Let <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struck">C </mrow> ( x

vilitatelp014

vilitatelp014

Answered question

2022-04-12

Let C ( x , y ) be a degree 2 extension of C ( x ) where y is a root of p ( Z ) = Z 2 + ( x 2 + 1 ).
Is it true that C ( x , y ) is not a rational function field? In other words, there is no z C ( x , y ) such that C ( x , y ) = C ( z )

Answer & Explanation

reflam2kfnr

reflam2kfnr

Beginner2022-04-13Added 16 answers

GeometricallyThe field C ( x , y ) is a rational function field because it is the field of rational functions of a smooth affine conic, namely the conic C A C 2 with equation y 2 + x 2 + 1 = 0.
Algebraically
More explicitly C ( x , y ) = C ( t ) with t = y i x :
Writing C ( x , y ) = C ( x ) [ Y ] ( Y 2 + x 2 + 1 ) an explicit isomorphism is given by
f : C ( x , y ) C ( t )
where f ( x ) = 2 i t 1 + t 2 and f ( y ) = i 1 t 2 1 + t 2 .
The inverse isomorphism is determined by f 1 ( t ) = y i x .
The link between both approaches
We fix the point P = ( 0 , i ) C and consider the line y = i + t x through P with slope t (notice that then t = y i x as above ).
That line cuts the conic C in P and in the point P t = ( 2 i t 1 + t 2 , i 1 t 2 1 + t 2 ) and this explains the formulas brutally parachuted in the previous section.

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