coproduct of noncommutative algebra and commutative algebras
up vote
1
down vote
favorite
I have read the book "Rings with generalized identities" and I understand that the free product of asociative unital algebras are the coproduct of them, but I can't understand why this reduces to the tensor product of them when alegbras are conmutative.
category-theory tensor-products noncommutative-algebra
asked Nov 16 at 4:25
GaSa
536
add a comment |
up vote
1
down vote
favorite
I have read the book "Rings with generalized identities" and I understand that the free product of asociative unital algebras are the coproduct of them, but I can't understand why this reduces to the tensor product of them when alegbras are conmutative.
category-theory tensor-products noncommutative-algebra
asked Nov 16 at 4:25
GaSa
536
3
The correct statement is that the coproduct in the category of commutative algebras is the tensor product. The coproduct of commutative algebras, in the category of noncommutative algebras, is still the free product.
– Qiaochu Yuan
Nov 16 at 5:11
@QiaochuYuan, you mean that coproduct in the category of algebras (commutative or noncommutative) is the free product, right? but the category of commutative algebras is a subcategory of category of algebras, so free product should reduce to tensor product taking commutative algebras as a subcategory or I'm wrong??
– GaSa
Nov 16 at 5:26
2
No, that's not what I mean. The inclusion of a subcategory into a category need not preserve coproducts. The coproduct of two commutative algebras is two different things depending on whether you take the coproduct in the category of commutative algebras or noncommutative algebras.
– Qiaochu Yuan
Nov 16 at 5:36
aahh ok, I understand now. Thanks for you time
– GaSa
Nov 16 at 18:43
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have read the book "Rings with generalized identities" and I understand that the free product of asociative unital algebras are the coproduct of them, but I can't understand why this reduces to the tensor product of them when alegbras are conmutative.
category-theory tensor-products noncommutative-algebra
asked Nov 16 at 4:25
GaSa
536
I have read the book "Rings with generalized identities" and I understand that the free product of asociative unital algebras are the coproduct of them, but I can't understand why this reduces to the tensor product of them when alegbras are conmutative.
category-theory tensor-products noncommutative-algebra
category-theory tensor-products noncommutative-algebra
asked Nov 16 at 4:25
GaSa
536
asked Nov 16 at 4:25
GaSa
536
asked Nov 16 at 4:25
GaSa
536
asked Nov 16 at 4:25
GaSa
536
asked Nov 16 at 4:25
GaSa
536
536
3
The correct statement is that the coproduct in the category of commutative algebras is the tensor product. The coproduct of commutative algebras, in the category of noncommutative algebras, is still the free product.
– Qiaochu Yuan
Nov 16 at 5:11
@QiaochuYuan, you mean that coproduct in the category of algebras (commutative or noncommutative) is the free product, right? but the category of commutative algebras is a subcategory of category of algebras, so free product should reduce to tensor product taking commutative algebras as a subcategory or I'm wrong??
– GaSa
Nov 16 at 5:26
2
No, that's not what I mean. The inclusion of a subcategory into a category need not preserve coproducts. The coproduct of two commutative algebras is two different things depending on whether you take the coproduct in the category of commutative algebras or noncommutative algebras.
– Qiaochu Yuan
Nov 16 at 5:36
aahh ok, I understand now. Thanks for you time
– GaSa
Nov 16 at 18:43
add a comment |
3
The correct statement is that the coproduct in the category of commutative algebras is the tensor product. The coproduct of commutative algebras, in the category of noncommutative algebras, is still the free product.
– Qiaochu Yuan
Nov 16 at 5:11
@QiaochuYuan, you mean that coproduct in the category of algebras (commutative or noncommutative) is the free product, right? but the category of commutative algebras is a subcategory of category of algebras, so free product should reduce to tensor product taking commutative algebras as a subcategory or I'm wrong??
– GaSa
Nov 16 at 5:26
2
No, that's not what I mean. The inclusion of a subcategory into a category need not preserve coproducts. The coproduct of two commutative algebras is two different things depending on whether you take the coproduct in the category of commutative algebras or noncommutative algebras.
– Qiaochu Yuan
Nov 16 at 5:36
aahh ok, I understand now. Thanks for you time
– GaSa
Nov 16 at 18:43
3
3
The correct statement is that the coproduct in the category of commutative algebras is the tensor product. The coproduct of commutative algebras, in the category of noncommutative algebras, is still the free product.
– Qiaochu Yuan
Nov 16 at 5:11
The correct statement is that the coproduct in the category of commutative algebras is the tensor product. The coproduct of commutative algebras, in the category of noncommutative algebras, is still the free product.
– Qiaochu Yuan
Nov 16 at 5:11
@QiaochuYuan, you mean that coproduct in the category of algebras (commutative or noncommutative) is the free product, right? but the category of commutative algebras is a subcategory of category of algebras, so free product should reduce to tensor product taking commutative algebras as a subcategory or I'm wrong??
– GaSa
Nov 16 at 5:26
@QiaochuYuan, you mean that coproduct in the category of algebras (commutative or noncommutative) is the free product, right? but the category of commutative algebras is a subcategory of category of algebras, so free product should reduce to tensor product taking commutative algebras as a subcategory or I'm wrong??
– GaSa
Nov 16 at 5:26
2
2
No, that's not what I mean. The inclusion of a subcategory into a category need not preserve coproducts. The coproduct of two commutative algebras is two different things depending on whether you take the coproduct in the category of commutative algebras or noncommutative algebras.
– Qiaochu Yuan
Nov 16 at 5:36
No, that's not what I mean. The inclusion of a subcategory into a category need not preserve coproducts. The coproduct of two commutative algebras is two different things depending on whether you take the coproduct in the category of commutative algebras or noncommutative algebras.
– Qiaochu Yuan
Nov 16 at 5:36
aahh ok, I understand now. Thanks for you time
– GaSa
Nov 16 at 18:43
aahh ok, I understand now. Thanks for you time
– GaSa
Nov 16 at 18:43
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000715%2fcoproduct-of-noncommutative-algebra-and-commutative-algebras%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
The correct statement is that the coproduct in the category of commutative algebras is the tensor product. The coproduct of commutative algebras, in the category of noncommutative algebras, is still the free product.
– Qiaochu Yuan
Nov 16 at 5:11
@QiaochuYuan, you mean that coproduct in the category of algebras (commutative or noncommutative) is the free product, right? but the category of commutative algebras is a subcategory of category of algebras, so free product should reduce to tensor product taking commutative algebras as a subcategory or I'm wrong??
– GaSa
Nov 16 at 5:26
2
No, that's not what I mean. The inclusion of a subcategory into a category need not preserve coproducts. The coproduct of two commutative algebras is two different things depending on whether you take the coproduct in the category of commutative algebras or noncommutative algebras.
– Qiaochu Yuan
Nov 16 at 5:36
aahh ok, I understand now. Thanks for you time
– GaSa
Nov 16 at 18:43