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I have a question for you.

Let $k$ be a field and $\pi \colon X \to Spec(k)$ a $k$-scheme. I would like to verify that there is a bijection between $X(k)$ and the set of $k$-rational points of $X$, where $X(k)$ is the set of sections of $\pi$. I have verified the inclusion $\subseteq$. Now, let $x \in X$ be a $k$-rational point; then we can construct a map $\sigma \colon Spec(k) \to X$. I claim that this map is a section of $\pi$; this means that $\pi \circ \sigma= {Id}_{Spec(k)}$. Topologically, the composition is clearly the identity on $Spec(k)$. But i cannot verify the same at the level of sheaves. I mean, i should verify (if i have understood well what i am doing) that the composition $O_{Spec(k)}(0)\to O_{X}(X)\to O_{Spec(k)}(0)$, induced by $\pi^{\#}$ and $\sigma^{\#}$ is the identity on $k$. But i can just see that it is an isomorphism. Why should it be exactly the identity map?

Thank you very much to everyone.

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    I'm having the same problem! http://math.stackexchange.com/questions/2147917/k-points-and-other-questions-regarding-algebraic-schemes2017-02-17
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    I think i solved the problem! The fact is that both the morphisms $O_{Spec(k)}(0) \to O_{X}(X) \to O_{Spec(k)}(0)$ are $k$-linear (they are morphisms of $k$-algebras; so their composition must be the identity on $k$. Thank you for your support :).2017-02-18

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