Let f be the rational function on P 2 given by f = x 1 / x 0 . Find the set of points where f is defiend and describe the corresponding regular function.My question: I know that the set of points where f is defined is { [ 1 : x 1 : x 2 ] ∈ P 2 }. Isn't the corresponding regular function also just f itself, or the regular function g = x 2 / x 0 (both works and both maps P 2 to A 1 )? So what is this problem trying to get at?Now think of this function as a rational map from P 2 to A 1 . Embed A 1 in P 1 , and let ϕ : P 2 → P 1 be the resultiong rational map. Find the set of points where ϕ is defined, and describe the corresponding morphism.My question: Since ϕ is f ∘ ϵ where ϵ is the embedding, why isn't ϕ defined on the same set as f is defined? I'm also not sure what I'm trying to solve for here.

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Let   f be the rational function on             P        2   given by   f  =      x    1        /        x    0  . Find the set of points where   f is defiend and describe the corresponding regular function.My question: I know that the set of points where   f is defined is   {  [  1  :      x    1    :      x    2    ]  ∈            P        2    }. Isn&#039;t the corresponding regular function also just   f itself, or the regular function   g  =      x    2        /        x    0   (both works and both maps             P        2   to             A        1  )? So what is this problem trying to get at?Now think of this function as a rational map from             P        2   to             A        1  . Embed             A        1   in             P        1  , and let   ϕ  :            P        2    →            P        1   be the resultiong rational map. Find the set of points where   ϕ is defined, and describe the corresponding morphism.My question: Since   ϕ is   f  ∘  ϵ where   ϵ is the embedding, why isn&#039;t   ϕ defined on the same set as   f is defined? I&#039;m also not sure what I&#039;m trying to solve for here.

Alexia Hart · Accepted Answer

(a) The rational function   f  =            x      1              x      0        ∈  k  (      x    0    ,      x    1    ,      x    2    ) is regular on the open subset       U    0    =  {  [      x    0    :      x    1    :      x    2    ]  ∈            P        2    |      x    0    ≠  0  ]  }  ⊂            P        2   and defines on that open subset the regular function      f    U    :  U  →            A        1    :  [      x    0    :      x    1    :      x    2    ]  ↦            x      1              x      0      (b) The rational map   ϕ  :            P        2    −  −  →            P        1   induced by   f is regular on the open subset   V  =            P        2    ∖  {  [  0  :  0  :  1  ]  }  ⊂            P        2   and defines on that open subset the regular function      f    V    :  V  →            P        1    :  [      x    0    :      x    1    :      x    2    ]  ↦  [      x    0    :      x    1    ](c) Of course we have   U  ⊂  V  ⊂            P        2   and thus letting  ϵ  :            A        1    ↪            P        1    :      x    1    ↦  [  1  :      x    1    ]be the canonical embedding we have the formula  ϵ  ∘      f    U    =      f    V    |  U  :  U  →            P        1  This is not very surprising: increasing the target of       f    U   from             A        1   to             P        1   allows       f    U  , which is regular on   U, to be extended to the function       f    V  , regular on the larger set   V.(d) Notice carefully that there is no way to extend the regular map      f    V    :  V  =            P        2    ∖  {  [  0  :  0  :  1  ]  }  →            P        1  to a regular map             P        2    →            P        1  : the point   [  0  :  0  :  1  ] is said to be a point of indeterminacy for the rational map   ϕ  :            P        2    −  −  →            P        1  .

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