What does this symbol mean in commutative algebra?












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$begingroup$


(Algebraic Geometry and commutative algebra - Bosch - page 16)



I could not find the symbol in the glossary at the end of the book. What does it mean ?



enter image description here



I cannot see it in my reference manual either (Commutative Algebra with a view towards algebraic geometry - Eisenbud)










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  • 2




    $begingroup$
    I suspect the symbol denotes a coproduct.
    $endgroup$
    – drhab
    Jan 12 at 9:42






  • 1




    $begingroup$
    As for the latex-symbol, see here.
    $endgroup$
    – Dietrich Burde
    Jan 12 at 9:48
















0












$begingroup$


(Algebraic Geometry and commutative algebra - Bosch - page 16)



I could not find the symbol in the glossary at the end of the book. What does it mean ?



enter image description here



I cannot see it in my reference manual either (Commutative Algebra with a view towards algebraic geometry - Eisenbud)










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I suspect the symbol denotes a coproduct.
    $endgroup$
    – drhab
    Jan 12 at 9:42






  • 1




    $begingroup$
    As for the latex-symbol, see here.
    $endgroup$
    – Dietrich Burde
    Jan 12 at 9:48














0












0








0





$begingroup$


(Algebraic Geometry and commutative algebra - Bosch - page 16)



I could not find the symbol in the glossary at the end of the book. What does it mean ?



enter image description here



I cannot see it in my reference manual either (Commutative Algebra with a view towards algebraic geometry - Eisenbud)










share|cite|improve this question









$endgroup$




(Algebraic Geometry and commutative algebra - Bosch - page 16)



I could not find the symbol in the glossary at the end of the book. What does it mean ?



enter image description here



I cannot see it in my reference manual either (Commutative Algebra with a view towards algebraic geometry - Eisenbud)







elementary-set-theory commutative-algebra






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 9:35









user3203476user3203476

746613




746613








  • 2




    $begingroup$
    I suspect the symbol denotes a coproduct.
    $endgroup$
    – drhab
    Jan 12 at 9:42






  • 1




    $begingroup$
    As for the latex-symbol, see here.
    $endgroup$
    – Dietrich Burde
    Jan 12 at 9:48














  • 2




    $begingroup$
    I suspect the symbol denotes a coproduct.
    $endgroup$
    – drhab
    Jan 12 at 9:42






  • 1




    $begingroup$
    As for the latex-symbol, see here.
    $endgroup$
    – Dietrich Burde
    Jan 12 at 9:48








2




2




$begingroup$
I suspect the symbol denotes a coproduct.
$endgroup$
– drhab
Jan 12 at 9:42




$begingroup$
I suspect the symbol denotes a coproduct.
$endgroup$
– drhab
Jan 12 at 9:42




1




1




$begingroup$
As for the latex-symbol, see here.
$endgroup$
– Dietrich Burde
Jan 12 at 9:48




$begingroup$
As for the latex-symbol, see here.
$endgroup$
– Dietrich Burde
Jan 12 at 9:48










1 Answer
1






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2












$begingroup$

Its the disjoint union from set theory. Each element of $text{Spec } R$ maps into a different set from the disjoint union.



In category theory this symbol can sometimes be a coproduct, but in this situation it’s not.



Added: Of course disjoint union is the coprduct in sets, but you hardly see all those rings/ $R$-modules and think “category of sets”. This tradition of confusing young scholars of scheme theory goes back, at least to Hartshorne’s book.






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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

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    2












    $begingroup$

    Its the disjoint union from set theory. Each element of $text{Spec } R$ maps into a different set from the disjoint union.



    In category theory this symbol can sometimes be a coproduct, but in this situation it’s not.



    Added: Of course disjoint union is the coprduct in sets, but you hardly see all those rings/ $R$-modules and think “category of sets”. This tradition of confusing young scholars of scheme theory goes back, at least to Hartshorne’s book.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Its the disjoint union from set theory. Each element of $text{Spec } R$ maps into a different set from the disjoint union.



      In category theory this symbol can sometimes be a coproduct, but in this situation it’s not.



      Added: Of course disjoint union is the coprduct in sets, but you hardly see all those rings/ $R$-modules and think “category of sets”. This tradition of confusing young scholars of scheme theory goes back, at least to Hartshorne’s book.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Its the disjoint union from set theory. Each element of $text{Spec } R$ maps into a different set from the disjoint union.



        In category theory this symbol can sometimes be a coproduct, but in this situation it’s not.



        Added: Of course disjoint union is the coprduct in sets, but you hardly see all those rings/ $R$-modules and think “category of sets”. This tradition of confusing young scholars of scheme theory goes back, at least to Hartshorne’s book.






        share|cite|improve this answer











        $endgroup$



        Its the disjoint union from set theory. Each element of $text{Spec } R$ maps into a different set from the disjoint union.



        In category theory this symbol can sometimes be a coproduct, but in this situation it’s not.



        Added: Of course disjoint union is the coprduct in sets, but you hardly see all those rings/ $R$-modules and think “category of sets”. This tradition of confusing young scholars of scheme theory goes back, at least to Hartshorne’s book.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 9:53

























        answered Jan 12 at 9:45









        BenBen

        3,756616




        3,756616






























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