Mixing
Does the category of artinian rings admit finite limits?
$begingroup$
Let $mathsf{Artin}$ be the category of artinian rings, viewed as a full subcategory of the category $mathsf{CRing}$ of rings. (Here "ring" means "commutative ring with one".)
Question 1. Does $mathsf{Artin}$ admit finite limits?
As $mathsf{Artin}$ has finite products, Question 1 is equivalent to
Question 2. Does $mathsf{Artin}$ admit equalizers?
A closely related question is
Question 3. Let $Ato B$ be the equalizer in $mathsf{CRing}$ of two morphisms from $B$ to $C$. Assume that $B$ and $C$ are artinian. Does this imply that $A$ is artinian?
Yes to Question 3 would imply yes to Questions 1 and 2. [Edit: I made a mistake when I claimed implicitly that yes to Question 3 would immediately imply yes to Questions 1 and 2. There was a subtlety I missed, but Jeremy's comment below his answer settles also Question 1 and 2. (I also missed that, sigh...)]
This answer of MooS implies that the category of noetherian rings does not admit finite limits.
commutative-algebra category-theory limits-colimits artinian
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
$endgroup$
add a comment |
$begingroup$
Let $mathsf{Artin}$ be the category of artinian rings, viewed as a full subcategory of the category $mathsf{CRing}$ of rings. (Here "ring" means "commutative ring with one".)
Question 1. Does $mathsf{Artin}$ admit finite limits?
As $mathsf{Artin}$ has finite products, Question 1 is equivalent to
Question 2. Does $mathsf{Artin}$ admit equalizers?
A closely related question is
Question 3. Let $Ato B$ be the equalizer in $mathsf{CRing}$ of two morphisms from $B$ to $C$. Assume that $B$ and $C$ are artinian. Does this imply that $A$ is artinian?
Yes to Question 3 would imply yes to Questions 1 and 2. [Edit: I made a mistake when I claimed implicitly that yes to Question 3 would immediately imply yes to Questions 1 and 2. There was a subtlety I missed, but Jeremy's comment below his answer settles also Question 1 and 2. (I also missed that, sigh...)]
This answer of MooS implies that the category of noetherian rings does not admit finite limits.
commutative-algebra category-theory limits-colimits artinian
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
$endgroup$
add a comment |
$begingroup$
Let $mathsf{Artin}$ be the category of artinian rings, viewed as a full subcategory of the category $mathsf{CRing}$ of rings. (Here "ring" means "commutative ring with one".)
Question 1. Does $mathsf{Artin}$ admit finite limits?
As $mathsf{Artin}$ has finite products, Question 1 is equivalent to
Question 2. Does $mathsf{Artin}$ admit equalizers?
A closely related question is
Question 3. Let $Ato B$ be the equalizer in $mathsf{CRing}$ of two morphisms from $B$ to $C$. Assume that $B$ and $C$ are artinian. Does this imply that $A$ is artinian?
Yes to Question 3 would imply yes to Questions 1 and 2. [Edit: I made a mistake when I claimed implicitly that yes to Question 3 would immediately imply yes to Questions 1 and 2. There was a subtlety I missed, but Jeremy's comment below his answer settles also Question 1 and 2. (I also missed that, sigh...)]
This answer of MooS implies that the category of noetherian rings does not admit finite limits.
commutative-algebra category-theory limits-colimits artinian
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
$endgroup$
Let $mathsf{Artin}$ be the category of artinian rings, viewed as a full subcategory of the category $mathsf{CRing}$ of rings. (Here "ring" means "commutative ring with one".)
Question 1. Does $mathsf{Artin}$ admit finite limits?
As $mathsf{Artin}$ has finite products, Question 1 is equivalent to
Question 2. Does $mathsf{Artin}$ admit equalizers?
A closely related question is
Question 3. Let $Ato B$ be the equalizer in $mathsf{CRing}$ of two morphisms from $B$ to $C$. Assume that $B$ and $C$ are artinian. Does this imply that $A$ is artinian?
Yes to Question 3 would imply yes to Questions 1 and 2. [Edit: I made a mistake when I claimed implicitly that yes to Question 3 would immediately imply yes to Questions 1 and 2. There was a subtlety I missed, but Jeremy's comment below his answer settles also Question 1 and 2. (I also missed that, sigh...)]
This answer of MooS implies that the category of noetherian rings does not admit finite limits.
commutative-algebra category-theory limits-colimits artinian
commutative-algebra category-theory limits-colimits artinian
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
edited Feb 1 at 14:21
Pierre-Yves Gaillard
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
asked Jan 31 at 13:21
Pierre-Yves GaillardPierre-Yves Gaillard
13.5k23184
13.5k23184
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There's an example in
Gilmer, Robert; Heinzer, William, Artinian subrings of a commutative ring, Trans. Am. Math. Soc. 336, No. 1, 295-310 (1993). ZBL0778.13012.
of two artinian subrings of a commutative artinian ring whose intersection is not artinian. Since the intersection is the pullback of the inclusion maps, this answers the question.
Let $R=mathbb{C}(x)[y]/(y^2)$. Then $R$ is a two-dimensional algebra over $mathbb{C}(x)$, and is therefore artinian.
Let $R_1=mathbb{C}(x^2)+ymathbb{C}(x)subset R$. Then $R_1$ is a three-dimensional algebra over $mathbb{C}(x^2)$, and is therefore artinian.
Let $R_2=mathbb{C}(x^2+x)+ymathbb{C}(x)subset R$. Then $R_2$ is a three-dimensional algebra over $mathbb{C}(x^2+x)$, and is therefore artinian.
But $mathbb{C}(x^2)capmathbb{C}(x^2+x)=mathbb{C}$, so $R_1cap R_2=mathbb{C}+ymathbb{C}(x)$, which is not artinian, or even noetherian, since $ymathbb{C}(x)$ is not finitely generated as an ideal.
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
1
$begingroup$
Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 12:13
1
$begingroup$
@Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $alpha:Ato R_1cap R_2$ through which every map from an artinian ring factors. But every element of $R_1cap R_2$ is in a fin.dim. and hence artinian $mathbb{C}$-subalgebra, so $alpha$ would have to be surjective. So $R_1cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian.
$endgroup$
– Jeremy Rickard
Feb 1 at 13:47
$begingroup$
You're perfectly right! I should have thought of this... I edited the question.
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 14:23
add a comment |
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$begingroup$
There's an example in
Gilmer, Robert; Heinzer, William, Artinian subrings of a commutative ring, Trans. Am. Math. Soc. 336, No. 1, 295-310 (1993). ZBL0778.13012.
of two artinian subrings of a commutative artinian ring whose intersection is not artinian. Since the intersection is the pullback of the inclusion maps, this answers the question.
Let $R=mathbb{C}(x)[y]/(y^2)$. Then $R$ is a two-dimensional algebra over $mathbb{C}(x)$, and is therefore artinian.
Let $R_1=mathbb{C}(x^2)+ymathbb{C}(x)subset R$. Then $R_1$ is a three-dimensional algebra over $mathbb{C}(x^2)$, and is therefore artinian.
Let $R_2=mathbb{C}(x^2+x)+ymathbb{C}(x)subset R$. Then $R_2$ is a three-dimensional algebra over $mathbb{C}(x^2+x)$, and is therefore artinian.
But $mathbb{C}(x^2)capmathbb{C}(x^2+x)=mathbb{C}$, so $R_1cap R_2=mathbb{C}+ymathbb{C}(x)$, which is not artinian, or even noetherian, since $ymathbb{C}(x)$ is not finitely generated as an ideal.
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
1
$begingroup$
Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 12:13
1
$begingroup$
@Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $alpha:Ato R_1cap R_2$ through which every map from an artinian ring factors. But every element of $R_1cap R_2$ is in a fin.dim. and hence artinian $mathbb{C}$-subalgebra, so $alpha$ would have to be surjective. So $R_1cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian.
$endgroup$
– Jeremy Rickard
Feb 1 at 13:47
$begingroup$
You're perfectly right! I should have thought of this... I edited the question.
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 14:23
add a comment |
$begingroup$
There's an example in
Gilmer, Robert; Heinzer, William, Artinian subrings of a commutative ring, Trans. Am. Math. Soc. 336, No. 1, 295-310 (1993). ZBL0778.13012.
of two artinian subrings of a commutative artinian ring whose intersection is not artinian. Since the intersection is the pullback of the inclusion maps, this answers the question.
Let $R=mathbb{C}(x)[y]/(y^2)$. Then $R$ is a two-dimensional algebra over $mathbb{C}(x)$, and is therefore artinian.
Let $R_1=mathbb{C}(x^2)+ymathbb{C}(x)subset R$. Then $R_1$ is a three-dimensional algebra over $mathbb{C}(x^2)$, and is therefore artinian.
Let $R_2=mathbb{C}(x^2+x)+ymathbb{C}(x)subset R$. Then $R_2$ is a three-dimensional algebra over $mathbb{C}(x^2+x)$, and is therefore artinian.
But $mathbb{C}(x^2)capmathbb{C}(x^2+x)=mathbb{C}$, so $R_1cap R_2=mathbb{C}+ymathbb{C}(x)$, which is not artinian, or even noetherian, since $ymathbb{C}(x)$ is not finitely generated as an ideal.
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
1
$begingroup$
Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 12:13
1
$begingroup$
@Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $alpha:Ato R_1cap R_2$ through which every map from an artinian ring factors. But every element of $R_1cap R_2$ is in a fin.dim. and hence artinian $mathbb{C}$-subalgebra, so $alpha$ would have to be surjective. So $R_1cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian.
$endgroup$
– Jeremy Rickard
Feb 1 at 13:47
$begingroup$
You're perfectly right! I should have thought of this... I edited the question.
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 14:23
add a comment |
$begingroup$
There's an example in
Gilmer, Robert; Heinzer, William, Artinian subrings of a commutative ring, Trans. Am. Math. Soc. 336, No. 1, 295-310 (1993). ZBL0778.13012.
of two artinian subrings of a commutative artinian ring whose intersection is not artinian. Since the intersection is the pullback of the inclusion maps, this answers the question.
Let $R=mathbb{C}(x)[y]/(y^2)$. Then $R$ is a two-dimensional algebra over $mathbb{C}(x)$, and is therefore artinian.
Let $R_1=mathbb{C}(x^2)+ymathbb{C}(x)subset R$. Then $R_1$ is a three-dimensional algebra over $mathbb{C}(x^2)$, and is therefore artinian.
Let $R_2=mathbb{C}(x^2+x)+ymathbb{C}(x)subset R$. Then $R_2$ is a three-dimensional algebra over $mathbb{C}(x^2+x)$, and is therefore artinian.
But $mathbb{C}(x^2)capmathbb{C}(x^2+x)=mathbb{C}$, so $R_1cap R_2=mathbb{C}+ymathbb{C}(x)$, which is not artinian, or even noetherian, since $ymathbb{C}(x)$ is not finitely generated as an ideal.
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
There's an example in
Gilmer, Robert; Heinzer, William, Artinian subrings of a commutative ring, Trans. Am. Math. Soc. 336, No. 1, 295-310 (1993). ZBL0778.13012.
of two artinian subrings of a commutative artinian ring whose intersection is not artinian. Since the intersection is the pullback of the inclusion maps, this answers the question.
Let $R=mathbb{C}(x)[y]/(y^2)$. Then $R$ is a two-dimensional algebra over $mathbb{C}(x)$, and is therefore artinian.
Let $R_1=mathbb{C}(x^2)+ymathbb{C}(x)subset R$. Then $R_1$ is a three-dimensional algebra over $mathbb{C}(x^2)$, and is therefore artinian.
Let $R_2=mathbb{C}(x^2+x)+ymathbb{C}(x)subset R$. Then $R_2$ is a three-dimensional algebra over $mathbb{C}(x^2+x)$, and is therefore artinian.
But $mathbb{C}(x^2)capmathbb{C}(x^2+x)=mathbb{C}$, so $R_1cap R_2=mathbb{C}+ymathbb{C}(x)$, which is not artinian, or even noetherian, since $ymathbb{C}(x)$ is not finitely generated as an ideal.
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
answered Feb 1 at 10:35
Jeremy RickardJeremy Rickard
16.9k11746
16.9k11746
1
$begingroup$
Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 12:13
1
$begingroup$
@Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $alpha:Ato R_1cap R_2$ through which every map from an artinian ring factors. But every element of $R_1cap R_2$ is in a fin.dim. and hence artinian $mathbb{C}$-subalgebra, so $alpha$ would have to be surjective. So $R_1cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian.
$endgroup$
– Jeremy Rickard
Feb 1 at 13:47
$begingroup$
You're perfectly right! I should have thought of this... I edited the question.
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 14:23
add a comment |
1
$begingroup$
Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 12:13
1
$begingroup$
@Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $alpha:Ato R_1cap R_2$ through which every map from an artinian ring factors. But every element of $R_1cap R_2$ is in a fin.dim. and hence artinian $mathbb{C}$-subalgebra, so $alpha$ would have to be surjective. So $R_1cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian.
$endgroup$
– Jeremy Rickard
Feb 1 at 13:47
$begingroup$
You're perfectly right! I should have thought of this... I edited the question.
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 14:23
1
1
$begingroup$
Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 12:13
$begingroup$
Thanks! I realized that I made a mistake in the statement of the question when I wrote "Yes to Question 3 would imply yes to Questions 1 and 2", because we must exclude the possibility that the pullback does exist in Artin but does not coincide with the pullback in CRing. I'm planning to remove Questions 1 and 2 (and change the title), but before doing so I wanted to ask you if you see a way of showing that the pullback does not exist in Artin (or of showing that it exists but differs from the other). - I just asked a related question: math.stackexchange.com/q/3096154/660
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 12:13
1
1
$begingroup$
@Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $alpha:Ato R_1cap R_2$ through which every map from an artinian ring factors. But every element of $R_1cap R_2$ is in a fin.dim. and hence artinian $mathbb{C}$-subalgebra, so $alpha$ would have to be surjective. So $R_1cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian.
$endgroup$
– Jeremy Rickard
Feb 1 at 13:47
$begingroup$
@Pierre-YvesGaillard Good point. No, in the example given, there is no pullback $A$ in Artin. If there were, there would be a map $alpha:Ato R_1cap R_2$ through which every map from an artinian ring factors. But every element of $R_1cap R_2$ is in a fin.dim. and hence artinian $mathbb{C}$-subalgebra, so $alpha$ would have to be surjective. So $R_1cap R_2$ would be a finitely generated $A$-module, and so would have to be artinian.
$endgroup$
– Jeremy Rickard
Feb 1 at 13:47
$begingroup$
You're perfectly right! I should have thought of this... I edited the question.
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 14:23
$begingroup$
You're perfectly right! I should have thought of this... I edited the question.
$endgroup$
– Pierre-Yves Gaillard
Feb 1 at 14:23
add a comment |
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