Sorry if the question is simple.

Why does not exist a surjective morphism (as morphism of schemes) from $\mathbb{A}^1$ onto a variety $Y$ with only two points?

I know that there does not exist such a morphism but I don't know why.

J.Harris says that in "Algebraic Geometry (A first course) in Lecture 10, Quotiens (pg 124)"

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    $\begingroup$ The image of a connected set is connected. $\endgroup$ – user171326 Jan 18 '17 at 10:12
  • $\begingroup$ Thank you, is simple and elegant. $\endgroup$ – V.Pérez Jan 18 '17 at 10:14
  • $\begingroup$ You're welcome ! $\endgroup$ – user171326 Jan 18 '17 at 10:19

Here $k$ is a field. Such a morphism is induced by a morphism of algebras $f:k\oplus k\rightarrow K[X]$,

the image of $f((1,1)=1$. you also have $f(1,0)f(0,1)=f(0,0)=0$. You can suppose $f(1,0)=0$, without restricting the generality. This implies $f(0,1)=1$. If $P$ is an irreducible polynomial, $f^{-1}((P))=(1,0)$ since $f(0,1)=1$ is not in $P$. and $(1,0)$ is the image of the map $Spec(k[X])\rightarrow Spec(k\oplus k)$ induced by $f$.


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