Timeline for reduced ⊗ reduced = reduced; what about connected?
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| Dec 22, 2009 at 9:37 | comment | added | Georges Elencwajg | Francisco, this is not true if k is not perfect. Take for k a field of characteristic p and let a be an element in some overfield such that a is not in k but a^p is in k. Then if A=B=k[a]=k(a), the k-algebra A \otimes B has the non-zero element (a \otimes 1 - 1 \otimes a) whose p-th power is zero.Hence the tensor product of A and B is NOT reduced. Shafarevich states explicitly that k is algebraically closed at the beginning of the paragraph you quote. | |
| Dec 22, 2009 at 8:20 | comment | added | darij grinberg | I don't know what an affine closed set is when $k$ is not algebraically closed. | |
| Dec 21, 2009 at 22:44 | history | answered | Hideyuki Kabayakawa | CC BY-SA 2.5 |