Find all $mathfrak{p}in operatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$...
$begingroup$
Find all $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$ where $zeta=e^{(ipi/3)}.$
I determined that the minimal polynomial is $f(x)=x^2-x+1$. Since $f(zeta)=0$ and $f$ is an irreducible quadratic it must be the minimal polynomial.
Then $mathbb{Z}[zeta]congmathbb{Z}[x]/(x^2-x+1)$. Since this is an integral extension of $mathbb{Z}$ it has the same Krull dimension, namely $1$. Primes of $mathbb{Z}[x]/(x^2-x+1)$ are primes of $mathbb{Z}[x]$ that contain $(x^2-x+1)$ so all primes in $mathbb{Z}[x]/(x^2-x+1)$ look like $(q,x^2-x+1)$ where $qinmathbb{Z}$ is prime. Thus there is only a single prime $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$, namely $(7,x^2-x+1)$, which is the ideal $(7)subset mathbb{Z}[zeta]$.
Somehow the wording of the problem "find all prime ideals and give generators" made me expect more... Is this correct? Many thanks for an answer.
ring-theory extension-field maximal-and-prime-ideals
edited Dec 17 '18 at 10:47
Namaste
1
asked May 23 '17 at 20:05
irhirh
304115
$endgroup$
add a comment |
$begingroup$
Find all $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$ where $zeta=e^{(ipi/3)}.$
I determined that the minimal polynomial is $f(x)=x^2-x+1$. Since $f(zeta)=0$ and $f$ is an irreducible quadratic it must be the minimal polynomial.
Then $mathbb{Z}[zeta]congmathbb{Z}[x]/(x^2-x+1)$. Since this is an integral extension of $mathbb{Z}$ it has the same Krull dimension, namely $1$. Primes of $mathbb{Z}[x]/(x^2-x+1)$ are primes of $mathbb{Z}[x]$ that contain $(x^2-x+1)$ so all primes in $mathbb{Z}[x]/(x^2-x+1)$ look like $(q,x^2-x+1)$ where $qinmathbb{Z}$ is prime. Thus there is only a single prime $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$, namely $(7,x^2-x+1)$, which is the ideal $(7)subset mathbb{Z}[zeta]$.
Somehow the wording of the problem "find all prime ideals and give generators" made me expect more... Is this correct? Many thanks for an answer.
ring-theory extension-field maximal-and-prime-ideals
edited Dec 17 '18 at 10:47
Namaste
1
asked May 23 '17 at 20:05
irhirh
304115
$endgroup$
1
$begingroup$
Modulo $7$ we have the factorization $x^2-x+1equiv (x-3)(x+2)$ so you have ideals like $(7,x-3)$ that contain $(7,x^2-x+1)$. Clearly $(7,x-3)$ is a prime ideal, but $(7,x^2-x+1)$ is not because of the above factorization.
$endgroup$
– Jyrki Lahtonen
May 23 '17 at 20:13
$begingroup$
Essentially, 7 is not prime in $mathbb{Z}[zeta]$, is the flaw in my reasoning.
$endgroup$
– irh
May 23 '17 at 21:07
$begingroup$
That's another correct way of looking at it! It happens that $Bbb{Z}[zeta]$ is a PID (it's actually a Euclidean domain), but that is usually not the case. This kind of questions are studied extensively in algebraic number theory.
$endgroup$
– Jyrki Lahtonen
May 24 '17 at 4:24
add a comment |
$begingroup$
Find all $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$ where $zeta=e^{(ipi/3)}.$
I determined that the minimal polynomial is $f(x)=x^2-x+1$. Since $f(zeta)=0$ and $f$ is an irreducible quadratic it must be the minimal polynomial.
Then $mathbb{Z}[zeta]congmathbb{Z}[x]/(x^2-x+1)$. Since this is an integral extension of $mathbb{Z}$ it has the same Krull dimension, namely $1$. Primes of $mathbb{Z}[x]/(x^2-x+1)$ are primes of $mathbb{Z}[x]$ that contain $(x^2-x+1)$ so all primes in $mathbb{Z}[x]/(x^2-x+1)$ look like $(q,x^2-x+1)$ where $qinmathbb{Z}$ is prime. Thus there is only a single prime $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$, namely $(7,x^2-x+1)$, which is the ideal $(7)subset mathbb{Z}[zeta]$.
Somehow the wording of the problem "find all prime ideals and give generators" made me expect more... Is this correct? Many thanks for an answer.
ring-theory extension-field maximal-and-prime-ideals
edited Dec 17 '18 at 10:47
Namaste
1
asked May 23 '17 at 20:05
irhirh
304115
$endgroup$
Find all $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$ where $zeta=e^{(ipi/3)}.$
I determined that the minimal polynomial is $f(x)=x^2-x+1$. Since $f(zeta)=0$ and $f$ is an irreducible quadratic it must be the minimal polynomial.
Then $mathbb{Z}[zeta]congmathbb{Z}[x]/(x^2-x+1)$. Since this is an integral extension of $mathbb{Z}$ it has the same Krull dimension, namely $1$. Primes of $mathbb{Z}[x]/(x^2-x+1)$ are primes of $mathbb{Z}[x]$ that contain $(x^2-x+1)$ so all primes in $mathbb{Z}[x]/(x^2-x+1)$ look like $(q,x^2-x+1)$ where $qinmathbb{Z}$ is prime. Thus there is only a single prime $mathfrak{p}inoperatorname{Spec}(mathbb{Z}[zeta])$ such that $mathfrak{p}capmathbb{Z}=7mathbb{Z}$, namely $(7,x^2-x+1)$, which is the ideal $(7)subset mathbb{Z}[zeta]$.
Somehow the wording of the problem "find all prime ideals and give generators" made me expect more... Is this correct? Many thanks for an answer.
ring-theory extension-field maximal-and-prime-ideals
ring-theory extension-field maximal-and-prime-ideals
edited Dec 17 '18 at 10:47
Namaste
1
asked May 23 '17 at 20:05
irhirh
304115
edited Dec 17 '18 at 10:47
Namaste
1
asked May 23 '17 at 20:05
irhirh
304115
edited Dec 17 '18 at 10:47
Namaste
1
edited Dec 17 '18 at 10:47
Namaste
1
edited Dec 17 '18 at 10:47
Namaste
1
1
asked May 23 '17 at 20:05
irhirh
304115
asked May 23 '17 at 20:05
irhirh
304115
asked May 23 '17 at 20:05
irhirh
304115
304115
1
$begingroup$
Modulo $7$ we have the factorization $x^2-x+1equiv (x-3)(x+2)$ so you have ideals like $(7,x-3)$ that contain $(7,x^2-x+1)$. Clearly $(7,x-3)$ is a prime ideal, but $(7,x^2-x+1)$ is not because of the above factorization.
$endgroup$
– Jyrki Lahtonen
May 23 '17 at 20:13
$begingroup$
Essentially, 7 is not prime in $mathbb{Z}[zeta]$, is the flaw in my reasoning.
$endgroup$
– irh
May 23 '17 at 21:07
$begingroup$
That's another correct way of looking at it! It happens that $Bbb{Z}[zeta]$ is a PID (it's actually a Euclidean domain), but that is usually not the case. This kind of questions are studied extensively in algebraic number theory.
$endgroup$
– Jyrki Lahtonen
May 24 '17 at 4:24
add a comment |
1
$begingroup$
Modulo $7$ we have the factorization $x^2-x+1equiv (x-3)(x+2)$ so you have ideals like $(7,x-3)$ that contain $(7,x^2-x+1)$. Clearly $(7,x-3)$ is a prime ideal, but $(7,x^2-x+1)$ is not because of the above factorization.
$endgroup$
– Jyrki Lahtonen
May 23 '17 at 20:13
$begingroup$
Essentially, 7 is not prime in $mathbb{Z}[zeta]$, is the flaw in my reasoning.
$endgroup$
– irh
May 23 '17 at 21:07
$begingroup$
That's another correct way of looking at it! It happens that $Bbb{Z}[zeta]$ is a PID (it's actually a Euclidean domain), but that is usually not the case. This kind of questions are studied extensively in algebraic number theory.
$endgroup$
– Jyrki Lahtonen
May 24 '17 at 4:24
1
1
$begingroup$
Modulo $7$ we have the factorization $x^2-x+1equiv (x-3)(x+2)$ so you have ideals like $(7,x-3)$ that contain $(7,x^2-x+1)$. Clearly $(7,x-3)$ is a prime ideal, but $(7,x^2-x+1)$ is not because of the above factorization.
$endgroup$
– Jyrki Lahtonen
May 23 '17 at 20:13
$begingroup$
Modulo $7$ we have the factorization $x^2-x+1equiv (x-3)(x+2)$ so you have ideals like $(7,x-3)$ that contain $(7,x^2-x+1)$. Clearly $(7,x-3)$ is a prime ideal, but $(7,x^2-x+1)$ is not because of the above factorization.
$endgroup$
– Jyrki Lahtonen
May 23 '17 at 20:13
$begingroup$
Essentially, 7 is not prime in $mathbb{Z}[zeta]$, is the flaw in my reasoning.
$endgroup$
– irh
May 23 '17 at 21:07
$begingroup$
Essentially, 7 is not prime in $mathbb{Z}[zeta]$, is the flaw in my reasoning.
$endgroup$
– irh
May 23 '17 at 21:07
$begingroup$
That's another correct way of looking at it! It happens that $Bbb{Z}[zeta]$ is a PID (it's actually a Euclidean domain), but that is usually not the case. This kind of questions are studied extensively in algebraic number theory.
$endgroup$
– Jyrki Lahtonen
May 24 '17 at 4:24
$begingroup$
That's another correct way of looking at it! It happens that $Bbb{Z}[zeta]$ is a PID (it's actually a Euclidean domain), but that is usually not the case. This kind of questions are studied extensively in algebraic number theory.
$endgroup$
– Jyrki Lahtonen
May 24 '17 at 4:24
add a comment |
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$begingroup$
Modulo $7$ we have the factorization $x^2-x+1equiv (x-3)(x+2)$ so you have ideals like $(7,x-3)$ that contain $(7,x^2-x+1)$. Clearly $(7,x-3)$ is a prime ideal, but $(7,x^2-x+1)$ is not because of the above factorization.
$endgroup$
– Jyrki Lahtonen
May 23 '17 at 20:13
$begingroup$
Essentially, 7 is not prime in $mathbb{Z}[zeta]$, is the flaw in my reasoning.
$endgroup$
– irh
May 23 '17 at 21:07
$begingroup$
That's another correct way of looking at it! It happens that $Bbb{Z}[zeta]$ is a PID (it's actually a Euclidean domain), but that is usually not the case. This kind of questions are studied extensively in algebraic number theory.
$endgroup$
– Jyrki Lahtonen
May 24 '17 at 4:24