Mixing
Ideals of every “dimension” in the ring Z[x]
$begingroup$
Let Z[x] be the ring of polynomials with integer coefficients.
I need help proving that given any natural number m there exists a collection of m polynomials such that the ideal generated by these is not generated by any collection of m-1 polynomials. I.e, there are ideals of every dimension in Z[x].
Where by "dimension" of an ideal J, I mean the least number of elements needed to generate J.
abstract-algebra ideals
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
$endgroup$
add a comment |
$begingroup$
Let Z[x] be the ring of polynomials with integer coefficients.
I need help proving that given any natural number m there exists a collection of m polynomials such that the ideal generated by these is not generated by any collection of m-1 polynomials. I.e, there are ideals of every dimension in Z[x].
Where by "dimension" of an ideal J, I mean the least number of elements needed to generate J.
abstract-algebra ideals
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
$endgroup$
2
$begingroup$
That's an unusual definition of dimension. Can you prove that if $M$ is a maximal ideal of the commutative ring $R$, $k=R/M$ is the quotient field, and $I$ is any ideal of $R$, then $I$ requires at least $m$ generators where $m=dim_k I/MI$?
$endgroup$
– Lord Shark the Unknown
Feb 3 at 5:14
add a comment |
$begingroup$
Let Z[x] be the ring of polynomials with integer coefficients.
I need help proving that given any natural number m there exists a collection of m polynomials such that the ideal generated by these is not generated by any collection of m-1 polynomials. I.e, there are ideals of every dimension in Z[x].
Where by "dimension" of an ideal J, I mean the least number of elements needed to generate J.
abstract-algebra ideals
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
$endgroup$
Let Z[x] be the ring of polynomials with integer coefficients.
I need help proving that given any natural number m there exists a collection of m polynomials such that the ideal generated by these is not generated by any collection of m-1 polynomials. I.e, there are ideals of every dimension in Z[x].
Where by "dimension" of an ideal J, I mean the least number of elements needed to generate J.
abstract-algebra ideals
abstract-algebra ideals
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
asked Feb 3 at 5:00
Manuel PicoManuel Pico
464
464
2
$begingroup$
That's an unusual definition of dimension. Can you prove that if $M$ is a maximal ideal of the commutative ring $R$, $k=R/M$ is the quotient field, and $I$ is any ideal of $R$, then $I$ requires at least $m$ generators where $m=dim_k I/MI$?
$endgroup$
– Lord Shark the Unknown
Feb 3 at 5:14
add a comment |
2
$begingroup$
That's an unusual definition of dimension. Can you prove that if $M$ is a maximal ideal of the commutative ring $R$, $k=R/M$ is the quotient field, and $I$ is any ideal of $R$, then $I$ requires at least $m$ generators where $m=dim_k I/MI$?
$endgroup$
– Lord Shark the Unknown
Feb 3 at 5:14
2
2
$begingroup$
That's an unusual definition of dimension. Can you prove that if $M$ is a maximal ideal of the commutative ring $R$, $k=R/M$ is the quotient field, and $I$ is any ideal of $R$, then $I$ requires at least $m$ generators where $m=dim_k I/MI$?
$endgroup$
– Lord Shark the Unknown
Feb 3 at 5:14
$begingroup$
That's an unusual definition of dimension. Can you prove that if $M$ is a maximal ideal of the commutative ring $R$, $k=R/M$ is the quotient field, and $I$ is any ideal of $R$, then $I$ requires at least $m$ generators where $m=dim_k I/MI$?
$endgroup$
– Lord Shark the Unknown
Feb 3 at 5:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Peeking at this answer , I'm going to venture the following guess: $(2^m,2^{m-1}x+2^{m-1},dots,2x^{m-1}+2)$.
On the other hand, this answer may be more to the point. Apparently we could do: $(p^n,p^{n-1}x,dots,px^{n-1},x^n)$, to get an ideal that can't be generated by fewer than $n+1$ elements.
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
Peeking at this answer , I'm going to venture the following guess: $(2^m,2^{m-1}x+2^{m-1},dots,2x^{m-1}+2)$.
On the other hand, this answer may be more to the point. Apparently we could do: $(p^n,p^{n-1}x,dots,px^{n-1},x^n)$, to get an ideal that can't be generated by fewer than $n+1$ elements.
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Peeking at this answer , I'm going to venture the following guess: $(2^m,2^{m-1}x+2^{m-1},dots,2x^{m-1}+2)$.
On the other hand, this answer may be more to the point. Apparently we could do: $(p^n,p^{n-1}x,dots,px^{n-1},x^n)$, to get an ideal that can't be generated by fewer than $n+1$ elements.
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
Peeking at this answer , I'm going to venture the following guess: $(2^m,2^{m-1}x+2^{m-1},dots,2x^{m-1}+2)$.
On the other hand, this answer may be more to the point. Apparently we could do: $(p^n,p^{n-1}x,dots,px^{n-1},x^n)$, to get an ideal that can't be generated by fewer than $n+1$ elements.
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
$endgroup$
Peeking at this answer , I'm going to venture the following guess: $(2^m,2^{m-1}x+2^{m-1},dots,2x^{m-1}+2)$.
On the other hand, this answer may be more to the point. Apparently we could do: $(p^n,p^{n-1}x,dots,px^{n-1},x^n)$, to get an ideal that can't be generated by fewer than $n+1$ elements.
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
edited Feb 3 at 6:23
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
answered Feb 3 at 6:03
Chris CusterChris Custer
14.4k3827
14.4k3827
add a comment |
add a comment |
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That's an unusual definition of dimension. Can you prove that if $M$ is a maximal ideal of the commutative ring $R$, $k=R/M$ is the quotient field, and $I$ is any ideal of $R$, then $I$ requires at least $m$ generators where $m=dim_k I/MI$?
$endgroup$
– Lord Shark the Unknown
Feb 3 at 5:14