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1. • CommentRowNumber1.
   • CommentAuthorJohn Baez
   • CommentTimeJul 18th 2010
   • (edited Jul 18th 2010)
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   Author: John Baez
   Format: MarkdownItexI'm so sick of making mistakes about separable algebras and their relation to Frobenius algebras that I wrote a page [[separable algebra]] and added more to the page [[Frobenius algebra]]. To make these pages make sense, I needed to create pages called [[semisimple algebra]], [[simple algebra]], and [[division algebra]]. Also [[projective module]]. I would love it if some experts on algebraic geometry vastly enhanced the little section about algebraic geometry in [[separable algebra]]. There's a question there, and also a very vague sentence about etale coverings.

   I’m so sick of making mistakes about separable algebras and their relation to Frobenius algebras that I wrote a page separable algebra and added more to the page Frobenius algebra. To make these pages make sense, I needed to create pages called semisimple algebra, simple algebra, and division algebra. Also projective module.

   I would love it if some experts on algebraic geometry vastly enhanced the little section about algebraic geometry in separable algebra. There’s a question there, and also a very vague sentence about etale coverings.

2. • CommentRowNumber2.
   • CommentAuthorJohn Baez
   • CommentTimeJul 19th 2010
   • PermaLink
   Author: John Baez
   Format: MarkdownItexOkay, Gerstenhaber just emailed me a new paper that talks about separable algebras, and it assures me that any separable algebra over any commutative ring $k$ which is projective as a $k$-module must be finitely generated, but this presumably means there are non-finitely-generated ones too, which sort of answers my question over at [[separable algebra]]. I'll polish that up. Btw, Eilenberg and Nakayama's old paper on Frobenius algebras is quite impressive. Eilenberg is famous for being terse and to the point, and here they define a Frobenius algebra as an algebra $A$ (over a field, let's say) such that $A \cong A^*$ as left $A$-modules. Then they move right on...

   Okay, Gerstenhaber just emailed me a new paper that talks about separable algebras, and it assures me that any separable algebra over any commutative ring $k$ which is projective as a $k$-module must be finitely generated, but this presumably means there are non-finitely-generated ones too, which sort of answers my question over at separable algebra. I’ll polish that up.

   Btw, Eilenberg and Nakayama’s old paper on Frobenius algebras is quite impressive. Eilenberg is famous for being terse and to the point, and here they define a Frobenius algebra as an algebra $A$ (over a field, let’s say) such that $A \cong A^*$ as left $A$-modules. Then they move right on…

3. • CommentRowNumber3.
   • CommentAuthorJohn Baez
   • CommentTimeJul 19th 2010
   • (edited Jul 19th 2010)
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   Author: John Baez
   Format: MarkdownItexOf course they are treating Frobeniuseanness as a property rather than a structure. (Here I am alluding to the title of Nakayama's older paper <i>On Frobeniusean Algebras</i>...)

   Of course they are treating Frobeniuseanness as a property rather than a structure.

   (Here I am alluding to the title of Nakayama’s older paper On Frobeniusean Algebras…)

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeJul 20th 2010
   • (edited Jul 20th 2010)
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex>[[division algebra]] Since a link there may have misled you, note that [[normed division algebra]] already exists.

   division algebra

   Since a link there may have misled you, note that normed division algebra already exists.

5. • CommentRowNumber5.
   • CommentAuthorJohn Baez
   • CommentTimeJul 20th 2010
   • PermaLink
   Author: John Baez
   Format: MarkdownItexYes, I didn't have the energy to write an article on normed division algebras, but someday one will exist... Thanks for making a proper bibliography in [[Frobenius algebra]]. I didn't know how! Just now I added a juicy example in this article explaining when a group algebra can be made into a special Frobenius algebra and when it can't.

   Yes, I didn’t have the energy to write an article on normed division algebras, but someday one will exist…

   Thanks for making a proper bibliography in Frobenius algebra. I didn’t know how!

   Just now I added a juicy example in this article explaining when a group algebra can be made into a special Frobenius algebra and when it can’t.

6. • CommentRowNumber6.
   • CommentAuthorJohn Baez
   • CommentTimeJul 20th 2010
   • PermaLink
   Author: John Baez
   Format: MarkdownItexToby wrote: >Since a link there may have misled you, not that normed division algebra already exists. Oh... by "not that" you meant "note that". Much better!

   Toby wrote:

   Since a link there may have misled you, not that normed division algebra already exists.

   Oh… by “not that” you meant “note that”. Much better!

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeJul 20th 2010
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   Author: TobyBartels
   Format: MarkdownItexHeh, heh, misled twice! (But now my original comment is fixed, to mislead future historians.)

   Heh, heh, misled twice!

   (But now my original comment is fixed, to mislead future historians.)

8. • CommentRowNumber8.
   • CommentAuthorTobyBartels
   • CommentTimeJul 20th 2010
   • (edited Jul 20th 2010)
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexI noticed that you edited the entry for Rosebrugh et al to say 2004 instead of 2005. The article is published in the proceedings of a 2004 conference, but those proceedings were themselves published in 2005. If you do keep it 2004, then you should change all of the citations within the article to say 2004. (Or just remove the date. I only put dates in the citations since there are two items by Joachim Kock.)

   I noticed that you edited the entry for Rosebrugh et al to say 2004 instead of 2005. The article is published in the proceedings of a 2004 conference, but those proceedings were themselves published in 2005. If you do keep it 2004, then you should change all of the citations within the article to say 2004. (Or just remove the date. I only put dates in the citations since there are two items by Joachim Kock.)

9. • CommentRowNumber9.
   • CommentAuthorJohn Baez
   • CommentTimeJul 20th 2010
   • PermaLink
   Author: John Baez
   Format: MarkdownItex>I noticed that you edited the entry for Rosebrugh et al to say 2004 instead of 2005. That was just a mistake, made for reasons too embarrassing to discuss. I fixed it.

   I noticed that you edited the entry for Rosebrugh et al to say 2004 instead of 2005.

   That was just a mistake, made for reasons too embarrassing to discuss. I fixed it.

10. • CommentRowNumber10.
    • CommentAuthorJohn Baez
    • CommentTimeJul 20th 2010
    • (edited Jul 20th 2010)
    • PermaLink
    Author: John Baez
    Format: MarkdownItexCan someone do a little sanity check on me? Any algebra $A$ is a left module of $A \otimes A^{op}$ in an obvious and standard way: $$ (b \otimes c) a = b a c $$ However, it is also a right module of $A \otimes A^{op}$, via $$a (b \otimes c) = c a b $$ <b>QUESTION</b>: Is it true that $A$ is flat as a right module of $A \otimes A^{op}$ if and only if it's flat as a left module of $A \otimes A^{op}$? I don't think you need to know what [[flat module|flat]] means to solve this puzzle! Rather, I want you guys to tell me if the category of right modules of $A \otimes A^{op}$ is equivalent to the category of left modules of $A \otimes A^{op}$, and then see if $A$ is sent to itself under this equivalence! You see, maybe $A$ is sent to $A^{op}$, in which case the answer to my question might be "no". And here's why I'm interested in: Aguiar says an algebra $A$ is separable if and only if it's projective as a left $A \otimes A^{op}$ module, which is true if and only if it's flat as a right $A \otimes A^{op}$ module. And I want to know if that left / right stuff is important or whether he's just being cute.

    Can someone do a little sanity check on me?

    Any algebra $A$ is a left module of $A \otimes A^{op}$ in an obvious and standard way:

    $$(b \otimes c) a = b a c$$

    However, it is also a right module of $A \otimes A^{op}$, via

    $$a (b \otimes c) = c a b$$

    QUESTION: Is it true that $A$ is flat as a right module of $A \otimes A^{op}$ if and only if it’s flat as a left module of $A \otimes A^{op}$?

    I don’t think you need to know what flat means to solve this puzzle! Rather, I want you guys to tell me if the category of right modules of $A \otimes A^{op}$ is equivalent to the category of left modules of $A \otimes A^{op}$, and then see if $A$ is sent to itself under this equivalence!

    You see, maybe $A$ is sent to $A^{op}$, in which case the answer to my question might be “no”.

    And here’s why I’m interested in: Aguiar says an algebra $A$ is separable if and only if it’s projective as a left $A \otimes A^{op}$ module, which is true if and only if it’s flat as a right $A \otimes A^{op}$ module. And I want to know if that left / right stuff is important or whether he’s just being cute.