Viewing algebraic varieties as Homomorphism
Let $k$ be an algebraically closed field, And let $V(mathfrak{a})$ be the algebraic variety generated by the ideal $mathfrak{a} subset k[x_1,cdots,x_n]$. I have read that we can identify $V(mathfrak{a})$ as
$V(mathfrak{a})=text{Hom}_{k-alg}( k[x_1,cdots,x_n]/ mathfrak{a} , k)$.
But I don't see how. I know that the points of $V(mathfrak{a})$ is in bijection with the maximal ideals of $k[x_1,cdots,x_n]/ mathfrak{a}$ by Hilbert's Nullstellenstaz, but I don't see how to view it as $k$-algebra homomorphism.
algebraic-geometry commutative-algebra
asked Dec 7 at 9:00
Ishigami
533212
add a comment |
Let $k$ be an algebraically closed field, And let $V(mathfrak{a})$ be the algebraic variety generated by the ideal $mathfrak{a} subset k[x_1,cdots,x_n]$. I have read that we can identify $V(mathfrak{a})$ as
$V(mathfrak{a})=text{Hom}_{k-alg}( k[x_1,cdots,x_n]/ mathfrak{a} , k)$.
But I don't see how. I know that the points of $V(mathfrak{a})$ is in bijection with the maximal ideals of $k[x_1,cdots,x_n]/ mathfrak{a}$ by Hilbert's Nullstellenstaz, but I don't see how to view it as $k$-algebra homomorphism.
algebraic-geometry commutative-algebra
asked Dec 7 at 9:00
Ishigami
533212
3
The kernal of a morphisn $k[x]/a to k$ is a maximal ideal.
– Youngsu
Dec 7 at 9:06
2
Adding to the above comment: An algebra homomorphism from a polynomial algebra is uniquely determined by where it sends the variables. Taking a quotient then introduces a suitable requirement that certain combinations of these values are 0.
– Tobias Kildetoft
Dec 7 at 9:35
Hi can you explain more I'm still a bit confused..
– Ishigami
Dec 7 at 14:19
I remember you that the homomorphism of $mathbb{K}$-algebras send the units in the units; so the homomorphism from $mathbb{K}[x_1,dots,x_n]/mathfrak{a}$ to $mathbb{K}$ are all "onto". From all this, using the comment of @Youngsu, you have the statement; for exact, $V(mathfrak{a})$ is in bijection with the set $Hom_{mathbb{K}-alg}(mathbb{K}[x_1,dots,x_n]/mathfrak{a},mathbb{K})$.
– Armando j18eos
Dec 13 at 13:47
add a comment |
Let $k$ be an algebraically closed field, And let $V(mathfrak{a})$ be the algebraic variety generated by the ideal $mathfrak{a} subset k[x_1,cdots,x_n]$. I have read that we can identify $V(mathfrak{a})$ as
$V(mathfrak{a})=text{Hom}_{k-alg}( k[x_1,cdots,x_n]/ mathfrak{a} , k)$.
But I don't see how. I know that the points of $V(mathfrak{a})$ is in bijection with the maximal ideals of $k[x_1,cdots,x_n]/ mathfrak{a}$ by Hilbert's Nullstellenstaz, but I don't see how to view it as $k$-algebra homomorphism.
algebraic-geometry commutative-algebra
asked Dec 7 at 9:00
Ishigami
533212
Let $k$ be an algebraically closed field, And let $V(mathfrak{a})$ be the algebraic variety generated by the ideal $mathfrak{a} subset k[x_1,cdots,x_n]$. I have read that we can identify $V(mathfrak{a})$ as
$V(mathfrak{a})=text{Hom}_{k-alg}( k[x_1,cdots,x_n]/ mathfrak{a} , k)$.
But I don't see how. I know that the points of $V(mathfrak{a})$ is in bijection with the maximal ideals of $k[x_1,cdots,x_n]/ mathfrak{a}$ by Hilbert's Nullstellenstaz, but I don't see how to view it as $k$-algebra homomorphism.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Dec 7 at 9:00
Ishigami
533212
asked Dec 7 at 9:00
Ishigami
533212
asked Dec 7 at 9:00
Ishigami
533212
asked Dec 7 at 9:00
Ishigami
533212
asked Dec 7 at 9:00
Ishigami
533212
533212
3
The kernal of a morphisn $k[x]/a to k$ is a maximal ideal.
– Youngsu
Dec 7 at 9:06
2
Adding to the above comment: An algebra homomorphism from a polynomial algebra is uniquely determined by where it sends the variables. Taking a quotient then introduces a suitable requirement that certain combinations of these values are 0.
– Tobias Kildetoft
Dec 7 at 9:35
Hi can you explain more I'm still a bit confused..
– Ishigami
Dec 7 at 14:19
I remember you that the homomorphism of $mathbb{K}$-algebras send the units in the units; so the homomorphism from $mathbb{K}[x_1,dots,x_n]/mathfrak{a}$ to $mathbb{K}$ are all "onto". From all this, using the comment of @Youngsu, you have the statement; for exact, $V(mathfrak{a})$ is in bijection with the set $Hom_{mathbb{K}-alg}(mathbb{K}[x_1,dots,x_n]/mathfrak{a},mathbb{K})$.
– Armando j18eos
Dec 13 at 13:47
add a comment |
3
The kernal of a morphisn $k[x]/a to k$ is a maximal ideal.
– Youngsu
Dec 7 at 9:06
2
Adding to the above comment: An algebra homomorphism from a polynomial algebra is uniquely determined by where it sends the variables. Taking a quotient then introduces a suitable requirement that certain combinations of these values are 0.
– Tobias Kildetoft
Dec 7 at 9:35
Hi can you explain more I'm still a bit confused..
– Ishigami
Dec 7 at 14:19
I remember you that the homomorphism of $mathbb{K}$-algebras send the units in the units; so the homomorphism from $mathbb{K}[x_1,dots,x_n]/mathfrak{a}$ to $mathbb{K}$ are all "onto". From all this, using the comment of @Youngsu, you have the statement; for exact, $V(mathfrak{a})$ is in bijection with the set $Hom_{mathbb{K}-alg}(mathbb{K}[x_1,dots,x_n]/mathfrak{a},mathbb{K})$.
– Armando j18eos
Dec 13 at 13:47
3
3
The kernal of a morphisn $k[x]/a to k$ is a maximal ideal.
– Youngsu
Dec 7 at 9:06
The kernal of a morphisn $k[x]/a to k$ is a maximal ideal.
– Youngsu
Dec 7 at 9:06
2
2
Adding to the above comment: An algebra homomorphism from a polynomial algebra is uniquely determined by where it sends the variables. Taking a quotient then introduces a suitable requirement that certain combinations of these values are 0.
– Tobias Kildetoft
Dec 7 at 9:35
Adding to the above comment: An algebra homomorphism from a polynomial algebra is uniquely determined by where it sends the variables. Taking a quotient then introduces a suitable requirement that certain combinations of these values are 0.
– Tobias Kildetoft
Dec 7 at 9:35
Hi can you explain more I'm still a bit confused..
– Ishigami
Dec 7 at 14:19
Hi can you explain more I'm still a bit confused..
– Ishigami
Dec 7 at 14:19
I remember you that the homomorphism of $mathbb{K}$-algebras send the units in the units; so the homomorphism from $mathbb{K}[x_1,dots,x_n]/mathfrak{a}$ to $mathbb{K}$ are all "onto". From all this, using the comment of @Youngsu, you have the statement; for exact, $V(mathfrak{a})$ is in bijection with the set $Hom_{mathbb{K}-alg}(mathbb{K}[x_1,dots,x_n]/mathfrak{a},mathbb{K})$.
– Armando j18eos
Dec 13 at 13:47
I remember you that the homomorphism of $mathbb{K}$-algebras send the units in the units; so the homomorphism from $mathbb{K}[x_1,dots,x_n]/mathfrak{a}$ to $mathbb{K}$ are all "onto". From all this, using the comment of @Youngsu, you have the statement; for exact, $V(mathfrak{a})$ is in bijection with the set $Hom_{mathbb{K}-alg}(mathbb{K}[x_1,dots,x_n]/mathfrak{a},mathbb{K})$.
– Armando j18eos
Dec 13 at 13:47
add a comment |
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3
The kernal of a morphisn $k[x]/a to k$ is a maximal ideal.
– Youngsu
Dec 7 at 9:06
2
Adding to the above comment: An algebra homomorphism from a polynomial algebra is uniquely determined by where it sends the variables. Taking a quotient then introduces a suitable requirement that certain combinations of these values are 0.
– Tobias Kildetoft
Dec 7 at 9:35
Hi can you explain more I'm still a bit confused..
– Ishigami
Dec 7 at 14:19
I remember you that the homomorphism of $mathbb{K}$-algebras send the units in the units; so the homomorphism from $mathbb{K}[x_1,dots,x_n]/mathfrak{a}$ to $mathbb{K}$ are all "onto". From all this, using the comment of @Youngsu, you have the statement; for exact, $V(mathfrak{a})$ is in bijection with the set $Hom_{mathbb{K}-alg}(mathbb{K}[x_1,dots,x_n]/mathfrak{a},mathbb{K})$.
– Armando j18eos
Dec 13 at 13:47