Mixing
Multiplicity and degree of irreducible projective subschemes.
$begingroup$
Suppose $X subset mathbb{P}^n$ is an irreducible projective scheme. Then its multiplicity $mu_X$ is defined as the length of the local ring $mathcal{O}_{X,eta}$ over itself, where $eta$ is the generic point of $X$.
The degree $d_X$ of $X$ is defined as $frac{c}{n!}$, where $c$ is the leading coefficient of the Hilbert polynomial $P_X$ of $X$, and $n$ is the degree of $P_X$.
I would like to show that $d_X = mu_X cdot d_{X_{red}}$, where $X_{red} subset X subset mathbb{P}^n$ is the reduced subscheme with the same set of underlying points, but I don't see how to relate the Hilbert polynomial to the multiplicity. So any ideas or hints would be appreciated.
One thought was that maybe even $P_X = mu_Xcdot P_{X_{red}}$ holds. But on further thought this is inplausible, because $X$ might have embedded components, which do not appear in $Y$, but contribute a part to $P_X$. But doing a (homogeneous) primary decomposition we can decompose $X = X_1 cup dots cup X_r$ scheme theoretically, and $X_1$ corresponds to the maximal component. Because $X_2, dots X_r$ do not contribute to the leading coefficient of $P_X$, we might assume wlog that $X$ does not have any embedded components. Does this imply $P_X = mu_X cdot P_{X_{red}}$?
algebraic-geometry projective-geometry schemes projective-schemes
asked Jan 31 at 15:18
red_trumpetred_trumpet
1,037319
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add a comment |
$begingroup$
Suppose $X subset mathbb{P}^n$ is an irreducible projective scheme. Then its multiplicity $mu_X$ is defined as the length of the local ring $mathcal{O}_{X,eta}$ over itself, where $eta$ is the generic point of $X$.
The degree $d_X$ of $X$ is defined as $frac{c}{n!}$, where $c$ is the leading coefficient of the Hilbert polynomial $P_X$ of $X$, and $n$ is the degree of $P_X$.
I would like to show that $d_X = mu_X cdot d_{X_{red}}$, where $X_{red} subset X subset mathbb{P}^n$ is the reduced subscheme with the same set of underlying points, but I don't see how to relate the Hilbert polynomial to the multiplicity. So any ideas or hints would be appreciated.
One thought was that maybe even $P_X = mu_Xcdot P_{X_{red}}$ holds. But on further thought this is inplausible, because $X$ might have embedded components, which do not appear in $Y$, but contribute a part to $P_X$. But doing a (homogeneous) primary decomposition we can decompose $X = X_1 cup dots cup X_r$ scheme theoretically, and $X_1$ corresponds to the maximal component. Because $X_2, dots X_r$ do not contribute to the leading coefficient of $P_X$, we might assume wlog that $X$ does not have any embedded components. Does this imply $P_X = mu_X cdot P_{X_{red}}$?
algebraic-geometry projective-geometry schemes projective-schemes
asked Jan 31 at 15:18
red_trumpetred_trumpet
1,037319
$endgroup$
add a comment |
$begingroup$
Suppose $X subset mathbb{P}^n$ is an irreducible projective scheme. Then its multiplicity $mu_X$ is defined as the length of the local ring $mathcal{O}_{X,eta}$ over itself, where $eta$ is the generic point of $X$.
The degree $d_X$ of $X$ is defined as $frac{c}{n!}$, where $c$ is the leading coefficient of the Hilbert polynomial $P_X$ of $X$, and $n$ is the degree of $P_X$.
I would like to show that $d_X = mu_X cdot d_{X_{red}}$, where $X_{red} subset X subset mathbb{P}^n$ is the reduced subscheme with the same set of underlying points, but I don't see how to relate the Hilbert polynomial to the multiplicity. So any ideas or hints would be appreciated.
One thought was that maybe even $P_X = mu_Xcdot P_{X_{red}}$ holds. But on further thought this is inplausible, because $X$ might have embedded components, which do not appear in $Y$, but contribute a part to $P_X$. But doing a (homogeneous) primary decomposition we can decompose $X = X_1 cup dots cup X_r$ scheme theoretically, and $X_1$ corresponds to the maximal component. Because $X_2, dots X_r$ do not contribute to the leading coefficient of $P_X$, we might assume wlog that $X$ does not have any embedded components. Does this imply $P_X = mu_X cdot P_{X_{red}}$?
algebraic-geometry projective-geometry schemes projective-schemes
asked Jan 31 at 15:18
red_trumpetred_trumpet
1,037319
$endgroup$
Suppose $X subset mathbb{P}^n$ is an irreducible projective scheme. Then its multiplicity $mu_X$ is defined as the length of the local ring $mathcal{O}_{X,eta}$ over itself, where $eta$ is the generic point of $X$.
The degree $d_X$ of $X$ is defined as $frac{c}{n!}$, where $c$ is the leading coefficient of the Hilbert polynomial $P_X$ of $X$, and $n$ is the degree of $P_X$.
I would like to show that $d_X = mu_X cdot d_{X_{red}}$, where $X_{red} subset X subset mathbb{P}^n$ is the reduced subscheme with the same set of underlying points, but I don't see how to relate the Hilbert polynomial to the multiplicity. So any ideas or hints would be appreciated.
One thought was that maybe even $P_X = mu_Xcdot P_{X_{red}}$ holds. But on further thought this is inplausible, because $X$ might have embedded components, which do not appear in $Y$, but contribute a part to $P_X$. But doing a (homogeneous) primary decomposition we can decompose $X = X_1 cup dots cup X_r$ scheme theoretically, and $X_1$ corresponds to the maximal component. Because $X_2, dots X_r$ do not contribute to the leading coefficient of $P_X$, we might assume wlog that $X$ does not have any embedded components. Does this imply $P_X = mu_X cdot P_{X_{red}}$?
algebraic-geometry projective-geometry schemes projective-schemes
algebraic-geometry projective-geometry schemes projective-schemes
asked Jan 31 at 15:18
red_trumpetred_trumpet
1,037319
asked Jan 31 at 15:18
red_trumpetred_trumpet
1,037319
edited Feb 1 at 14:52
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asked Jan 31 at 15:18
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asked Jan 31 at 15:18
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asked Jan 31 at 15:18
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1 Answer
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$begingroup$
We will use Proposition I 7.4 from Hartshorne's Algebraic geometry (p. 50), that is:
Let $M$ be a finitely generated graded module over a noetherian graded ring
$S$. Then there exists a filtration $0 = M^0 subset M^1 subset dots subset M^r = M$ by graded submodules, such that for each $i$, $M^i / M^{i+1} cong (S/mathfrak{p}_i)(l_i)$, where $mathfrak{p}_i$ is a graded homogeneous prime
ideal of $S$, and $l_i in mathbb{Z}$. The filtration is not unique, but for
any such filtration we have:
- if $mathfrak{p}$ is a homogeneous prime ideal of $S$, then $mathfrak{p} supset text{Ann }M Leftrightarrow mathfrak{p} supset mathfrak{p_i}$
for some $i$. In particular, the minimal elements of the set ${mathfrak{p}_1, dots,mathfrak{p}_r}$ are just the minimal primes of $M$, i.e. the primes
which are minimal containing $text{Ann }M$.
- for each minimal prime of $M$, the number of times which $mathfrak{p}$ occurs in the set ${mathfrak{p_1},dots,mathfrak{p}_r}$ is equal to the length of $M_mathfrak{p}$ over the local ring $S_mathfrak{p}$ (and hence is independent of the filtration).
To compare the Hilbert polynomial of $X$ and $X_{red}$, we consider the homomorphism on homogeneous coordinate rings
$$ S(X) = k[x_0,dots,x_n]/I rightarrow k[x_0,dots,x_n]/sqrt{I} = S(X_{red}).$$
Applying the proposition, we see that $P_X(l) = sum_i dim(S(X)/mathfrak{p}_i)_{l_i + l}$ for $l gg 0$. Only the minimal primes of $S(X)$ contribute to the leading coefficient of $P_X$, because the degree of the Hilbert polynomial equals the dimension of the scheme. The only minimal prime is the nilradical, so we see that the leading coefficient is a multiple of the leading coefficient of $S(X_{red})$, noting that the shift $l mapsto l + l_i$ does not change the leading coefficient of a polynomial.
According to part 2. of the proposition, the factor of the degree equals $text{length}_{S_mathfrak{p}} M_mathfrak{p}$. But following the proof of the proposition, one can check that this also holds if we make to homogeneous localisation, i.e. take $S_{(mathfrak{p})}$ and $M_{(mathfrak{p})}$ instead.
I think this works without mentioning primary decompositions, though the connection would be interesting.
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
We will use Proposition I 7.4 from Hartshorne's Algebraic geometry (p. 50), that is:
Let $M$ be a finitely generated graded module over a noetherian graded ring
$S$. Then there exists a filtration $0 = M^0 subset M^1 subset dots subset M^r = M$ by graded submodules, such that for each $i$, $M^i / M^{i+1} cong (S/mathfrak{p}_i)(l_i)$, where $mathfrak{p}_i$ is a graded homogeneous prime
ideal of $S$, and $l_i in mathbb{Z}$. The filtration is not unique, but for
any such filtration we have:
- if $mathfrak{p}$ is a homogeneous prime ideal of $S$, then $mathfrak{p} supset text{Ann }M Leftrightarrow mathfrak{p} supset mathfrak{p_i}$
for some $i$. In particular, the minimal elements of the set ${mathfrak{p}_1, dots,mathfrak{p}_r}$ are just the minimal primes of $M$, i.e. the primes
which are minimal containing $text{Ann }M$.
- for each minimal prime of $M$, the number of times which $mathfrak{p}$ occurs in the set ${mathfrak{p_1},dots,mathfrak{p}_r}$ is equal to the length of $M_mathfrak{p}$ over the local ring $S_mathfrak{p}$ (and hence is independent of the filtration).
To compare the Hilbert polynomial of $X$ and $X_{red}$, we consider the homomorphism on homogeneous coordinate rings
$$ S(X) = k[x_0,dots,x_n]/I rightarrow k[x_0,dots,x_n]/sqrt{I} = S(X_{red}).$$
Applying the proposition, we see that $P_X(l) = sum_i dim(S(X)/mathfrak{p}_i)_{l_i + l}$ for $l gg 0$. Only the minimal primes of $S(X)$ contribute to the leading coefficient of $P_X$, because the degree of the Hilbert polynomial equals the dimension of the scheme. The only minimal prime is the nilradical, so we see that the leading coefficient is a multiple of the leading coefficient of $S(X_{red})$, noting that the shift $l mapsto l + l_i$ does not change the leading coefficient of a polynomial.
According to part 2. of the proposition, the factor of the degree equals $text{length}_{S_mathfrak{p}} M_mathfrak{p}$. But following the proof of the proposition, one can check that this also holds if we make to homogeneous localisation, i.e. take $S_{(mathfrak{p})}$ and $M_{(mathfrak{p})}$ instead.
I think this works without mentioning primary decompositions, though the connection would be interesting.
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
$endgroup$
add a comment |
$begingroup$
We will use Proposition I 7.4 from Hartshorne's Algebraic geometry (p. 50), that is:
Let $M$ be a finitely generated graded module over a noetherian graded ring
$S$. Then there exists a filtration $0 = M^0 subset M^1 subset dots subset M^r = M$ by graded submodules, such that for each $i$, $M^i / M^{i+1} cong (S/mathfrak{p}_i)(l_i)$, where $mathfrak{p}_i$ is a graded homogeneous prime
ideal of $S$, and $l_i in mathbb{Z}$. The filtration is not unique, but for
any such filtration we have:
- if $mathfrak{p}$ is a homogeneous prime ideal of $S$, then $mathfrak{p} supset text{Ann }M Leftrightarrow mathfrak{p} supset mathfrak{p_i}$
for some $i$. In particular, the minimal elements of the set ${mathfrak{p}_1, dots,mathfrak{p}_r}$ are just the minimal primes of $M$, i.e. the primes
which are minimal containing $text{Ann }M$.
- for each minimal prime of $M$, the number of times which $mathfrak{p}$ occurs in the set ${mathfrak{p_1},dots,mathfrak{p}_r}$ is equal to the length of $M_mathfrak{p}$ over the local ring $S_mathfrak{p}$ (and hence is independent of the filtration).
To compare the Hilbert polynomial of $X$ and $X_{red}$, we consider the homomorphism on homogeneous coordinate rings
$$ S(X) = k[x_0,dots,x_n]/I rightarrow k[x_0,dots,x_n]/sqrt{I} = S(X_{red}).$$
Applying the proposition, we see that $P_X(l) = sum_i dim(S(X)/mathfrak{p}_i)_{l_i + l}$ for $l gg 0$. Only the minimal primes of $S(X)$ contribute to the leading coefficient of $P_X$, because the degree of the Hilbert polynomial equals the dimension of the scheme. The only minimal prime is the nilradical, so we see that the leading coefficient is a multiple of the leading coefficient of $S(X_{red})$, noting that the shift $l mapsto l + l_i$ does not change the leading coefficient of a polynomial.
According to part 2. of the proposition, the factor of the degree equals $text{length}_{S_mathfrak{p}} M_mathfrak{p}$. But following the proof of the proposition, one can check that this also holds if we make to homogeneous localisation, i.e. take $S_{(mathfrak{p})}$ and $M_{(mathfrak{p})}$ instead.
I think this works without mentioning primary decompositions, though the connection would be interesting.
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
$endgroup$
add a comment |
$begingroup$
We will use Proposition I 7.4 from Hartshorne's Algebraic geometry (p. 50), that is:
Let $M$ be a finitely generated graded module over a noetherian graded ring
$S$. Then there exists a filtration $0 = M^0 subset M^1 subset dots subset M^r = M$ by graded submodules, such that for each $i$, $M^i / M^{i+1} cong (S/mathfrak{p}_i)(l_i)$, where $mathfrak{p}_i$ is a graded homogeneous prime
ideal of $S$, and $l_i in mathbb{Z}$. The filtration is not unique, but for
any such filtration we have:
- if $mathfrak{p}$ is a homogeneous prime ideal of $S$, then $mathfrak{p} supset text{Ann }M Leftrightarrow mathfrak{p} supset mathfrak{p_i}$
for some $i$. In particular, the minimal elements of the set ${mathfrak{p}_1, dots,mathfrak{p}_r}$ are just the minimal primes of $M$, i.e. the primes
which are minimal containing $text{Ann }M$.
- for each minimal prime of $M$, the number of times which $mathfrak{p}$ occurs in the set ${mathfrak{p_1},dots,mathfrak{p}_r}$ is equal to the length of $M_mathfrak{p}$ over the local ring $S_mathfrak{p}$ (and hence is independent of the filtration).
To compare the Hilbert polynomial of $X$ and $X_{red}$, we consider the homomorphism on homogeneous coordinate rings
$$ S(X) = k[x_0,dots,x_n]/I rightarrow k[x_0,dots,x_n]/sqrt{I} = S(X_{red}).$$
Applying the proposition, we see that $P_X(l) = sum_i dim(S(X)/mathfrak{p}_i)_{l_i + l}$ for $l gg 0$. Only the minimal primes of $S(X)$ contribute to the leading coefficient of $P_X$, because the degree of the Hilbert polynomial equals the dimension of the scheme. The only minimal prime is the nilradical, so we see that the leading coefficient is a multiple of the leading coefficient of $S(X_{red})$, noting that the shift $l mapsto l + l_i$ does not change the leading coefficient of a polynomial.
According to part 2. of the proposition, the factor of the degree equals $text{length}_{S_mathfrak{p}} M_mathfrak{p}$. But following the proof of the proposition, one can check that this also holds if we make to homogeneous localisation, i.e. take $S_{(mathfrak{p})}$ and $M_{(mathfrak{p})}$ instead.
I think this works without mentioning primary decompositions, though the connection would be interesting.
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
$endgroup$
We will use Proposition I 7.4 from Hartshorne's Algebraic geometry (p. 50), that is:
Let $M$ be a finitely generated graded module over a noetherian graded ring
$S$. Then there exists a filtration $0 = M^0 subset M^1 subset dots subset M^r = M$ by graded submodules, such that for each $i$, $M^i / M^{i+1} cong (S/mathfrak{p}_i)(l_i)$, where $mathfrak{p}_i$ is a graded homogeneous prime
ideal of $S$, and $l_i in mathbb{Z}$. The filtration is not unique, but for
any such filtration we have:
- if $mathfrak{p}$ is a homogeneous prime ideal of $S$, then $mathfrak{p} supset text{Ann }M Leftrightarrow mathfrak{p} supset mathfrak{p_i}$
for some $i$. In particular, the minimal elements of the set ${mathfrak{p}_1, dots,mathfrak{p}_r}$ are just the minimal primes of $M$, i.e. the primes
which are minimal containing $text{Ann }M$.
- for each minimal prime of $M$, the number of times which $mathfrak{p}$ occurs in the set ${mathfrak{p_1},dots,mathfrak{p}_r}$ is equal to the length of $M_mathfrak{p}$ over the local ring $S_mathfrak{p}$ (and hence is independent of the filtration).
To compare the Hilbert polynomial of $X$ and $X_{red}$, we consider the homomorphism on homogeneous coordinate rings
$$ S(X) = k[x_0,dots,x_n]/I rightarrow k[x_0,dots,x_n]/sqrt{I} = S(X_{red}).$$
Applying the proposition, we see that $P_X(l) = sum_i dim(S(X)/mathfrak{p}_i)_{l_i + l}$ for $l gg 0$. Only the minimal primes of $S(X)$ contribute to the leading coefficient of $P_X$, because the degree of the Hilbert polynomial equals the dimension of the scheme. The only minimal prime is the nilradical, so we see that the leading coefficient is a multiple of the leading coefficient of $S(X_{red})$, noting that the shift $l mapsto l + l_i$ does not change the leading coefficient of a polynomial.
According to part 2. of the proposition, the factor of the degree equals $text{length}_{S_mathfrak{p}} M_mathfrak{p}$. But following the proof of the proposition, one can check that this also holds if we make to homogeneous localisation, i.e. take $S_{(mathfrak{p})}$ and $M_{(mathfrak{p})}$ instead.
I think this works without mentioning primary decompositions, though the connection would be interesting.
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
answered Feb 1 at 14:51
red_trumpetred_trumpet
1,037319
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add a comment |
add a comment |
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