Tensor Product of C with itself over R.

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SUMMARY

The discussion focuses on proving that the tensor product of the complex numbers C with itself over the reals R, denoted as C⊗C, is equivalent to C². The equivalence is established through a series of transformations: C⊗C = C⊗(R[T]/) = C[T]/ = C. The main challenge presented is demonstrating that C⊗(R[T]/) equals C[T]/, utilizing the universal property of the tensor product and defining a bilinear map from C×R[T]/ to C[T]/.

PREREQUISITES
  • Understanding of tensor products, specifically C⊗C over R.
  • Familiarity with the universal property of tensor products.
  • Knowledge of polynomial rings, particularly R[T]/.
  • Concept of complexification in linear algebra.
NEXT STEPS
  • Study the properties of tensor products in linear algebra.
  • Learn about the universal property of tensor products in detail.
  • Explore the structure of polynomial rings and their quotients, focusing on R[T]/.
  • Investigate complexification techniques and their applications in vector spaces.
USEFUL FOR

Mathematicians, particularly those specializing in linear algebra and complex analysis, as well as students seeking to deepen their understanding of tensor products and complexification methods.

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I am trying to prove that C\otimesC (taken over R) is equal to C^2. The method I have seen is to show the following equivalences:
C\otimesC = C\otimes(R[T]/<T^2+1>) = C[T]/<T^2+1> = C.
(All tensor products taken over R).

The only part I am having trouble with is showing that C\otimes(R[T]/<T^2+1>) = C[T]/<T^2+1>. I have tried to show this by using the universal property of the tensor product. First I defined a bilinear map from CXR[T]/<T^2+1> to C[T]/<T^2+1> by sending an element (z,f(T)) to zf(t) (where I am omitting the cosets). This then induces a linear map from the tensor product. However, I cannot seem to find an inverse for this map.

Any help would be greatly appreciated.
 
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