Mixing
![Distinct integer solutions to the equation $(x^2+y^2+z^2)/(x+y+z)=2n$ [closed]](https://lh4.googleusercontent.com/-4VFp_N4zzm0/AAAAAAAAAAI/AAAAAAAAAAA/AKxrwcYPDOG6HxhEfFGDodekA3e7f_Y8Rg/s72-c-mo/photo.jpg?sz=32)
How to understand the notation $S(-3)$?
$begingroup$
I asked a question about linear maps between free modules. I have a related question.
Let $S=k[x_1, ldots, x_n]$ where $k$ is a field. Then $S$ is a graded ring with usual grading given by $deg x_i = 1$. By definition, $S(a)$ is a graded module defined by $S(a)_d = S_{a+d}$. What is the basis of $S(a)$? For example, what is the basis of $S(-3)$?
Thank you very much.
algebraic-geometry commutative-algebra
asked Jan 20 at 13:04
LJRLJR
6,63141849
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add a comment |
$begingroup$
I asked a question about linear maps between free modules. I have a related question.
Let $S=k[x_1, ldots, x_n]$ where $k$ is a field. Then $S$ is a graded ring with usual grading given by $deg x_i = 1$. By definition, $S(a)$ is a graded module defined by $S(a)_d = S_{a+d}$. What is the basis of $S(a)$? For example, what is the basis of $S(-3)$?
Thank you very much.
algebraic-geometry commutative-algebra
asked Jan 20 at 13:04
LJRLJR
6,63141849
$endgroup$
add a comment |
$begingroup$
I asked a question about linear maps between free modules. I have a related question.
Let $S=k[x_1, ldots, x_n]$ where $k$ is a field. Then $S$ is a graded ring with usual grading given by $deg x_i = 1$. By definition, $S(a)$ is a graded module defined by $S(a)_d = S_{a+d}$. What is the basis of $S(a)$? For example, what is the basis of $S(-3)$?
Thank you very much.
algebraic-geometry commutative-algebra
asked Jan 20 at 13:04
LJRLJR
6,63141849
$endgroup$
I asked a question about linear maps between free modules. I have a related question.
Let $S=k[x_1, ldots, x_n]$ where $k$ is a field. Then $S$ is a graded ring with usual grading given by $deg x_i = 1$. By definition, $S(a)$ is a graded module defined by $S(a)_d = S_{a+d}$. What is the basis of $S(a)$? For example, what is the basis of $S(-3)$?
Thank you very much.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Jan 20 at 13:04
LJRLJR
6,63141849
asked Jan 20 at 13:04
LJRLJR
6,63141849
edited Jan 22 at 5:28
LJR
asked Jan 20 at 13:04
LJRLJR
6,63141849
asked Jan 20 at 13:04
LJRLJR
6,63141849
asked Jan 20 at 13:04
LJRLJR
6,63141849
6,63141849
add a comment |
add a comment |
1 Answer
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The basis of $S(a)$ is formed by all monomials $x^i = x_1^{i_1}cdots x_n^{i_n}$, but the degree of $x^i$ is equal to $i_1 + dots + i_n - a$.
answered Jan 20 at 13:24
SashaSasha
5,108139
$endgroup$
$begingroup$
This is a basis as a $k$-vector space, right? Because as an $S$-module, $S(a)$ has rank $1$.
$endgroup$
– André 3000
Jan 20 at 13:25
1
$begingroup$
Sure. As an $S$-module the basis is given by one element (monomial 1) in degree $-a$.
$endgroup$
– Sasha
Jan 20 at 14:27
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The basis of $S(a)$ is formed by all monomials $x^i = x_1^{i_1}cdots x_n^{i_n}$, but the degree of $x^i$ is equal to $i_1 + dots + i_n - a$.
answered Jan 20 at 13:24
SashaSasha
5,108139
$endgroup$
$begingroup$
This is a basis as a $k$-vector space, right? Because as an $S$-module, $S(a)$ has rank $1$.
$endgroup$
– André 3000
Jan 20 at 13:25
1
$begingroup$
Sure. As an $S$-module the basis is given by one element (monomial 1) in degree $-a$.
$endgroup$
– Sasha
Jan 20 at 14:27
add a comment |
$begingroup$
The basis of $S(a)$ is formed by all monomials $x^i = x_1^{i_1}cdots x_n^{i_n}$, but the degree of $x^i$ is equal to $i_1 + dots + i_n - a$.
answered Jan 20 at 13:24
SashaSasha
5,108139
$endgroup$
$begingroup$
This is a basis as a $k$-vector space, right? Because as an $S$-module, $S(a)$ has rank $1$.
$endgroup$
– André 3000
Jan 20 at 13:25
1
$begingroup$
Sure. As an $S$-module the basis is given by one element (monomial 1) in degree $-a$.
$endgroup$
– Sasha
Jan 20 at 14:27
add a comment |
$begingroup$
The basis of $S(a)$ is formed by all monomials $x^i = x_1^{i_1}cdots x_n^{i_n}$, but the degree of $x^i$ is equal to $i_1 + dots + i_n - a$.
answered Jan 20 at 13:24
SashaSasha
5,108139
$endgroup$
The basis of $S(a)$ is formed by all monomials $x^i = x_1^{i_1}cdots x_n^{i_n}$, but the degree of $x^i$ is equal to $i_1 + dots + i_n - a$.
answered Jan 20 at 13:24
SashaSasha
5,108139
answered Jan 20 at 13:24
SashaSasha
5,108139
answered Jan 20 at 13:24
SashaSasha
5,108139
answered Jan 20 at 13:24
SashaSasha
5,108139
5,108139
$begingroup$
This is a basis as a $k$-vector space, right? Because as an $S$-module, $S(a)$ has rank $1$.
$endgroup$
– André 3000
Jan 20 at 13:25
1
$begingroup$
Sure. As an $S$-module the basis is given by one element (monomial 1) in degree $-a$.
$endgroup$
– Sasha
Jan 20 at 14:27
add a comment |
$begingroup$
This is a basis as a $k$-vector space, right? Because as an $S$-module, $S(a)$ has rank $1$.
$endgroup$
– André 3000
Jan 20 at 13:25
1
$begingroup$
Sure. As an $S$-module the basis is given by one element (monomial 1) in degree $-a$.
$endgroup$
– Sasha
Jan 20 at 14:27
$begingroup$
This is a basis as a $k$-vector space, right? Because as an $S$-module, $S(a)$ has rank $1$.
$endgroup$
– André 3000
Jan 20 at 13:25
$begingroup$
This is a basis as a $k$-vector space, right? Because as an $S$-module, $S(a)$ has rank $1$.
$endgroup$
– André 3000
Jan 20 at 13:25
1
1
$begingroup$
Sure. As an $S$-module the basis is given by one element (monomial 1) in degree $-a$.
$endgroup$
– Sasha
Jan 20 at 14:27
$begingroup$
Sure. As an $S$-module the basis is given by one element (monomial 1) in degree $-a$.
$endgroup$
– Sasha
Jan 20 at 14:27
add a comment |
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