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Timeline for Homomorphism between exterior powers of a free module of finite rank

Current License: CC BY-SA 2.5

14 events
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Dec 16, 2009 at 19:55 vote accept Hideyuki Kabayakawa
Dec 10, 2009 at 15:22 comment added Alicia Garcia-Raboso @Mariano: From the beginning I was thinking of commutative rings. I guess I got confused... Edited now. Thanks!
Dec 10, 2009 at 15:21 history edited Alicia Garcia-Raboso CC BY-SA 2.5
R need to be commutative
Dec 10, 2009 at 14:59 comment added Mariano Suárez-Álvarez If the ring is not commutative, there is no exterior power to speak of...
Dec 10, 2009 at 14:43 comment added Alicia Garcia-Raboso @Sammy: I edited my answer to include the IBN property.
Dec 10, 2009 at 14:43 history edited Alicia Garcia-Raboso CC BY-SA 2.5
Added IBN
Dec 9, 2009 at 23:13 vote accept Hideyuki Kabayakawa
Dec 9, 2009 at 23:13
Dec 8, 2009 at 19:59 comment added Alicia Garcia-Raboso @Sammy: thanks for pointing that out!
Dec 8, 2009 at 19:49 comment added Sammy Black @Alberto: You need the Invariant Basis Number property to hold for the ring $R$. Commutative rings, noetherian rings, have IBN.
Dec 8, 2009 at 17:09 vote accept Hideyuki Kabayakawa
Dec 9, 2009 at 23:13
Dec 8, 2009 at 13:33 comment added Alicia Garcia-Raboso If R is a field, then M and the Hom are vector spaces of the same dimension. To specify an isomorphism, you should give the image of a basis of M, which must be a basis of the Hom. The argument is the same in the general case: you should think of free modules over any ring as if they were vector spaces, since the concept of a basis works.
Dec 8, 2009 at 12:52 comment added Hideyuki Kabayakawa Why is it an isomorphism? You don´t specify inverse morphism
Dec 8, 2009 at 1:05 history edited Alicia Garcia-Raboso CC BY-SA 2.5
edited body
Dec 8, 2009 at 1:00 history answered Alicia Garcia-Raboso CC BY-SA 2.5