Mixing
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Proof of Hilbert's Theorem
$begingroup$
Here is Hilbert's theorem from Eisenbud:
There are a few things I don't understand:
(1) Why does $H_M(s)=0$ for $s$ large in the case of graded vector spaces?
(2) What is the point of working with twists here? Is it used somehow that the maps between twisted objects have degree $0$ (map homogeneous elements to homogeneous elements of the same degree)?
(3) How does the equality (second display) follow from taking components of degree $s$? That is, why is the sum of dimensions of terms 1 and 3 equal the sum of dimensions of terms 2 and 4?
(4) I don't quite understand why $K$ and $M/x_rM$ are finitely generated modules over $k[x_1,dots,x_{r-1}]$. Well, a f.g. module over a Noetherian ring is Noetherian, and $K$ is a submodule of a f.g. Noetherian module, so it's finitely generated. And the quotient of a Noetherian module is f.g. But why are they modules over $k[x_1,dots,x_{r-1}]$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra proof-explanation
edited Jan 26 at 21:04
user26857
39.4k124183
asked Jan 26 at 20:17
user437309user437309
766314
$endgroup$
|
show 3 more comments
$begingroup$
Here is Hilbert's theorem from Eisenbud:
There are a few things I don't understand:
(1) Why does $H_M(s)=0$ for $s$ large in the case of graded vector spaces?
(2) What is the point of working with twists here? Is it used somehow that the maps between twisted objects have degree $0$ (map homogeneous elements to homogeneous elements of the same degree)?
(3) How does the equality (second display) follow from taking components of degree $s$? That is, why is the sum of dimensions of terms 1 and 3 equal the sum of dimensions of terms 2 and 4?
(4) I don't quite understand why $K$ and $M/x_rM$ are finitely generated modules over $k[x_1,dots,x_{r-1}]$. Well, a f.g. module over a Noetherian ring is Noetherian, and $K$ is a submodule of a f.g. Noetherian module, so it's finitely generated. And the quotient of a Noetherian module is f.g. But why are they modules over $k[x_1,dots,x_{r-1}]$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra proof-explanation
edited Jan 26 at 21:04
user26857
39.4k124183
asked Jan 26 at 20:17
user437309user437309
766314
$endgroup$
$begingroup$
(1) Can have a finitely dimensional graded vector space infinitely many non-zero graded pieces?
$endgroup$
– user26857
Jan 26 at 20:58
$begingroup$
(2) Yes, the maps should be graded.
$endgroup$
– user26857
Jan 26 at 20:59
2
$begingroup$
(4) If $M$ is a module and $I$ an ideal of $R$, then $M/IM$ is an $R/I$-module. Can you see that $x_rK=0$ and $x_r(M/x_rM)=0$?
$endgroup$
– user26857
Jan 26 at 21:02
1
$begingroup$
Yes, maps that preserve the degree. Do you know the definition of a morphism of graded modules?
$endgroup$
– user26857
Jan 26 at 21:05
1
$begingroup$
@user26857 I see, so loosely speaking, since we are working with graded modules, all maps between them should be from the appropriate category.
$endgroup$
– user437309
Jan 26 at 21:07
|
show 3 more comments
$begingroup$
Here is Hilbert's theorem from Eisenbud:
There are a few things I don't understand:
(1) Why does $H_M(s)=0$ for $s$ large in the case of graded vector spaces?
(2) What is the point of working with twists here? Is it used somehow that the maps between twisted objects have degree $0$ (map homogeneous elements to homogeneous elements of the same degree)?
(3) How does the equality (second display) follow from taking components of degree $s$? That is, why is the sum of dimensions of terms 1 and 3 equal the sum of dimensions of terms 2 and 4?
(4) I don't quite understand why $K$ and $M/x_rM$ are finitely generated modules over $k[x_1,dots,x_{r-1}]$. Well, a f.g. module over a Noetherian ring is Noetherian, and $K$ is a submodule of a f.g. Noetherian module, so it's finitely generated. And the quotient of a Noetherian module is f.g. But why are they modules over $k[x_1,dots,x_{r-1}]$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra proof-explanation
edited Jan 26 at 21:04
user26857
39.4k124183
asked Jan 26 at 20:17
user437309user437309
766314
$endgroup$
Here is Hilbert's theorem from Eisenbud:
There are a few things I don't understand:
(1) Why does $H_M(s)=0$ for $s$ large in the case of graded vector spaces?
(2) What is the point of working with twists here? Is it used somehow that the maps between twisted objects have degree $0$ (map homogeneous elements to homogeneous elements of the same degree)?
(3) How does the equality (second display) follow from taking components of degree $s$? That is, why is the sum of dimensions of terms 1 and 3 equal the sum of dimensions of terms 2 and 4?
(4) I don't quite understand why $K$ and $M/x_rM$ are finitely generated modules over $k[x_1,dots,x_{r-1}]$. Well, a f.g. module over a Noetherian ring is Noetherian, and $K$ is a submodule of a f.g. Noetherian module, so it's finitely generated. And the quotient of a Noetherian module is f.g. But why are they modules over $k[x_1,dots,x_{r-1}]$?
abstract-algebra algebraic-geometry ring-theory commutative-algebra proof-explanation
abstract-algebra algebraic-geometry ring-theory commutative-algebra proof-explanation
edited Jan 26 at 21:04
user26857
39.4k124183
asked Jan 26 at 20:17
user437309user437309
766314
edited Jan 26 at 21:04
user26857
39.4k124183
asked Jan 26 at 20:17
user437309user437309
766314
edited Jan 26 at 21:04
user26857
39.4k124183
edited Jan 26 at 21:04
user26857
39.4k124183
edited Jan 26 at 21:04
user26857
39.4k124183
39.4k124183
asked Jan 26 at 20:17
user437309user437309
766314
asked Jan 26 at 20:17
user437309user437309
766314
asked Jan 26 at 20:17
user437309user437309
766314
766314
$begingroup$
(1) Can have a finitely dimensional graded vector space infinitely many non-zero graded pieces?
$endgroup$
– user26857
Jan 26 at 20:58
$begingroup$
(2) Yes, the maps should be graded.
$endgroup$
– user26857
Jan 26 at 20:59
2
$begingroup$
(4) If $M$ is a module and $I$ an ideal of $R$, then $M/IM$ is an $R/I$-module. Can you see that $x_rK=0$ and $x_r(M/x_rM)=0$?
$endgroup$
– user26857
Jan 26 at 21:02
1
$begingroup$
Yes, maps that preserve the degree. Do you know the definition of a morphism of graded modules?
$endgroup$
– user26857
Jan 26 at 21:05
1
$begingroup$
@user26857 I see, so loosely speaking, since we are working with graded modules, all maps between them should be from the appropriate category.
$endgroup$
– user437309
Jan 26 at 21:07
|
show 3 more comments
$begingroup$
(1) Can have a finitely dimensional graded vector space infinitely many non-zero graded pieces?
$endgroup$
– user26857
Jan 26 at 20:58
$begingroup$
(2) Yes, the maps should be graded.
$endgroup$
– user26857
Jan 26 at 20:59
2
$begingroup$
(4) If $M$ is a module and $I$ an ideal of $R$, then $M/IM$ is an $R/I$-module. Can you see that $x_rK=0$ and $x_r(M/x_rM)=0$?
$endgroup$
– user26857
Jan 26 at 21:02
1
$begingroup$
Yes, maps that preserve the degree. Do you know the definition of a morphism of graded modules?
$endgroup$
– user26857
Jan 26 at 21:05
1
$begingroup$
@user26857 I see, so loosely speaking, since we are working with graded modules, all maps between them should be from the appropriate category.
$endgroup$
– user437309
Jan 26 at 21:07
$begingroup$
(1) Can have a finitely dimensional graded vector space infinitely many non-zero graded pieces?
$endgroup$
– user26857
Jan 26 at 20:58
$begingroup$
(1) Can have a finitely dimensional graded vector space infinitely many non-zero graded pieces?
$endgroup$
– user26857
Jan 26 at 20:58
$begingroup$
(2) Yes, the maps should be graded.
$endgroup$
– user26857
Jan 26 at 20:59
$begingroup$
(2) Yes, the maps should be graded.
$endgroup$
– user26857
Jan 26 at 20:59
2
2
$begingroup$
(4) If $M$ is a module and $I$ an ideal of $R$, then $M/IM$ is an $R/I$-module. Can you see that $x_rK=0$ and $x_r(M/x_rM)=0$?
$endgroup$
– user26857
Jan 26 at 21:02
$begingroup$
(4) If $M$ is a module and $I$ an ideal of $R$, then $M/IM$ is an $R/I$-module. Can you see that $x_rK=0$ and $x_r(M/x_rM)=0$?
$endgroup$
– user26857
Jan 26 at 21:02
1
1
$begingroup$
Yes, maps that preserve the degree. Do you know the definition of a morphism of graded modules?
$endgroup$
– user26857
Jan 26 at 21:05
$begingroup$
Yes, maps that preserve the degree. Do you know the definition of a morphism of graded modules?
$endgroup$
– user26857
Jan 26 at 21:05
1
1
$begingroup$
@user26857 I see, so loosely speaking, since we are working with graded modules, all maps between them should be from the appropriate category.
$endgroup$
– user437309
Jan 26 at 21:07
$begingroup$
@user26857 I see, so loosely speaking, since we are working with graded modules, all maps between them should be from the appropriate category.
$endgroup$
– user437309
Jan 26 at 21:07
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
(1) A finite dimensional graded vector space has a finite basis. The degrees of these basis elements is a finite list of integers. Letting $B$ be the maximum degree, then $H_M(s) = 0$ for all $s > B$ since there are no elements of degree larger than $B$.
(2) Yes, that's right: a morphism of graded modules must preserve degrees, and that's what twisting accomplishes.
(3) This is just a statement about exact sequences of vector spaces, namely that they have Euler characteristic $0$. (See this answer.) This is really just a highbrow version of the rank-nullity theorem from linear algebra. Rank-nullity says that for a vector space $V$ with a subspace $W$, $dim(V/W) = dim(V) - dim(W)$, i.e., $dim(W) - dim(V) + dim(V/W) = 0$. We can express this as a short exact sequence of vector spaces $0 to W to V to V/W to 0$. One can generalize this result and show that given any exact sequence of vector spaces $0 to V_1 to V_2 to cdots to V_n to 0$, we have $sum_{i=1}^n (-1)^i dim(V_i) = 0$.
Applying this to the exact sequence
$$
0 to K(-1) to M(-1) to M to M/x_r M to 0
$$
we get
begin{align*}
0 &= H_{K(-1)}(s) - H_{M(-1)}(s) + H_M(s) - H_{M/x_r M}(s) = H_{K}(s-1) - H_{M}(s-1) + H_M(s) - H_{M/x_r M}(s)
end{align*}
and rearranging yields the result.
(4) We can consider any $k[x_1, ldots, x_r]$-module as a $k[x_1, ldots, x_{r-1}]$-module by considering the inclusion $k[x_1, ldots, x_{r-1}] hookrightarrow k[x_1, ldots, x_r]$. The key observation here is that $x_r$ acts as $0$ on both $K$ and $M/x_r M$, so they are still finitely generated over the subring $k[x_1, ldots, x_{r-1}]$.
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
$endgroup$
add a comment |
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$begingroup$
(1) A finite dimensional graded vector space has a finite basis. The degrees of these basis elements is a finite list of integers. Letting $B$ be the maximum degree, then $H_M(s) = 0$ for all $s > B$ since there are no elements of degree larger than $B$.
(2) Yes, that's right: a morphism of graded modules must preserve degrees, and that's what twisting accomplishes.
(3) This is just a statement about exact sequences of vector spaces, namely that they have Euler characteristic $0$. (See this answer.) This is really just a highbrow version of the rank-nullity theorem from linear algebra. Rank-nullity says that for a vector space $V$ with a subspace $W$, $dim(V/W) = dim(V) - dim(W)$, i.e., $dim(W) - dim(V) + dim(V/W) = 0$. We can express this as a short exact sequence of vector spaces $0 to W to V to V/W to 0$. One can generalize this result and show that given any exact sequence of vector spaces $0 to V_1 to V_2 to cdots to V_n to 0$, we have $sum_{i=1}^n (-1)^i dim(V_i) = 0$.
Applying this to the exact sequence
$$
0 to K(-1) to M(-1) to M to M/x_r M to 0
$$
we get
begin{align*}
0 &= H_{K(-1)}(s) - H_{M(-1)}(s) + H_M(s) - H_{M/x_r M}(s) = H_{K}(s-1) - H_{M}(s-1) + H_M(s) - H_{M/x_r M}(s)
end{align*}
and rearranging yields the result.
(4) We can consider any $k[x_1, ldots, x_r]$-module as a $k[x_1, ldots, x_{r-1}]$-module by considering the inclusion $k[x_1, ldots, x_{r-1}] hookrightarrow k[x_1, ldots, x_r]$. The key observation here is that $x_r$ acts as $0$ on both $K$ and $M/x_r M$, so they are still finitely generated over the subring $k[x_1, ldots, x_{r-1}]$.
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
$endgroup$
add a comment |
$begingroup$
(1) A finite dimensional graded vector space has a finite basis. The degrees of these basis elements is a finite list of integers. Letting $B$ be the maximum degree, then $H_M(s) = 0$ for all $s > B$ since there are no elements of degree larger than $B$.
(2) Yes, that's right: a morphism of graded modules must preserve degrees, and that's what twisting accomplishes.
(3) This is just a statement about exact sequences of vector spaces, namely that they have Euler characteristic $0$. (See this answer.) This is really just a highbrow version of the rank-nullity theorem from linear algebra. Rank-nullity says that for a vector space $V$ with a subspace $W$, $dim(V/W) = dim(V) - dim(W)$, i.e., $dim(W) - dim(V) + dim(V/W) = 0$. We can express this as a short exact sequence of vector spaces $0 to W to V to V/W to 0$. One can generalize this result and show that given any exact sequence of vector spaces $0 to V_1 to V_2 to cdots to V_n to 0$, we have $sum_{i=1}^n (-1)^i dim(V_i) = 0$.
Applying this to the exact sequence
$$
0 to K(-1) to M(-1) to M to M/x_r M to 0
$$
we get
begin{align*}
0 &= H_{K(-1)}(s) - H_{M(-1)}(s) + H_M(s) - H_{M/x_r M}(s) = H_{K}(s-1) - H_{M}(s-1) + H_M(s) - H_{M/x_r M}(s)
end{align*}
and rearranging yields the result.
(4) We can consider any $k[x_1, ldots, x_r]$-module as a $k[x_1, ldots, x_{r-1}]$-module by considering the inclusion $k[x_1, ldots, x_{r-1}] hookrightarrow k[x_1, ldots, x_r]$. The key observation here is that $x_r$ acts as $0$ on both $K$ and $M/x_r M$, so they are still finitely generated over the subring $k[x_1, ldots, x_{r-1}]$.
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
$endgroup$
add a comment |
$begingroup$
(1) A finite dimensional graded vector space has a finite basis. The degrees of these basis elements is a finite list of integers. Letting $B$ be the maximum degree, then $H_M(s) = 0$ for all $s > B$ since there are no elements of degree larger than $B$.
(2) Yes, that's right: a morphism of graded modules must preserve degrees, and that's what twisting accomplishes.
(3) This is just a statement about exact sequences of vector spaces, namely that they have Euler characteristic $0$. (See this answer.) This is really just a highbrow version of the rank-nullity theorem from linear algebra. Rank-nullity says that for a vector space $V$ with a subspace $W$, $dim(V/W) = dim(V) - dim(W)$, i.e., $dim(W) - dim(V) + dim(V/W) = 0$. We can express this as a short exact sequence of vector spaces $0 to W to V to V/W to 0$. One can generalize this result and show that given any exact sequence of vector spaces $0 to V_1 to V_2 to cdots to V_n to 0$, we have $sum_{i=1}^n (-1)^i dim(V_i) = 0$.
Applying this to the exact sequence
$$
0 to K(-1) to M(-1) to M to M/x_r M to 0
$$
we get
begin{align*}
0 &= H_{K(-1)}(s) - H_{M(-1)}(s) + H_M(s) - H_{M/x_r M}(s) = H_{K}(s-1) - H_{M}(s-1) + H_M(s) - H_{M/x_r M}(s)
end{align*}
and rearranging yields the result.
(4) We can consider any $k[x_1, ldots, x_r]$-module as a $k[x_1, ldots, x_{r-1}]$-module by considering the inclusion $k[x_1, ldots, x_{r-1}] hookrightarrow k[x_1, ldots, x_r]$. The key observation here is that $x_r$ acts as $0$ on both $K$ and $M/x_r M$, so they are still finitely generated over the subring $k[x_1, ldots, x_{r-1}]$.
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
$endgroup$
(1) A finite dimensional graded vector space has a finite basis. The degrees of these basis elements is a finite list of integers. Letting $B$ be the maximum degree, then $H_M(s) = 0$ for all $s > B$ since there are no elements of degree larger than $B$.
(2) Yes, that's right: a morphism of graded modules must preserve degrees, and that's what twisting accomplishes.
(3) This is just a statement about exact sequences of vector spaces, namely that they have Euler characteristic $0$. (See this answer.) This is really just a highbrow version of the rank-nullity theorem from linear algebra. Rank-nullity says that for a vector space $V$ with a subspace $W$, $dim(V/W) = dim(V) - dim(W)$, i.e., $dim(W) - dim(V) + dim(V/W) = 0$. We can express this as a short exact sequence of vector spaces $0 to W to V to V/W to 0$. One can generalize this result and show that given any exact sequence of vector spaces $0 to V_1 to V_2 to cdots to V_n to 0$, we have $sum_{i=1}^n (-1)^i dim(V_i) = 0$.
Applying this to the exact sequence
$$
0 to K(-1) to M(-1) to M to M/x_r M to 0
$$
we get
begin{align*}
0 &= H_{K(-1)}(s) - H_{M(-1)}(s) + H_M(s) - H_{M/x_r M}(s) = H_{K}(s-1) - H_{M}(s-1) + H_M(s) - H_{M/x_r M}(s)
end{align*}
and rearranging yields the result.
(4) We can consider any $k[x_1, ldots, x_r]$-module as a $k[x_1, ldots, x_{r-1}]$-module by considering the inclusion $k[x_1, ldots, x_{r-1}] hookrightarrow k[x_1, ldots, x_r]$. The key observation here is that $x_r$ acts as $0$ on both $K$ and $M/x_r M$, so they are still finitely generated over the subring $k[x_1, ldots, x_{r-1}]$.
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
answered Jan 26 at 21:06
André 3000André 3000
12.8k22243
12.8k22243
add a comment |
add a comment |
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$begingroup$
(1) Can have a finitely dimensional graded vector space infinitely many non-zero graded pieces?
$endgroup$
– user26857
Jan 26 at 20:58
$begingroup$
(2) Yes, the maps should be graded.
$endgroup$
– user26857
Jan 26 at 20:59
2
$begingroup$
(4) If $M$ is a module and $I$ an ideal of $R$, then $M/IM$ is an $R/I$-module. Can you see that $x_rK=0$ and $x_r(M/x_rM)=0$?
$endgroup$
– user26857
Jan 26 at 21:02
1
$begingroup$
Yes, maps that preserve the degree. Do you know the definition of a morphism of graded modules?
$endgroup$
– user26857
Jan 26 at 21:05
1
$begingroup$
@user26857 I see, so loosely speaking, since we are working with graded modules, all maps between them should be from the appropriate category.
$endgroup$
– user437309
Jan 26 at 21:07