Posted on 16 Aug 2022, this text provides information on Syllabus Queries related to Course Queries. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.
My syllabus on algebraic geometry states the following:
''Let ϕ:X→Yϕ:X→Y be a surjective morphism of complete non-singular curves. Then XX is the normalization of YY in the function field of XX. In particular, the morphism ϕϕ is finite. ''
How do we know that XX is the normalization of YY?
''A normalization of YY in k(X)k(X) is a finite surjective morphism π:X→Yπ:X→Y such that XX is normal and the extension of function fields corresponding to ππ is the extension k(Y)⊃k(X)k(Y)⊃k(X). ''
I understand that ϕϕ is surjective and XX is normal (ie. OX,POX,P is integrally closed ∀P∈X∀P∈X) because it is non-singular, but how do we know that the extension of function fields is exactly
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manpreet
Best Answer
2 years ago
My syllabus on algebraic geometry states the following:
''Let ϕ:X→Yϕ:X→Y be a surjective morphism of complete non-singular curves. Then XX is the normalization of YY in the function field of XX. In particular, the morphism ϕϕ is finite. ''
How do we know that XX is the normalization of YY?
The definition of normalization we use is the following:
''A normalization of YY in k(X)k(X) is a finite surjective morphism π:X→Yπ:X→Y such that XX is normal and the extension of function fields corresponding to ππ is the extension k(Y)⊃k(X)k(Y)⊃k(X). ''
I understand that ϕϕ is surjective and XX is normal (ie. OX,POX,P is integrally closed ∀P∈X∀P∈X) because it is non-singular, but how do we know that the extension of function fields is exactly
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