mathbbF_p[X]to mathcalO_L/I$. This morphism is surjective since $mathcalO_L=mathbbZ[alpha]$ . Its kernel is generated by a monic polynomial. Done.
2) If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, and $p$ is totally ramified in $K_2$, and $I=mathfrakp_2mathcalO_L.$ where $mathfrakp_2$ is the unique prime ideal of $mathcalO_K_2$ lygin above $p$. In this case, one may show that $mathcalO_L/Isimeq mathbbF_p[X]/(overlinemu_alpha_1,mathbbQ)$.
Nevertheless, I suspect that I am missing obvious counterexamples...
About 2), i wonder if it is a particular case of 1), so there is a side question:
Side question. If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, is $mathcalO_L=mathbbZ[alpha]$ for some $alpha$ ? For example, is $alpha=alpha_1+alpha_2$ working ?
Any thoughts ?
Greg
nt.number-theory
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $L$ be a number field, let $p$ be a prime number, and let $I$ be a ideal of $mathcalO_L$ containing $p$. I am not assuming that $mathcalO_L$ or that $I$ is prime. The quotient ring $mathcalO_L/I$ has a natural structure of $mathbbF_p$-algebra.
Question. Do we have an isomorphism of $mathbbF_p$-algebras $$mathcalO_L/Isimeq mathbbF_p[X]/(f)$$ for some nonzero monic polynomial $f$ ?
I know that the answer if YES in several cases.
1) For example, it is true if $mathcalO_L=mathbbZ[alpha]$ for some $alpha$.
Indeed, since $I$ contains $p$, evaluation at $alpha$ induces a morphism of $mathbbF_p$-algebras $mathbbF_p[X]to mathcalO_L/I$. This morphism is surjective since $mathcalO_L=mathbbZ[alpha]$ . Its kernel is generated by a monic polynomial. Done.
2) If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, and $p$ is totally ramified in $K_2$, and $I=mathfrakp_2mathcalO_L.$ where $mathfrakp_2$ is the unique prime ideal of $mathcalO_K_2$ lygin above $p$. In this case, one may show that $mathcalO_L/Isimeq mathbbF_p[X]/(overlinemu_alpha_1,mathbbQ)$.
Nevertheless, I suspect that I am missing obvious counterexamples...
About 2), i wonder if it is a particular case of 1), so there is a side question:
Side question. If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, is $mathcalO_L=mathbbZ[alpha]$ for some $alpha$ ? For example, is $alpha=alpha_1+alpha_2$ working ?
Any thoughts ?
Greg
nt.number-theory
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
$endgroup$
4
$begingroup$
Here is one counterexample. Let $K$ be $mathbbQ$. Let $L$ be $mathbbQ[s,t]/langle s^2+s+2,t^2+t+4 rangle$. Let $p$ be $2$, and let $I$ be $2mathcalO_L$. Then $mathcalO_L/I$ equals $mathbbF_2[s,t]/langle s^2+s,t^2+t rangle$. This is isomorphic as an $mathbbF_2$-algebra to a product of four copies of $mathbbF_2$. Yet a monic polynomial has at most two $mathbbF_2$-rational points.
$endgroup$
– Jason Starr
8 hours ago
add a comment |
$begingroup$
Let $L$ be a number field, let $p$ be a prime number, and let $I$ be a ideal of $mathcalO_L$ containing $p$. I am not assuming that $mathcalO_L$ or that $I$ is prime. The quotient ring $mathcalO_L/I$ has a natural structure of $mathbbF_p$-algebra.
Question. Do we have an isomorphism of $mathbbF_p$-algebras $$mathcalO_L/Isimeq mathbbF_p[X]/(f)$$ for some nonzero monic polynomial $f$ ?
I know that the answer if YES in several cases.
1) For example, it is true if $mathcalO_L=mathbbZ[alpha]$ for some $alpha$.
Indeed, since $I$ contains $p$, evaluation at $alpha$ induces a morphism of $mathbbF_p$-algebras $mathbbF_p[X]to mathcalO_L/I$. This morphism is surjective since $mathcalO_L=mathbbZ[alpha]$ . Its kernel is generated by a monic polynomial. Done.
2) If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, and $p$ is totally ramified in $K_2$, and $I=mathfrakp_2mathcalO_L.$ where $mathfrakp_2$ is the unique prime ideal of $mathcalO_K_2$ lygin above $p$. In this case, one may show that $mathcalO_L/Isimeq mathbbF_p[X]/(overlinemu_alpha_1,mathbbQ)$.
Nevertheless, I suspect that I am missing obvious counterexamples...
About 2), i wonder if it is a particular case of 1), so there is a side question:
Side question. If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, is $mathcalO_L=mathbbZ[alpha]$ for some $alpha$ ? For example, is $alpha=alpha_1+alpha_2$ working ?
Any thoughts ?
Greg
nt.number-theory
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
$endgroup$
Let $L$ be a number field, let $p$ be a prime number, and let $I$ be a ideal of $mathcalO_L$ containing $p$. I am not assuming that $mathcalO_L$ or that $I$ is prime. The quotient ring $mathcalO_L/I$ has a natural structure of $mathbbF_p$-algebra.
Question. Do we have an isomorphism of $mathbbF_p$-algebras $$mathcalO_L/Isimeq mathbbF_p[X]/(f)$$ for some nonzero monic polynomial $f$ ?
I know that the answer if YES in several cases.
1) For example, it is true if $mathcalO_L=mathbbZ[alpha]$ for some $alpha$.
Indeed, since $I$ contains $p$, evaluation at $alpha$ induces a morphism of $mathbbF_p$-algebras $mathbbF_p[X]to mathcalO_L/I$. This morphism is surjective since $mathcalO_L=mathbbZ[alpha]$ . Its kernel is generated by a monic polynomial. Done.
2) If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, and $p$ is totally ramified in $K_2$, and $I=mathfrakp_2mathcalO_L.$ where $mathfrakp_2$ is the unique prime ideal of $mathcalO_K_2$ lygin above $p$. In this case, one may show that $mathcalO_L/Isimeq mathbbF_p[X]/(overlinemu_alpha_1,mathbbQ)$.
Nevertheless, I suspect that I am missing obvious counterexamples...
About 2), i wonder if it is a particular case of 1), so there is a side question:
Side question. If $L=K_1K_2,$ where $mathcalO_K_i=mathbbZ[alpha_i]$, and the discriminants of $K_1$ and $K_2$ are coprime, is $mathcalO_L=mathbbZ[alpha]$ for some $alpha$ ? For example, is $alpha=alpha_1+alpha_2$ working ?
Any thoughts ?
Greg
nt.number-theory
nt.number-theory
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
edited 4 hours ago
GreginGre
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
asked 8 hours ago
GreginGreGreginGre
6163 silver badges8 bronze badges
6163 silver badges8 bronze badges
4
$begingroup$
Here is one counterexample. Let $K$ be $mathbbQ$. Let $L$ be $mathbbQ[s,t]/langle s^2+s+2,t^2+t+4 rangle$. Let $p$ be $2$, and let $I$ be $2mathcalO_L$. Then $mathcalO_L/I$ equals $mathbbF_2[s,t]/langle s^2+s,t^2+t rangle$. This is isomorphic as an $mathbbF_2$-algebra to a product of four copies of $mathbbF_2$. Yet a monic polynomial has at most two $mathbbF_2$-rational points.
$endgroup$
– Jason Starr
8 hours ago
add a comment |
4
$begingroup$
Here is one counterexample. Let $K$ be $mathbbQ$. Let $L$ be $mathbbQ[s,t]/langle s^2+s+2,t^2+t+4 rangle$. Let $p$ be $2$, and let $I$ be $2mathcalO_L$. Then $mathcalO_L/I$ equals $mathbbF_2[s,t]/langle s^2+s,t^2+t rangle$. This is isomorphic as an $mathbbF_2$-algebra to a product of four copies of $mathbbF_2$. Yet a monic polynomial has at most two $mathbbF_2$-rational points.
$endgroup$
– Jason Starr
8 hours ago
4
4
$begingroup$
Here is one counterexample. Let $K$ be $mathbbQ$. Let $L$ be $mathbbQ[s,t]/langle s^2+s+2,t^2+t+4 rangle$. Let $p$ be $2$, and let $I$ be $2mathcalO_L$. Then $mathcalO_L/I$ equals $mathbbF_2[s,t]/langle s^2+s,t^2+t rangle$. This is isomorphic as an $mathbbF_2$-algebra to a product of four copies of $mathbbF_2$. Yet a monic polynomial has at most two $mathbbF_2$-rational points.
$endgroup$
– Jason Starr
8 hours ago
$begingroup$
Here is one counterexample. Let $K$ be $mathbbQ$. Let $L$ be $mathbbQ[s,t]/langle s^2+s+2,t^2+t+4 rangle$. Let $p$ be $2$, and let $I$ be $2mathcalO_L$. Then $mathcalO_L/I$ equals $mathbbF_2[s,t]/langle s^2+s,t^2+t rangle$. This is isomorphic as an $mathbbF_2$-algebra to a product of four copies of $mathbbF_2$. Yet a monic polynomial has at most two $mathbbF_2$-rational points.
$endgroup$
– Jason Starr
8 hours ago
add a comment |
1 Answer
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$begingroup$
The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $mathcal O_L/(p) cong mathbf F_p^r$ but if $mathcal O_L/(p) cong mathbf F_p[X]/(f)$ for monic $f$ in $mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $mathcal O_L$ is not $mathbf Z[alpha]$ for some $alpha$ ($L$ is not monogenic).
Dedekind found the first example of this: $L = mathbf Q(theta)$ where $theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $mathcal O_L/(2) cong mathbf F_2^3$ and $mathcal O_L/(2) notcong mathbf F_2[X]/(f)$ for any $f$ in $mathbf F_2[X]$. Also $mathcal O_L not= mathbf Z[alpha]$ for some $alpha$ in $mathcal O_L$.
answered 6 hours ago
KConradKConrad
31.1k5 gold badges134 silver badges207 bronze badges
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$begingroup$
The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $mathcal O_L/(p) cong mathbf F_p^r$ but if $mathcal O_L/(p) cong mathbf F_p[X]/(f)$ for monic $f$ in $mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $mathcal O_L$ is not $mathbf Z[alpha]$ for some $alpha$ ($L$ is not monogenic).
Dedekind found the first example of this: $L = mathbf Q(theta)$ where $theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $mathcal O_L/(2) cong mathbf F_2^3$ and $mathcal O_L/(2) notcong mathbf F_2[X]/(f)$ for any $f$ in $mathbf F_2[X]$. Also $mathcal O_L not= mathbf Z[alpha]$ for some $alpha$ in $mathcal O_L$.
answered 6 hours ago
KConradKConrad
31.1k5 gold badges134 silver badges207 bronze badges
$endgroup$
add a comment |
$begingroup$
The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $mathcal O_L/(p) cong mathbf F_p^r$ but if $mathcal O_L/(p) cong mathbf F_p[X]/(f)$ for monic $f$ in $mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $mathcal O_L$ is not $mathbf Z[alpha]$ for some $alpha$ ($L$ is not monogenic).
Dedekind found the first example of this: $L = mathbf Q(theta)$ where $theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $mathcal O_L/(2) cong mathbf F_2^3$ and $mathcal O_L/(2) notcong mathbf F_2[X]/(f)$ for any $f$ in $mathbf F_2[X]$. Also $mathcal O_L not= mathbf Z[alpha]$ for some $alpha$ in $mathcal O_L$.
answered 6 hours ago
KConradKConrad
31.1k5 gold badges134 silver badges207 bronze badges
$endgroup$
add a comment |
$begingroup$
The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $mathcal O_L/(p) cong mathbf F_p^r$ but if $mathcal O_L/(p) cong mathbf F_p[X]/(f)$ for monic $f$ in $mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $mathcal O_L$ is not $mathbf Z[alpha]$ for some $alpha$ ($L$ is not monogenic).
Dedekind found the first example of this: $L = mathbf Q(theta)$ where $theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $mathcal O_L/(2) cong mathbf F_2^3$ and $mathcal O_L/(2) notcong mathbf F_2[X]/(f)$ for any $f$ in $mathbf F_2[X]$. Also $mathcal O_L not= mathbf Z[alpha]$ for some $alpha$ in $mathcal O_L$.
answered 6 hours ago
KConradKConrad
31.1k5 gold badges134 silver badges207 bronze badges
$endgroup$
The answer to your main question is no. Let $I = (p)$ where $p$ splits completely in $L$ and $r := [L:mathbf Q] > p$. (Example: cubic field in which $p = 2$ splits completely.) Then $mathcal O_L/(p) cong mathbf F_p^r$ but if $mathcal O_L/(p) cong mathbf F_p[X]/(f)$ for monic $f$ in $mathbf F_p[X]$ then $f$ is a product of $r$ distinct monic linear factors, which is impossible for $r > p$. This proves $mathcal O_L$ is not $mathbf Z[alpha]$ for some $alpha$ ($L$ is not monogenic).
Dedekind found the first example of this: $L = mathbf Q(theta)$ where $theta$ is a root of $X^3-X^2-2X-8$. This is a cubic field in which $2$ splits completely. Thus $mathcal O_L/(2) cong mathbf F_2^3$ and $mathcal O_L/(2) notcong mathbf F_2[X]/(f)$ for any $f$ in $mathbf F_2[X]$. Also $mathcal O_L not= mathbf Z[alpha]$ for some $alpha$ in $mathcal O_L$.
answered 6 hours ago
KConradKConrad
31.1k5 gold badges134 silver badges207 bronze badges
edited 6 hours ago
answered 6 hours ago
KConradKConrad
31.1k5 gold badges134 silver badges207 bronze badges
answered 6 hours ago
KConradKConrad
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answered 6 hours ago
KConradKConrad
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31.1k5 gold badges134 silver badges207 bronze badges
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4
$begingroup$
Here is one counterexample. Let $K$ be $mathbbQ$. Let $L$ be $mathbbQ[s,t]/langle s^2+s+2,t^2+t+4 rangle$. Let $p$ be $2$, and let $I$ be $2mathcalO_L$. Then $mathcalO_L/I$ equals $mathbbF_2[s,t]/langle s^2+s,t^2+t rangle$. This is isomorphic as an $mathbbF_2$-algebra to a product of four copies of $mathbbF_2$. Yet a monic polynomial has at most two $mathbbF_2$-rational points.
$endgroup$
– Jason Starr
8 hours ago