partial derivative of a polynomial belongs to a maximal ideal
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
asked Nov 24 at 1:30
user8891548
513
add a comment |
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
asked Nov 24 at 1:30
user8891548
513
@baharampuri The $f$ seems not contained in $mathfrak m$.
– user8891548
Nov 24 at 7:09
Sorry missed that.
– baharampuri
Nov 24 at 7:22
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
– KReiser
Nov 24 at 8:06
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
– user8891548
Nov 24 at 8:45
add a comment |
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
asked Nov 24 at 1:30
user8891548
513
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 24 at 1:30
user8891548
513
asked Nov 24 at 1:30
user8891548
513
edited Nov 24 at 6:51
asked Nov 24 at 1:30
user8891548
513
asked Nov 24 at 1:30
user8891548
513
asked Nov 24 at 1:30
user8891548
513
513
@baharampuri The $f$ seems not contained in $mathfrak m$.
– user8891548
Nov 24 at 7:09
Sorry missed that.
– baharampuri
Nov 24 at 7:22
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
– KReiser
Nov 24 at 8:06
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
– user8891548
Nov 24 at 8:45
add a comment |
@baharampuri The $f$ seems not contained in $mathfrak m$.
– user8891548
Nov 24 at 7:09
Sorry missed that.
– baharampuri
Nov 24 at 7:22
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
– KReiser
Nov 24 at 8:06
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
– user8891548
Nov 24 at 8:45
@baharampuri The $f$ seems not contained in $mathfrak m$.
– user8891548
Nov 24 at 7:09
@baharampuri The $f$ seems not contained in $mathfrak m$.
– user8891548
Nov 24 at 7:09
Sorry missed that.
– baharampuri
Nov 24 at 7:22
Sorry missed that.
– baharampuri
Nov 24 at 7:22
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
– KReiser
Nov 24 at 8:06
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
– KReiser
Nov 24 at 8:06
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
– user8891548
Nov 24 at 8:45
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
– user8891548
Nov 24 at 8:45
add a comment |
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@baharampuri The $f$ seems not contained in $mathfrak m$.
– user8891548
Nov 24 at 7:09
Sorry missed that.
– baharampuri
Nov 24 at 7:22
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
– KReiser
Nov 24 at 8:06
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
– user8891548
Nov 24 at 8:45