Mixing
How to find this ring homomorphism and the required kernel?
$begingroup$
I have been wasting my time since morning in this very 'easy looking' thing:
Let k be algebraically closed field, $ V subset mathbb{A}^n$ be a non empty variety. $I(V)$ be the ideal of $V$, set of all polynomials in $k[x_1,...x_n]$ that vanish on $V$.
$mathscr{F}(V,k)$ be the set of all functions from $V$ to $k$, we call a function in this ring of functions, a polynomial function of there is a $F in k[x_1, ...x_n]$ such that $ f(a_1,...a_n)= F(a_1...a_n)$. for all $(a_1...a_n)in V$.
And we can identify the domain $k[x_1,...x_n]/I(V)$ with the subring of $mathscr{F}(V,k)$ consisting of polynomial functioning $V$.
I can see why that would be true, but the best exercise is to show that the map that associates each $F in k[x_1,...x_n]$ to a polynomial function in $mathscr{F}(V,k)$ is a ring homomorphism and that it's kernel is $I(V)$.
And I'm lost on how to do that. I can't even think of what a suitable map might and the ones I came up with did not give me the required kernel...
Can anyone give me hints? I would very much appreciate it.
Thanks a bunch in advance!
Source: W. Fulton's Algebraic Curves, Chapter 2, Section 2.1
EDIT: I think I have solved it, can't believe I'm so dumb. Here, $f $ is infact restriction of $F$ on $V$, so kernel consists all polynomials which are zero on $V$, i.e., $I(V)$.
algebraic-geometry algebraic-curves
asked Jan 26 at 14:08
clearclear
1,44121130
$endgroup$
add a comment |
$begingroup$
I have been wasting my time since morning in this very 'easy looking' thing:
Let k be algebraically closed field, $ V subset mathbb{A}^n$ be a non empty variety. $I(V)$ be the ideal of $V$, set of all polynomials in $k[x_1,...x_n]$ that vanish on $V$.
$mathscr{F}(V,k)$ be the set of all functions from $V$ to $k$, we call a function in this ring of functions, a polynomial function of there is a $F in k[x_1, ...x_n]$ such that $ f(a_1,...a_n)= F(a_1...a_n)$. for all $(a_1...a_n)in V$.
And we can identify the domain $k[x_1,...x_n]/I(V)$ with the subring of $mathscr{F}(V,k)$ consisting of polynomial functioning $V$.
I can see why that would be true, but the best exercise is to show that the map that associates each $F in k[x_1,...x_n]$ to a polynomial function in $mathscr{F}(V,k)$ is a ring homomorphism and that it's kernel is $I(V)$.
And I'm lost on how to do that. I can't even think of what a suitable map might and the ones I came up with did not give me the required kernel...
Can anyone give me hints? I would very much appreciate it.
Thanks a bunch in advance!
Source: W. Fulton's Algebraic Curves, Chapter 2, Section 2.1
EDIT: I think I have solved it, can't believe I'm so dumb. Here, $f $ is infact restriction of $F$ on $V$, so kernel consists all polynomials which are zero on $V$, i.e., $I(V)$.
algebraic-geometry algebraic-curves
asked Jan 26 at 14:08
clearclear
1,44121130
$endgroup$
add a comment |
$begingroup$
I have been wasting my time since morning in this very 'easy looking' thing:
Let k be algebraically closed field, $ V subset mathbb{A}^n$ be a non empty variety. $I(V)$ be the ideal of $V$, set of all polynomials in $k[x_1,...x_n]$ that vanish on $V$.
$mathscr{F}(V,k)$ be the set of all functions from $V$ to $k$, we call a function in this ring of functions, a polynomial function of there is a $F in k[x_1, ...x_n]$ such that $ f(a_1,...a_n)= F(a_1...a_n)$. for all $(a_1...a_n)in V$.
And we can identify the domain $k[x_1,...x_n]/I(V)$ with the subring of $mathscr{F}(V,k)$ consisting of polynomial functioning $V$.
I can see why that would be true, but the best exercise is to show that the map that associates each $F in k[x_1,...x_n]$ to a polynomial function in $mathscr{F}(V,k)$ is a ring homomorphism and that it's kernel is $I(V)$.
And I'm lost on how to do that. I can't even think of what a suitable map might and the ones I came up with did not give me the required kernel...
Can anyone give me hints? I would very much appreciate it.
Thanks a bunch in advance!
Source: W. Fulton's Algebraic Curves, Chapter 2, Section 2.1
EDIT: I think I have solved it, can't believe I'm so dumb. Here, $f $ is infact restriction of $F$ on $V$, so kernel consists all polynomials which are zero on $V$, i.e., $I(V)$.
algebraic-geometry algebraic-curves
asked Jan 26 at 14:08
clearclear
1,44121130
$endgroup$
I have been wasting my time since morning in this very 'easy looking' thing:
Let k be algebraically closed field, $ V subset mathbb{A}^n$ be a non empty variety. $I(V)$ be the ideal of $V$, set of all polynomials in $k[x_1,...x_n]$ that vanish on $V$.
$mathscr{F}(V,k)$ be the set of all functions from $V$ to $k$, we call a function in this ring of functions, a polynomial function of there is a $F in k[x_1, ...x_n]$ such that $ f(a_1,...a_n)= F(a_1...a_n)$. for all $(a_1...a_n)in V$.
And we can identify the domain $k[x_1,...x_n]/I(V)$ with the subring of $mathscr{F}(V,k)$ consisting of polynomial functioning $V$.
I can see why that would be true, but the best exercise is to show that the map that associates each $F in k[x_1,...x_n]$ to a polynomial function in $mathscr{F}(V,k)$ is a ring homomorphism and that it's kernel is $I(V)$.
And I'm lost on how to do that. I can't even think of what a suitable map might and the ones I came up with did not give me the required kernel...
Can anyone give me hints? I would very much appreciate it.
Thanks a bunch in advance!
Source: W. Fulton's Algebraic Curves, Chapter 2, Section 2.1
EDIT: I think I have solved it, can't believe I'm so dumb. Here, $f $ is infact restriction of $F$ on $V$, so kernel consists all polynomials which are zero on $V$, i.e., $I(V)$.
algebraic-geometry algebraic-curves
algebraic-geometry algebraic-curves
asked Jan 26 at 14:08
clearclear
1,44121130
asked Jan 26 at 14:08
clearclear
1,44121130
edited Jan 26 at 14:49
clear
asked Jan 26 at 14:08
clearclear
1,44121130
asked Jan 26 at 14:08
clearclear
1,44121130
asked Jan 26 at 14:08
clearclear
1,44121130
1,44121130
add a comment |
add a comment |
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