Mixing
the ring of integers: prove that (2) is a prime ideal and that it is a pid
$begingroup$
Consider a real root $alpha$ of $f(X)=X^3-3X+1$. Consider the ring of integers $A_K$ for $K=mathbf{Q}[alpha]$. I showed that the ideal $(1+alpha)$ is prime in $A_K$ and that $A_K=mathbf{Z}[alpha]+(alpha +1)A_K$. It is clear that the discriminant of $mathbf{Z}[alpha]$ is $3^4$. How can I prove that $A_K=mathbf{Z}[alpha]$ and that the ideal $(2)$ is prime in $A_K$ and $A_K$ is a PID?
Thanks everyone for the help.
algebraic-number-theory integer-rings
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
$endgroup$
add a comment |
$begingroup$
Consider a real root $alpha$ of $f(X)=X^3-3X+1$. Consider the ring of integers $A_K$ for $K=mathbf{Q}[alpha]$. I showed that the ideal $(1+alpha)$ is prime in $A_K$ and that $A_K=mathbf{Z}[alpha]+(alpha +1)A_K$. It is clear that the discriminant of $mathbf{Z}[alpha]$ is $3^4$. How can I prove that $A_K=mathbf{Z}[alpha]$ and that the ideal $(2)$ is prime in $A_K$ and $A_K$ is a PID?
Thanks everyone for the help.
algebraic-number-theory integer-rings
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
$endgroup$
$begingroup$
$f(X-1)$ is Eisenstein at $3$, and $f$ is irreducible modulo $2$. Then use Minkowski bounds.
$endgroup$
– franz lemmermeyer
Jan 13 at 16:41
$begingroup$
Thanks @franzlemmermeyer for the precious hints! Can you give me some suggestions to prove the equality $A_K=mathbf{Z}[alpha]$?
$endgroup$
– Lei Feima
Jan 14 at 8:21
add a comment |
$begingroup$
Consider a real root $alpha$ of $f(X)=X^3-3X+1$. Consider the ring of integers $A_K$ for $K=mathbf{Q}[alpha]$. I showed that the ideal $(1+alpha)$ is prime in $A_K$ and that $A_K=mathbf{Z}[alpha]+(alpha +1)A_K$. It is clear that the discriminant of $mathbf{Z}[alpha]$ is $3^4$. How can I prove that $A_K=mathbf{Z}[alpha]$ and that the ideal $(2)$ is prime in $A_K$ and $A_K$ is a PID?
Thanks everyone for the help.
algebraic-number-theory integer-rings
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
$endgroup$
Consider a real root $alpha$ of $f(X)=X^3-3X+1$. Consider the ring of integers $A_K$ for $K=mathbf{Q}[alpha]$. I showed that the ideal $(1+alpha)$ is prime in $A_K$ and that $A_K=mathbf{Z}[alpha]+(alpha +1)A_K$. It is clear that the discriminant of $mathbf{Z}[alpha]$ is $3^4$. How can I prove that $A_K=mathbf{Z}[alpha]$ and that the ideal $(2)$ is prime in $A_K$ and $A_K$ is a PID?
Thanks everyone for the help.
algebraic-number-theory integer-rings
algebraic-number-theory integer-rings
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
asked Jan 13 at 15:43
Lei FeimaLei Feima
867
867
$begingroup$
$f(X-1)$ is Eisenstein at $3$, and $f$ is irreducible modulo $2$. Then use Minkowski bounds.
$endgroup$
– franz lemmermeyer
Jan 13 at 16:41
$begingroup$
Thanks @franzlemmermeyer for the precious hints! Can you give me some suggestions to prove the equality $A_K=mathbf{Z}[alpha]$?
$endgroup$
– Lei Feima
Jan 14 at 8:21
add a comment |
$begingroup$
$f(X-1)$ is Eisenstein at $3$, and $f$ is irreducible modulo $2$. Then use Minkowski bounds.
$endgroup$
– franz lemmermeyer
Jan 13 at 16:41
$begingroup$
Thanks @franzlemmermeyer for the precious hints! Can you give me some suggestions to prove the equality $A_K=mathbf{Z}[alpha]$?
$endgroup$
– Lei Feima
Jan 14 at 8:21
$begingroup$
$f(X-1)$ is Eisenstein at $3$, and $f$ is irreducible modulo $2$. Then use Minkowski bounds.
$endgroup$
– franz lemmermeyer
Jan 13 at 16:41
$begingroup$
$f(X-1)$ is Eisenstein at $3$, and $f$ is irreducible modulo $2$. Then use Minkowski bounds.
$endgroup$
– franz lemmermeyer
Jan 13 at 16:41
$begingroup$
Thanks @franzlemmermeyer for the precious hints! Can you give me some suggestions to prove the equality $A_K=mathbf{Z}[alpha]$?
$endgroup$
– Lei Feima
Jan 14 at 8:21
$begingroup$
Thanks @franzlemmermeyer for the precious hints! Can you give me some suggestions to prove the equality $A_K=mathbf{Z}[alpha]$?
$endgroup$
– Lei Feima
Jan 14 at 8:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $f(x) = x^n + a_{n-1}x^{n-1} + ldots + a_0$ be Eisenstein at $p$
(all $a_j$ are integers divisible by $p$ and $p^2 nmid a_0$) and let $alpha$ be a root of $f$. Then $p$ does not divide the index of $alpha$, that is, if
$$ xi = frac{r_1 + r_2 alpha + ldots +r_{n-1}alpha^{n-1}}p $$
is an algebraic integer, then $p$ divides all the $r_j$. The proof is simple:
$xi alpha$ is an algebraic integer, but $alpha^n$ is divisible by $p$,
hence
$$ xi_1 = frac{r_1 alpha + ldots +r_{n-2}alpha^{n-2}}p $$
is an algebraic integer. Repeating this step the claim follows.
In your case, $f(X-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$, and since
the discriminant of this polynomial is a power of $3$, we find that ${1, alpha, alpha^2}$ must be an integral basis.
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
$endgroup$
$begingroup$
Thank you for the proof. Do you think that it is correct the following way: I know that $A_K 0mathbf{Z}[alpha]+(alpha+1)A_K$ and that $3^m A_K subset mathbf{Z}[alpha] subset A_K$, so $A_K=mathbf{Z}[alpha]+(alpha +1)A_K subset mathbf{Z}[alpha]+3A_K subset mathbf{Z}[alpha]+mathbf{Z}[alpha]subset mathbf{Z}[alpha]$? Thanks
$endgroup$
– Lei Feima
Jan 18 at 14:48
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
oldest
votes
$begingroup$
Let $f(x) = x^n + a_{n-1}x^{n-1} + ldots + a_0$ be Eisenstein at $p$
(all $a_j$ are integers divisible by $p$ and $p^2 nmid a_0$) and let $alpha$ be a root of $f$. Then $p$ does not divide the index of $alpha$, that is, if
$$ xi = frac{r_1 + r_2 alpha + ldots +r_{n-1}alpha^{n-1}}p $$
is an algebraic integer, then $p$ divides all the $r_j$. The proof is simple:
$xi alpha$ is an algebraic integer, but $alpha^n$ is divisible by $p$,
hence
$$ xi_1 = frac{r_1 alpha + ldots +r_{n-2}alpha^{n-2}}p $$
is an algebraic integer. Repeating this step the claim follows.
In your case, $f(X-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$, and since
the discriminant of this polynomial is a power of $3$, we find that ${1, alpha, alpha^2}$ must be an integral basis.
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
$endgroup$
$begingroup$
Thank you for the proof. Do you think that it is correct the following way: I know that $A_K 0mathbf{Z}[alpha]+(alpha+1)A_K$ and that $3^m A_K subset mathbf{Z}[alpha] subset A_K$, so $A_K=mathbf{Z}[alpha]+(alpha +1)A_K subset mathbf{Z}[alpha]+3A_K subset mathbf{Z}[alpha]+mathbf{Z}[alpha]subset mathbf{Z}[alpha]$? Thanks
$endgroup$
– Lei Feima
Jan 18 at 14:48
add a comment |
$begingroup$
Let $f(x) = x^n + a_{n-1}x^{n-1} + ldots + a_0$ be Eisenstein at $p$
(all $a_j$ are integers divisible by $p$ and $p^2 nmid a_0$) and let $alpha$ be a root of $f$. Then $p$ does not divide the index of $alpha$, that is, if
$$ xi = frac{r_1 + r_2 alpha + ldots +r_{n-1}alpha^{n-1}}p $$
is an algebraic integer, then $p$ divides all the $r_j$. The proof is simple:
$xi alpha$ is an algebraic integer, but $alpha^n$ is divisible by $p$,
hence
$$ xi_1 = frac{r_1 alpha + ldots +r_{n-2}alpha^{n-2}}p $$
is an algebraic integer. Repeating this step the claim follows.
In your case, $f(X-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$, and since
the discriminant of this polynomial is a power of $3$, we find that ${1, alpha, alpha^2}$ must be an integral basis.
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
$endgroup$
$begingroup$
Thank you for the proof. Do you think that it is correct the following way: I know that $A_K 0mathbf{Z}[alpha]+(alpha+1)A_K$ and that $3^m A_K subset mathbf{Z}[alpha] subset A_K$, so $A_K=mathbf{Z}[alpha]+(alpha +1)A_K subset mathbf{Z}[alpha]+3A_K subset mathbf{Z}[alpha]+mathbf{Z}[alpha]subset mathbf{Z}[alpha]$? Thanks
$endgroup$
– Lei Feima
Jan 18 at 14:48
add a comment |
$begingroup$
Let $f(x) = x^n + a_{n-1}x^{n-1} + ldots + a_0$ be Eisenstein at $p$
(all $a_j$ are integers divisible by $p$ and $p^2 nmid a_0$) and let $alpha$ be a root of $f$. Then $p$ does not divide the index of $alpha$, that is, if
$$ xi = frac{r_1 + r_2 alpha + ldots +r_{n-1}alpha^{n-1}}p $$
is an algebraic integer, then $p$ divides all the $r_j$. The proof is simple:
$xi alpha$ is an algebraic integer, but $alpha^n$ is divisible by $p$,
hence
$$ xi_1 = frac{r_1 alpha + ldots +r_{n-2}alpha^{n-2}}p $$
is an algebraic integer. Repeating this step the claim follows.
In your case, $f(X-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$, and since
the discriminant of this polynomial is a power of $3$, we find that ${1, alpha, alpha^2}$ must be an integral basis.
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
$endgroup$
Let $f(x) = x^n + a_{n-1}x^{n-1} + ldots + a_0$ be Eisenstein at $p$
(all $a_j$ are integers divisible by $p$ and $p^2 nmid a_0$) and let $alpha$ be a root of $f$. Then $p$ does not divide the index of $alpha$, that is, if
$$ xi = frac{r_1 + r_2 alpha + ldots +r_{n-1}alpha^{n-1}}p $$
is an algebraic integer, then $p$ divides all the $r_j$. The proof is simple:
$xi alpha$ is an algebraic integer, but $alpha^n$ is divisible by $p$,
hence
$$ xi_1 = frac{r_1 alpha + ldots +r_{n-2}alpha^{n-2}}p $$
is an algebraic integer. Repeating this step the claim follows.
In your case, $f(X-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$, and since
the discriminant of this polynomial is a power of $3$, we find that ${1, alpha, alpha^2}$ must be an integral basis.
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
answered Jan 14 at 15:45
franz lemmermeyerfranz lemmermeyer
7,13922047
7,13922047
$begingroup$
Thank you for the proof. Do you think that it is correct the following way: I know that $A_K 0mathbf{Z}[alpha]+(alpha+1)A_K$ and that $3^m A_K subset mathbf{Z}[alpha] subset A_K$, so $A_K=mathbf{Z}[alpha]+(alpha +1)A_K subset mathbf{Z}[alpha]+3A_K subset mathbf{Z}[alpha]+mathbf{Z}[alpha]subset mathbf{Z}[alpha]$? Thanks
$endgroup$
– Lei Feima
Jan 18 at 14:48
add a comment |
$begingroup$
Thank you for the proof. Do you think that it is correct the following way: I know that $A_K 0mathbf{Z}[alpha]+(alpha+1)A_K$ and that $3^m A_K subset mathbf{Z}[alpha] subset A_K$, so $A_K=mathbf{Z}[alpha]+(alpha +1)A_K subset mathbf{Z}[alpha]+3A_K subset mathbf{Z}[alpha]+mathbf{Z}[alpha]subset mathbf{Z}[alpha]$? Thanks
$endgroup$
– Lei Feima
Jan 18 at 14:48
$begingroup$
Thank you for the proof. Do you think that it is correct the following way: I know that $A_K 0mathbf{Z}[alpha]+(alpha+1)A_K$ and that $3^m A_K subset mathbf{Z}[alpha] subset A_K$, so $A_K=mathbf{Z}[alpha]+(alpha +1)A_K subset mathbf{Z}[alpha]+3A_K subset mathbf{Z}[alpha]+mathbf{Z}[alpha]subset mathbf{Z}[alpha]$? Thanks
$endgroup$
– Lei Feima
Jan 18 at 14:48
$begingroup$
Thank you for the proof. Do you think that it is correct the following way: I know that $A_K 0mathbf{Z}[alpha]+(alpha+1)A_K$ and that $3^m A_K subset mathbf{Z}[alpha] subset A_K$, so $A_K=mathbf{Z}[alpha]+(alpha +1)A_K subset mathbf{Z}[alpha]+3A_K subset mathbf{Z}[alpha]+mathbf{Z}[alpha]subset mathbf{Z}[alpha]$? Thanks
$endgroup$
– Lei Feima
Jan 18 at 14:48
add a comment |
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$begingroup$
$f(X-1)$ is Eisenstein at $3$, and $f$ is irreducible modulo $2$. Then use Minkowski bounds.
$endgroup$
– franz lemmermeyer
Jan 13 at 16:41
$begingroup$
Thanks @franzlemmermeyer for the precious hints! Can you give me some suggestions to prove the equality $A_K=mathbf{Z}[alpha]$?
$endgroup$
– Lei Feima
Jan 14 at 8:21