Dimension of an ideal of a coordinate ring
I am struggling to understand the following point in a paper (the paper is this one, concretely section 4.2).
Let's consider a generic quintic in $mathbb{P}^{4}$. Then its coordinate ring would be: $A=frac{mathbb{C}left [ x_{0},...,x_{4} right ] }{p}$ where $x_{i}$ are projective coordinates and $p$ is the defining quintic polynomial.
Now, (because other considerations explained in the paper) let $f_{1i}$, with $i=1,...,7$ be an ideal of $A$, formed by seven polynomials in $mathbb{P}^{4}$. The idea is to particularize that ideal $f_{1i}$ in such a way that its dimension is $1$ at degree $4$. The text gives this choice as an example:
$f_{1i}=(40x_{3}+94x_{4}, 117x_{3}+119x_{4}, 449x_{3}+464x_{4}+266x_{0}+195x_{1}+173x_{2}, 306x_{2}, 273x_{3},259x_{3}+291x_{4},76x_{3}+98x_{2})$
My idea of the dimension of an ideal $I$ in a field $k$ at certain degree $s$ is that it corresponds to the dimension of $I_{leq s}$ as a vector space over $k$, being $I_{leq s}$ the set of polynomials in $I$ of total degree $leq s$.
It seems clear that this is not the definition of dimension that it is being used here. Can someone explain why the dimension of that choice of $f_{1i}$ is $1$ at degree $4$ (or point out what I am misunderstanding)?
algebraic-geometry commutative-algebra ideals
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
add a comment |
I am struggling to understand the following point in a paper (the paper is this one, concretely section 4.2).
Let's consider a generic quintic in $mathbb{P}^{4}$. Then its coordinate ring would be: $A=frac{mathbb{C}left [ x_{0},...,x_{4} right ] }{p}$ where $x_{i}$ are projective coordinates and $p$ is the defining quintic polynomial.
Now, (because other considerations explained in the paper) let $f_{1i}$, with $i=1,...,7$ be an ideal of $A$, formed by seven polynomials in $mathbb{P}^{4}$. The idea is to particularize that ideal $f_{1i}$ in such a way that its dimension is $1$ at degree $4$. The text gives this choice as an example:
$f_{1i}=(40x_{3}+94x_{4}, 117x_{3}+119x_{4}, 449x_{3}+464x_{4}+266x_{0}+195x_{1}+173x_{2}, 306x_{2}, 273x_{3},259x_{3}+291x_{4},76x_{3}+98x_{2})$
My idea of the dimension of an ideal $I$ in a field $k$ at certain degree $s$ is that it corresponds to the dimension of $I_{leq s}$ as a vector space over $k$, being $I_{leq s}$ the set of polynomials in $I$ of total degree $leq s$.
It seems clear that this is not the definition of dimension that it is being used here. Can someone explain why the dimension of that choice of $f_{1i}$ is $1$ at degree $4$ (or point out what I am misunderstanding)?
algebraic-geometry commutative-algebra ideals
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
add a comment |
I am struggling to understand the following point in a paper (the paper is this one, concretely section 4.2).
Let's consider a generic quintic in $mathbb{P}^{4}$. Then its coordinate ring would be: $A=frac{mathbb{C}left [ x_{0},...,x_{4} right ] }{p}$ where $x_{i}$ are projective coordinates and $p$ is the defining quintic polynomial.
Now, (because other considerations explained in the paper) let $f_{1i}$, with $i=1,...,7$ be an ideal of $A$, formed by seven polynomials in $mathbb{P}^{4}$. The idea is to particularize that ideal $f_{1i}$ in such a way that its dimension is $1$ at degree $4$. The text gives this choice as an example:
$f_{1i}=(40x_{3}+94x_{4}, 117x_{3}+119x_{4}, 449x_{3}+464x_{4}+266x_{0}+195x_{1}+173x_{2}, 306x_{2}, 273x_{3},259x_{3}+291x_{4},76x_{3}+98x_{2})$
My idea of the dimension of an ideal $I$ in a field $k$ at certain degree $s$ is that it corresponds to the dimension of $I_{leq s}$ as a vector space over $k$, being $I_{leq s}$ the set of polynomials in $I$ of total degree $leq s$.
It seems clear that this is not the definition of dimension that it is being used here. Can someone explain why the dimension of that choice of $f_{1i}$ is $1$ at degree $4$ (or point out what I am misunderstanding)?
algebraic-geometry commutative-algebra ideals
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
I am struggling to understand the following point in a paper (the paper is this one, concretely section 4.2).
Let's consider a generic quintic in $mathbb{P}^{4}$. Then its coordinate ring would be: $A=frac{mathbb{C}left [ x_{0},...,x_{4} right ] }{p}$ where $x_{i}$ are projective coordinates and $p$ is the defining quintic polynomial.
Now, (because other considerations explained in the paper) let $f_{1i}$, with $i=1,...,7$ be an ideal of $A$, formed by seven polynomials in $mathbb{P}^{4}$. The idea is to particularize that ideal $f_{1i}$ in such a way that its dimension is $1$ at degree $4$. The text gives this choice as an example:
$f_{1i}=(40x_{3}+94x_{4}, 117x_{3}+119x_{4}, 449x_{3}+464x_{4}+266x_{0}+195x_{1}+173x_{2}, 306x_{2}, 273x_{3},259x_{3}+291x_{4},76x_{3}+98x_{2})$
My idea of the dimension of an ideal $I$ in a field $k$ at certain degree $s$ is that it corresponds to the dimension of $I_{leq s}$ as a vector space over $k$, being $I_{leq s}$ the set of polynomials in $I$ of total degree $leq s$.
It seems clear that this is not the definition of dimension that it is being used here. Can someone explain why the dimension of that choice of $f_{1i}$ is $1$ at degree $4$ (or point out what I am misunderstanding)?
algebraic-geometry commutative-algebra ideals
algebraic-geometry commutative-algebra ideals
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
edited Nov 29 '18 at 8:48
Chequez
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
asked Nov 29 '18 at 3:28
ChequezChequez
8652511
8652511
add a comment |
add a comment |
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