Mixing
Criteria for Ideal of a Hopf Algebra to Yield Closed Subgroup
Clash Royale CLAN TAG#URR8PPP
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Let $k$ be a commutative ring with unity, and let $A$ be the $k$-algebra of an affine group scheme over $k,$ endowed with its usual structure as a Hopf algebra over $k.$ In Waterhouse's textbook, An Introduction to Affine Group Schemes, he discusses the criteria that an ideal $I$ of $A$ must satisfy in order for $A/I$ to represent a closed subgroup scheme. These conditions are
1) If $Delta$ is the comultiplication, then $Delta(I) subseteq A otimes_k I + I otimes_k A$.
2) If $S$ is the antipode, then $S(I) subseteq I$.
3) If $epsilon$ is the augmentation, then $epsilon(I) = 0.$
I understand that 3) is both necessary and sufficient for the unit element(s) to belong to the subgroup scheme. I also understand that 1) is sufficient for closure under multiplication for the subgroup scheme and that 2) is sufficient for closure under inversion for the subgroup scheme.
I am guessing that 2) is necessary for closure under inversion because we can just consider the canonical projection $A to A/I,$ which is a map with kernel equal to $I.$
However, I am confused as to why 1) is necessary for closure under multiplication.
That is, if it is true that for every $k$-algebra $R$ and for every $f, g : A to R$ with $f(I) = 0 = g(I)$ that we have $left((f otimes g) circ Deltaright) = 0$, then why must condition 1) follow?
Thank you very much.
algebraic-groups affine-schemes
asked Jul 22 at 2:29
Mishel Skenderi
1398
 |Â
show 1 more comment
up vote
1
down vote
favorite
Let $k$ be a commutative ring with unity, and let $A$ be the $k$-algebra of an affine group scheme over $k,$ endowed with its usual structure as a Hopf algebra over $k.$ In Waterhouse's textbook, An Introduction to Affine Group Schemes, he discusses the criteria that an ideal $I$ of $A$ must satisfy in order for $A/I$ to represent a closed subgroup scheme. These conditions are
1) If $Delta$ is the comultiplication, then $Delta(I) subseteq A otimes_k I + I otimes_k A$.
2) If $S$ is the antipode, then $S(I) subseteq I$.
3) If $epsilon$ is the augmentation, then $epsilon(I) = 0.$
I understand that 3) is both necessary and sufficient for the unit element(s) to belong to the subgroup scheme. I also understand that 1) is sufficient for closure under multiplication for the subgroup scheme and that 2) is sufficient for closure under inversion for the subgroup scheme.
I am guessing that 2) is necessary for closure under inversion because we can just consider the canonical projection $A to A/I,$ which is a map with kernel equal to $I.$
However, I am confused as to why 1) is necessary for closure under multiplication.
That is, if it is true that for every $k$-algebra $R$ and for every $f, g : A to R$ with $f(I) = 0 = g(I)$ that we have $left((f otimes g) circ Deltaright) = 0$, then why must condition 1) follow?
Thank you very much.
algebraic-groups affine-schemes
asked Jul 22 at 2:29
Mishel Skenderi
1398
I think that I can do sort of the same thing. Namely let $f = g = pi : A to A/I.$ Then $(pi otimes_k pi) circ Delta$ vanishes on $I.$ Hence, $Delta(I) subseteq ker(pi otimes_k pi).$ Now, one must show $ker(pi otimes_k pi) subseteq A otimes_k I + I otimes_k A.$ Unfortunately, I am not sure how to show this last step. (The reverse inclusion is clear to me, but that is not what is needed.)
– Mishel Skenderi
Jul 22 at 2:57
We want $Delta$ to descend to an algebra homorphism $A/I to A/I otimes A/I$. So I think it must boil down to: suppose $M' subset M$, and $N' subset N$ are $k$-modules. Then $M/M' otimes N/N' simeq Motimes N / (M'otimes N + M otimes N')$. Agree?
– peter a g
Jul 22 at 3:57
Namely, the iso above seems correct to me - or am I being stupid?
– peter a g
Jul 22 at 3:59
I think I see what you mean...but is it clear to you that equality holds for the kernel in the map of my comment? I can’t figure out how to prove the other inclusion...
– Mishel Skenderi
Jul 22 at 15:57
Just to acknowledge that I saw your comment: I'll try to get back to you on this tonight or tomorrow night...
– peter a g
Jul 22 at 16:09
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $k$ be a commutative ring with unity, and let $A$ be the $k$-algebra of an affine group scheme over $k,$ endowed with its usual structure as a Hopf algebra over $k.$ In Waterhouse's textbook, An Introduction to Affine Group Schemes, he discusses the criteria that an ideal $I$ of $A$ must satisfy in order for $A/I$ to represent a closed subgroup scheme. These conditions are
1) If $Delta$ is the comultiplication, then $Delta(I) subseteq A otimes_k I + I otimes_k A$.
2) If $S$ is the antipode, then $S(I) subseteq I$.
3) If $epsilon$ is the augmentation, then $epsilon(I) = 0.$
I understand that 3) is both necessary and sufficient for the unit element(s) to belong to the subgroup scheme. I also understand that 1) is sufficient for closure under multiplication for the subgroup scheme and that 2) is sufficient for closure under inversion for the subgroup scheme.
I am guessing that 2) is necessary for closure under inversion because we can just consider the canonical projection $A to A/I,$ which is a map with kernel equal to $I.$
However, I am confused as to why 1) is necessary for closure under multiplication.
That is, if it is true that for every $k$-algebra $R$ and for every $f, g : A to R$ with $f(I) = 0 = g(I)$ that we have $left((f otimes g) circ Deltaright) = 0$, then why must condition 1) follow?
Thank you very much.
algebraic-groups affine-schemes
asked Jul 22 at 2:29
Mishel Skenderi
1398
Let $k$ be a commutative ring with unity, and let $A$ be the $k$-algebra of an affine group scheme over $k,$ endowed with its usual structure as a Hopf algebra over $k.$ In Waterhouse's textbook, An Introduction to Affine Group Schemes, he discusses the criteria that an ideal $I$ of $A$ must satisfy in order for $A/I$ to represent a closed subgroup scheme. These conditions are
1) If $Delta$ is the comultiplication, then $Delta(I) subseteq A otimes_k I + I otimes_k A$.
2) If $S$ is the antipode, then $S(I) subseteq I$.
3) If $epsilon$ is the augmentation, then $epsilon(I) = 0.$
I understand that 3) is both necessary and sufficient for the unit element(s) to belong to the subgroup scheme. I also understand that 1) is sufficient for closure under multiplication for the subgroup scheme and that 2) is sufficient for closure under inversion for the subgroup scheme.
I am guessing that 2) is necessary for closure under inversion because we can just consider the canonical projection $A to A/I,$ which is a map with kernel equal to $I.$
However, I am confused as to why 1) is necessary for closure under multiplication.
That is, if it is true that for every $k$-algebra $R$ and for every $f, g : A to R$ with $f(I) = 0 = g(I)$ that we have $left((f otimes g) circ Deltaright) = 0$, then why must condition 1) follow?
Thank you very much.
algebraic-groups affine-schemes
asked Jul 22 at 2:29
Mishel Skenderi
1398
asked Jul 22 at 2:29
Mishel Skenderi
1398
asked Jul 22 at 2:29
Mishel Skenderi
1398
asked Jul 22 at 2:29
Mishel Skenderi
1398
1398
I think that I can do sort of the same thing. Namely let $f = g = pi : A to A/I.$ Then $(pi otimes_k pi) circ Delta$ vanishes on $I.$ Hence, $Delta(I) subseteq ker(pi otimes_k pi).$ Now, one must show $ker(pi otimes_k pi) subseteq A otimes_k I + I otimes_k A.$ Unfortunately, I am not sure how to show this last step. (The reverse inclusion is clear to me, but that is not what is needed.)
– Mishel Skenderi
Jul 22 at 2:57
We want $Delta$ to descend to an algebra homorphism $A/I to A/I otimes A/I$. So I think it must boil down to: suppose $M' subset M$, and $N' subset N$ are $k$-modules. Then $M/M' otimes N/N' simeq Motimes N / (M'otimes N + M otimes N')$. Agree?
– peter a g
Jul 22 at 3:57
Namely, the iso above seems correct to me - or am I being stupid?
– peter a g
Jul 22 at 3:59
I think I see what you mean...but is it clear to you that equality holds for the kernel in the map of my comment? I can’t figure out how to prove the other inclusion...
– Mishel Skenderi
Jul 22 at 15:57
Just to acknowledge that I saw your comment: I'll try to get back to you on this tonight or tomorrow night...
– peter a g
Jul 22 at 16:09
 |Â
show 1 more comment
I think that I can do sort of the same thing. Namely let $f = g = pi : A to A/I.$ Then $(pi otimes_k pi) circ Delta$ vanishes on $I.$ Hence, $Delta(I) subseteq ker(pi otimes_k pi).$ Now, one must show $ker(pi otimes_k pi) subseteq A otimes_k I + I otimes_k A.$ Unfortunately, I am not sure how to show this last step. (The reverse inclusion is clear to me, but that is not what is needed.)
– Mishel Skenderi
Jul 22 at 2:57
We want $Delta$ to descend to an algebra homorphism $A/I to A/I otimes A/I$. So I think it must boil down to: suppose $M' subset M$, and $N' subset N$ are $k$-modules. Then $M/M' otimes N/N' simeq Motimes N / (M'otimes N + M otimes N')$. Agree?
– peter a g
Jul 22 at 3:57
Namely, the iso above seems correct to me - or am I being stupid?
– peter a g
Jul 22 at 3:59
I think I see what you mean...but is it clear to you that equality holds for the kernel in the map of my comment? I can’t figure out how to prove the other inclusion...
– Mishel Skenderi
Jul 22 at 15:57
Just to acknowledge that I saw your comment: I'll try to get back to you on this tonight or tomorrow night...
– peter a g
Jul 22 at 16:09
I think that I can do sort of the same thing. Namely let $f = g = pi : A to A/I.$ Then $(pi otimes_k pi) circ Delta$ vanishes on $I.$ Hence, $Delta(I) subseteq ker(pi otimes_k pi).$ Now, one must show $ker(pi otimes_k pi) subseteq A otimes_k I + I otimes_k A.$ Unfortunately, I am not sure how to show this last step. (The reverse inclusion is clear to me, but that is not what is needed.)
– Mishel Skenderi
Jul 22 at 2:57
I think that I can do sort of the same thing. Namely let $f = g = pi : A to A/I.$ Then $(pi otimes_k pi) circ Delta$ vanishes on $I.$ Hence, $Delta(I) subseteq ker(pi otimes_k pi).$ Now, one must show $ker(pi otimes_k pi) subseteq A otimes_k I + I otimes_k A.$ Unfortunately, I am not sure how to show this last step. (The reverse inclusion is clear to me, but that is not what is needed.)
– Mishel Skenderi
Jul 22 at 2:57
We want $Delta$ to descend to an algebra homorphism $A/I to A/I otimes A/I$. So I think it must boil down to: suppose $M' subset M$, and $N' subset N$ are $k$-modules. Then $M/M' otimes N/N' simeq Motimes N / (M'otimes N + M otimes N')$. Agree?
– peter a g
Jul 22 at 3:57
We want $Delta$ to descend to an algebra homorphism $A/I to A/I otimes A/I$. So I think it must boil down to: suppose $M' subset M$, and $N' subset N$ are $k$-modules. Then $M/M' otimes N/N' simeq Motimes N / (M'otimes N + M otimes N')$. Agree?
– peter a g
Jul 22 at 3:57
Namely, the iso above seems correct to me - or am I being stupid?
– peter a g
Jul 22 at 3:59
Namely, the iso above seems correct to me - or am I being stupid?
– peter a g
Jul 22 at 3:59
I think I see what you mean...but is it clear to you that equality holds for the kernel in the map of my comment? I can’t figure out how to prove the other inclusion...
– Mishel Skenderi
Jul 22 at 15:57
I think I see what you mean...but is it clear to you that equality holds for the kernel in the map of my comment? I can’t figure out how to prove the other inclusion...
– Mishel Skenderi
Jul 22 at 15:57
Just to acknowledge that I saw your comment: I'll try to get back to you on this tonight or tomorrow night...
– peter a g
Jul 22 at 16:09
Just to acknowledge that I saw your comment: I'll try to get back to you on this tonight or tomorrow night...
– peter a g
Jul 22 at 16:09
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
As per the comments: we want $Delta colon A to Aotimes A $ to descend to $Delta colon A/I to A/I otimes A/I$.
Condition 1) implies this (and is equivalent to the descent of $Delta$) if the natural map
$$ Aotimes A / (Aotimes I + I otimes A) to A/I otimes A/I,$$
defined and determined by $$ aotimes bmapsto overline a otimes overline b,$$
is an isomorphism, where $overline x = x + I$.
More generally, and for clarity, again as in the comment section, if $M'subset M$, $N'subset N$ are $k$-modules, we want to show that the natural map
$$ Motimes N / (M'otimes N + M otimes N') to M/M' otimes N/N',$$
defined/determined by
$$ motimes n + (M'otimes N + M otimes N') to overline m otimes overline n$$
is an isomorphism (and $overline m = m + M'$, and similarly for $n$).
Just to give this last map a name, call it $alpha$.
To construct an inverse map $beta$:
Fix $m in M$. Then there is a $k$-linear map
$$phi_m colon N to Motimes N / (M'otimes N + M otimes N') ,$$
defined by
$$phi_mcolon n to m otimes n + (M'otimes N + M otimes N').$$
Clearly, $phi_m$ vanishes on $N'$, so $phi_m$ descends to a $k$-linear map
- call it $phi_m$ again - abuse of notation:
$$ phi_m colon bar n to motimes n + (M'otimes N + M otimes N').$$
Now, $m to phi_m$ is $k$-linear, and depends only on $bar m$, so we obtain a $k$-bilinear map
$$ M/M' times N/N' to M otimes N / (M'otimes N + M otimes N'),$$
with
$$ (bar m,bar n) to (motimes n) + (M'otimes N + M otimes N').$$
Therefore, by definition (defining property of tensor products) one has the desired $k$-linear map $beta$
$$ M/M' otimes N/N' to M otimes N / (M'otimes N + M otimes N').$$
On generators, the two maps $alpha$ and $beta$ are inverses, so $alpha$ and $beta$ are $k$-module inverses.
answered Jul 24 at 1:49
peter a g
2,9801614
I see, thank you. I was hoping to somehow get away with showing that it is injective (and thus an isomorphism) without explicitly constructing an inverse.
– Mishel Skenderi
Jul 24 at 1:59
Yes - I figured as much... Enjoy the book. I remember it being v. good (I have to admit that I didn't read the smoothness section, and am sorry that I never did).
– peter a g
Jul 24 at 2:01
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As per the comments: we want $Delta colon A to Aotimes A $ to descend to $Delta colon A/I to A/I otimes A/I$.
Condition 1) implies this (and is equivalent to the descent of $Delta$) if the natural map
$$ Aotimes A / (Aotimes I + I otimes A) to A/I otimes A/I,$$
defined and determined by $$ aotimes bmapsto overline a otimes overline b,$$
is an isomorphism, where $overline x = x + I$.
More generally, and for clarity, again as in the comment section, if $M'subset M$, $N'subset N$ are $k$-modules, we want to show that the natural map
$$ Motimes N / (M'otimes N + M otimes N') to M/M' otimes N/N',$$
defined/determined by
$$ motimes n + (M'otimes N + M otimes N') to overline m otimes overline n$$
is an isomorphism (and $overline m = m + M'$, and similarly for $n$).
Just to give this last map a name, call it $alpha$.
To construct an inverse map $beta$:
Fix $m in M$. Then there is a $k$-linear map
$$phi_m colon N to Motimes N / (M'otimes N + M otimes N') ,$$
defined by
$$phi_mcolon n to m otimes n + (M'otimes N + M otimes N').$$
Clearly, $phi_m$ vanishes on $N'$, so $phi_m$ descends to a $k$-linear map
- call it $phi_m$ again - abuse of notation:
$$ phi_m colon bar n to motimes n + (M'otimes N + M otimes N').$$
Now, $m to phi_m$ is $k$-linear, and depends only on $bar m$, so we obtain a $k$-bilinear map
$$ M/M' times N/N' to M otimes N / (M'otimes N + M otimes N'),$$
with
$$ (bar m,bar n) to (motimes n) + (M'otimes N + M otimes N').$$
Therefore, by definition (defining property of tensor products) one has the desired $k$-linear map $beta$
$$ M/M' otimes N/N' to M otimes N / (M'otimes N + M otimes N').$$
On generators, the two maps $alpha$ and $beta$ are inverses, so $alpha$ and $beta$ are $k$-module inverses.
answered Jul 24 at 1:49
peter a g
2,9801614
I see, thank you. I was hoping to somehow get away with showing that it is injective (and thus an isomorphism) without explicitly constructing an inverse.
– Mishel Skenderi
Jul 24 at 1:59
Yes - I figured as much... Enjoy the book. I remember it being v. good (I have to admit that I didn't read the smoothness section, and am sorry that I never did).
– peter a g
Jul 24 at 2:01
add a comment |Â
up vote
1
down vote
accepted
As per the comments: we want $Delta colon A to Aotimes A $ to descend to $Delta colon A/I to A/I otimes A/I$.
Condition 1) implies this (and is equivalent to the descent of $Delta$) if the natural map
$$ Aotimes A / (Aotimes I + I otimes A) to A/I otimes A/I,$$
defined and determined by $$ aotimes bmapsto overline a otimes overline b,$$
is an isomorphism, where $overline x = x + I$.
More generally, and for clarity, again as in the comment section, if $M'subset M$, $N'subset N$ are $k$-modules, we want to show that the natural map
$$ Motimes N / (M'otimes N + M otimes N') to M/M' otimes N/N',$$
defined/determined by
$$ motimes n + (M'otimes N + M otimes N') to overline m otimes overline n$$
is an isomorphism (and $overline m = m + M'$, and similarly for $n$).
Just to give this last map a name, call it $alpha$.
To construct an inverse map $beta$:
Fix $m in M$. Then there is a $k$-linear map
$$phi_m colon N to Motimes N / (M'otimes N + M otimes N') ,$$
defined by
$$phi_mcolon n to m otimes n + (M'otimes N + M otimes N').$$
Clearly, $phi_m$ vanishes on $N'$, so $phi_m$ descends to a $k$-linear map
- call it $phi_m$ again - abuse of notation:
$$ phi_m colon bar n to motimes n + (M'otimes N + M otimes N').$$
Now, $m to phi_m$ is $k$-linear, and depends only on $bar m$, so we obtain a $k$-bilinear map
$$ M/M' times N/N' to M otimes N / (M'otimes N + M otimes N'),$$
with
$$ (bar m,bar n) to (motimes n) + (M'otimes N + M otimes N').$$
Therefore, by definition (defining property of tensor products) one has the desired $k$-linear map $beta$
$$ M/M' otimes N/N' to M otimes N / (M'otimes N + M otimes N').$$
On generators, the two maps $alpha$ and $beta$ are inverses, so $alpha$ and $beta$ are $k$-module inverses.
answered Jul 24 at 1:49
peter a g
2,9801614
I see, thank you. I was hoping to somehow get away with showing that it is injective (and thus an isomorphism) without explicitly constructing an inverse.
– Mishel Skenderi
Jul 24 at 1:59
Yes - I figured as much... Enjoy the book. I remember it being v. good (I have to admit that I didn't read the smoothness section, and am sorry that I never did).
– peter a g
Jul 24 at 2:01
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As per the comments: we want $Delta colon A to Aotimes A $ to descend to $Delta colon A/I to A/I otimes A/I$.
Condition 1) implies this (and is equivalent to the descent of $Delta$) if the natural map
$$ Aotimes A / (Aotimes I + I otimes A) to A/I otimes A/I,$$
defined and determined by $$ aotimes bmapsto overline a otimes overline b,$$
is an isomorphism, where $overline x = x + I$.
More generally, and for clarity, again as in the comment section, if $M'subset M$, $N'subset N$ are $k$-modules, we want to show that the natural map
$$ Motimes N / (M'otimes N + M otimes N') to M/M' otimes N/N',$$
defined/determined by
$$ motimes n + (M'otimes N + M otimes N') to overline m otimes overline n$$
is an isomorphism (and $overline m = m + M'$, and similarly for $n$).
Just to give this last map a name, call it $alpha$.
To construct an inverse map $beta$:
Fix $m in M$. Then there is a $k$-linear map
$$phi_m colon N to Motimes N / (M'otimes N + M otimes N') ,$$
defined by
$$phi_mcolon n to m otimes n + (M'otimes N + M otimes N').$$
Clearly, $phi_m$ vanishes on $N'$, so $phi_m$ descends to a $k$-linear map
- call it $phi_m$ again - abuse of notation:
$$ phi_m colon bar n to motimes n + (M'otimes N + M otimes N').$$
Now, $m to phi_m$ is $k$-linear, and depends only on $bar m$, so we obtain a $k$-bilinear map
$$ M/M' times N/N' to M otimes N / (M'otimes N + M otimes N'),$$
with
$$ (bar m,bar n) to (motimes n) + (M'otimes N + M otimes N').$$
Therefore, by definition (defining property of tensor products) one has the desired $k$-linear map $beta$
$$ M/M' otimes N/N' to M otimes N / (M'otimes N + M otimes N').$$
On generators, the two maps $alpha$ and $beta$ are inverses, so $alpha$ and $beta$ are $k$-module inverses.
answered Jul 24 at 1:49
peter a g
2,9801614
As per the comments: we want $Delta colon A to Aotimes A $ to descend to $Delta colon A/I to A/I otimes A/I$.
Condition 1) implies this (and is equivalent to the descent of $Delta$) if the natural map
$$ Aotimes A / (Aotimes I + I otimes A) to A/I otimes A/I,$$
defined and determined by $$ aotimes bmapsto overline a otimes overline b,$$
is an isomorphism, where $overline x = x + I$.
More generally, and for clarity, again as in the comment section, if $M'subset M$, $N'subset N$ are $k$-modules, we want to show that the natural map
$$ Motimes N / (M'otimes N + M otimes N') to M/M' otimes N/N',$$
defined/determined by
$$ motimes n + (M'otimes N + M otimes N') to overline m otimes overline n$$
is an isomorphism (and $overline m = m + M'$, and similarly for $n$).
Just to give this last map a name, call it $alpha$.
To construct an inverse map $beta$:
Fix $m in M$. Then there is a $k$-linear map
$$phi_m colon N to Motimes N / (M'otimes N + M otimes N') ,$$
defined by
$$phi_mcolon n to m otimes n + (M'otimes N + M otimes N').$$
Clearly, $phi_m$ vanishes on $N'$, so $phi_m$ descends to a $k$-linear map
- call it $phi_m$ again - abuse of notation:
$$ phi_m colon bar n to motimes n + (M'otimes N + M otimes N').$$
Now, $m to phi_m$ is $k$-linear, and depends only on $bar m$, so we obtain a $k$-bilinear map
$$ M/M' times N/N' to M otimes N / (M'otimes N + M otimes N'),$$
with
$$ (bar m,bar n) to (motimes n) + (M'otimes N + M otimes N').$$
Therefore, by definition (defining property of tensor products) one has the desired $k$-linear map $beta$
$$ M/M' otimes N/N' to M otimes N / (M'otimes N + M otimes N').$$
On generators, the two maps $alpha$ and $beta$ are inverses, so $alpha$ and $beta$ are $k$-module inverses.
answered Jul 24 at 1:49
peter a g
2,9801614
answered Jul 24 at 1:49
peter a g
2,9801614
answered Jul 24 at 1:49
peter a g
2,9801614
answered Jul 24 at 1:49
peter a g
2,9801614
2,9801614
I see, thank you. I was hoping to somehow get away with showing that it is injective (and thus an isomorphism) without explicitly constructing an inverse.
– Mishel Skenderi
Jul 24 at 1:59
Yes - I figured as much... Enjoy the book. I remember it being v. good (I have to admit that I didn't read the smoothness section, and am sorry that I never did).
– peter a g
Jul 24 at 2:01
add a comment |Â
I see, thank you. I was hoping to somehow get away with showing that it is injective (and thus an isomorphism) without explicitly constructing an inverse.
– Mishel Skenderi
Jul 24 at 1:59
Yes - I figured as much... Enjoy the book. I remember it being v. good (I have to admit that I didn't read the smoothness section, and am sorry that I never did).
– peter a g
Jul 24 at 2:01
I see, thank you. I was hoping to somehow get away with showing that it is injective (and thus an isomorphism) without explicitly constructing an inverse.
– Mishel Skenderi
Jul 24 at 1:59
I see, thank you. I was hoping to somehow get away with showing that it is injective (and thus an isomorphism) without explicitly constructing an inverse.
– Mishel Skenderi
Jul 24 at 1:59
Yes - I figured as much... Enjoy the book. I remember it being v. good (I have to admit that I didn't read the smoothness section, and am sorry that I never did).
– peter a g
Jul 24 at 2:01
Yes - I figured as much... Enjoy the book. I remember it being v. good (I have to admit that I didn't read the smoothness section, and am sorry that I never did).
– peter a g
Jul 24 at 2:01
add a comment |Â
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I think that I can do sort of the same thing. Namely let $f = g = pi : A to A/I.$ Then $(pi otimes_k pi) circ Delta$ vanishes on $I.$ Hence, $Delta(I) subseteq ker(pi otimes_k pi).$ Now, one must show $ker(pi otimes_k pi) subseteq A otimes_k I + I otimes_k A.$ Unfortunately, I am not sure how to show this last step. (The reverse inclusion is clear to me, but that is not what is needed.)
– Mishel Skenderi
Jul 22 at 2:57
We want $Delta$ to descend to an algebra homorphism $A/I to A/I otimes A/I$. So I think it must boil down to: suppose $M' subset M$, and $N' subset N$ are $k$-modules. Then $M/M' otimes N/N' simeq Motimes N / (M'otimes N + M otimes N')$. Agree?
– peter a g
Jul 22 at 3:57
Namely, the iso above seems correct to me - or am I being stupid?
– peter a g
Jul 22 at 3:59
I think I see what you mean...but is it clear to you that equality holds for the kernel in the map of my comment? I can’t figure out how to prove the other inclusion...
– Mishel Skenderi
Jul 22 at 15:57
Just to acknowledge that I saw your comment: I'll try to get back to you on this tonight or tomorrow night...
– peter a g
Jul 22 at 16:09